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Introduction………………………………………………………………………………..3 Chapter I. From history complex numbers……………………………………………………4 Chapter II. Fundamentals of the complex number method……………………………………6 Chapter III. Geometry of a triangle in complex numbers…………………......12 Chapter IV. Solution Unified State Exam problems and various Olympiads using the complex number method…………………………………………………………………....20 Conclusion…………………………………… …………………………………….24 Bibliography………………………………………………………………..25
Introduction
The great importance of complex numbers in mathematics and its applications is widely known. The algebra of complex numbers can be successfully used in elementary geometry, trigonometry, theory of motion and similarities, as well as in electrical engineering, various mechanical and physical problems. In planimetry, the method of complex numbers allows you to solve problems by direct calculation using ready-made formulas. This is the simplicity of this method, compared to vector and coordinate methods, by the method of geometric transformations, requiring students to have considerable intelligence and lengthy searches. For several millennia, the triangle has been a symbol of geometry. You can even say that a triangle is an atom of geometry. Any polygon can be divided into triangles, and the study of its properties comes down to studying the properties of the triangles of its components. Let's look at how the complex number method works when proving the properties of a triangle from school course planimetry, as well as for solving problems C-4 of the Unified State Exam. 2
Chapter I. From the history of complex numbers,,
For the first time, apparently, imaginary quantities were mentioned in the famous work “Great Art, or About algebraic rules» Cardano (1545), as part of a formal solution to the problem of calculating two numbers that add up to 10 and when multiplied give 40. For this problem, he obtained a quadratic equation for one of the terms, and found its roots: 5 + √ − 15 and 5 − √ − 15 . In a commentary to the decision, he wrote: “these the most complex quantities useless, although very clever" and "Arithmetic considerations become more and more elusive, reaching a limit as subtle as it is useless." The possibility of using imaginary quantities when solving a cubic equation, in the so-called irreducible case (when the real roots of the polynomial are expressed through cube roots of imaginary quantities), was first described by Bombelli (1572). He was the first to describe the rules of addition, subtraction, multiplication and division of complex numbers, but still considered them a useless and cunning “invention”. Expressions representable in the form a + b √ − 1, appearing when solving quadratic and cubic equations, began to be called “imaginary” in XVI-XVII centuries at the instigation of Descartes, who called them that, rejecting their reality, and for many other major scientists XVII centuries, the nature and right to the existence of imaginary quantities seemed very doubtful, just as they considered doubtful at that time irrational numbers, and even negative values. Despite this, mathematicians boldly applied formal methods algebras of real quantities and to complex ones, obtained correct real results even from intermediate complex ones, and this could not but begin to inspire confidence. For a long time it was unclear whether all operations on complex numbers lead to complex or real results, or whether, for example, extracting a root could lead to the discovery of some other new type of numbers. The problem of expressing roots of degree n from given number was solved in the works of Moivre (1707) and Cotes (1722). The symbol for denoting the imaginary unit was proposed by Euler (1777, published 1794), who took the first letter of the Latin word for this. imaginarius - imaginary. He also extended all standard functions, including the logarithm, to the complex domain. Euler also expressed the idea in 1751 that the field of complex numbers is algebraically closed. D'Alembert (1747) came to the same conclusion, but the first rigorous proof of this fact belongs to Gauss (1799). It was Gauss who coined the term "complex number" into widespread use in 1831, although the term had previously been used in the same sense by the French mathematician Lazare Carnot in 1803. 3
The arithmetic (standard) model of complex numbers as pairs of real numbers was constructed by Hamilton (1837); this proved the consistency of their properties. Much earlier, in 1685, in his work “Algebra,” Wallis (England) showed that complex roots a quadratic equation with real coefficients can be represented geometrically, by points on a plane. But it went unnoticed. The next time a geometric interpretation of complex numbers and operations on them appeared in the work of Wessel (1799). The modern geometric representation, sometimes called the “Argand diagram,” came into use after the publication of J. R. Argand’s work in 1806 and 1814, which independently repeated Wessel’s conclusions. The terms “modulus”, “argument” and “conjugate number” were introduced by Cauchy. Thus, it was discovered that complex numbers are also suitable for pure execution. algebraic operations addition, subtraction, multiplication and division of vectors on the plane, which greatly changed vector algebra. 4
Chapter II. Basics of the Complex Number Method
[
1
]
,
[2], [3] [4] Geometric interpretation of complex numbers Length of a segment Given a rectangular Cartesian system coordinates on the plane, the complex number z = x+iy (i 2 = -1) can be one-to-one associated with the point M of the plane with coordinates x, y (Fig. 1): z = x + iy ↔M (x, y ) ↔M (z) . The number z is then called the complex coordinate of the point M. Since the set of points of the Euclidean plane is in a one-to-one correspondence with the set of complex numbers, this plane is also called the plane of complex numbers. The origin O of the Cartesian coordinate system is called the initial or zero point of the plane of complex numbers. When = 0 the number z is real. Real numbers are represented by points on the x-axis, which is why it is called the real axis. At x=0, the number z is purely imaginary: z=iy. Imaginary numbers are represented by points on the y-axis, which is why it is called the imaginary axis. Zero is both a real and purely imaginary number. The distance from the beginning of the O plane to the point M(z) is called the modulus of the complex number z and is denoted by |z| or r: | z | = r = | OM | = √ x 2 + y 2 If φ is the oriented angle formed by the vector ⃗ OM with the x axis, then by definition of the sine and cosine function sin φ = y r, cos φ = x r 5
whence x = r cos φ, y = r sin φ, and therefore z = r (cos φ + sin φ). This representation of a complex number z is called its
trigonometry
cheskoe
form. The original representation z=x+iy is called
algebraic
form of this number. At trigonometric representation the angle is called the argument of a complex number and is also denoted by arg z: φ = arg z If a complex number z = x + iy is given, then the number ´ z = x − iy is called
complex conjugate
(or simply
conjugate
) to this number z. Then, obviously, the number z is also conjugate to the number ´ z. The points M(z) and M 1 (´ z) are symmetrical about the x axis. From the equality z = ´ z it follows that y = 0 and vice versa. It means that
a number equal to
to its conjugate is real and vice versa.
Points with complex coordinates z and -z are symmetrical with respect to the initial point O. Points with complex coordinates z and − ´z are symmetrical with respect to the y-axis. From the equality z = ´ z it follows that x = 0 and vice versa. Therefore, the condition z =− ´ z is a criterion for a purely imaginary number. For any number z, obviously | z | = | ´ z | =¿− z ∨¿∨−´ z ∨¿ .
Sum and product
two conjugate complex numbers are real numbers: z + ´ z = 2 z, z ´ z = x 2 + y 2 =¿ z 2 ∨¿. A number conjugate to a sum, product, or quotient of complex 6
numbers are, respectively, the sum, product or quotient of numbers conjugate to given complex numbers: ´ z 1 + z 2 = ´ z 1 + ´ z 2 ; ´ z 1 z 2 = ´ z 1 ´ z 2 ; ´ z 1: z 2 = ´ z 1: ´ z 2 These equalities can be easily verified using formulas for operations on complex numbers. If a and b are the complex coordinates of points A and B, respectively, then the number c = a + b is the coordinate of point C, such that ⃗ OC = ⃗ OA + ⃗ OB (Fig. 3). A complex number d = a − b corresponds to a point D such that ⃗ OD = ⃗ OA − ⃗ OB . The distance between points A and B is | ⃗BA | = | ⃗ OD | =¿ a − b ∨¿: ¿ AB ∨¿∨ a − b ∨¿ (1) Since ¿ z ∨ 2 = z ´ z , then ¿ AB ∨ 2 =(a − b) (´ a − ´ b) . (2)
The equation
z ´ z = r 2
defines a circle with center
About radius
r.
The relation AC CB = λ, (λ ≠ − 1) in which point C divides this segment AB, is expressed through the complex coordinates of these points as follows: λ = c − a b − c, λ = ´ λ, from where c = a + λb 1 + λ (3) For λ = 1, point C is the midpoint of the segment AB, and vice versa. Then: c = 1 2 (a + b) (4) Multiplication of complex numbers Multiplication of complex numbers is performed according to the formula, That is, | a b | = | a || b | , and 7
Parallelism and perpendicularity Collinearity of three points Let points A(a) and B(b) be given on the plane of complex numbers. Vectors ⃗ OA and ⃗ OB are co-directed if and only if arg a = arg b, i.e. when arg a – arg b=arg a b =0 (when dividing complex numbers, the argument of the divisor is subtracted from the argument of the dividend). It is also obvious that these vectors are directed in opposite directions if and only if arg a - arg b= arg a b = ± π. Complex numbers with arguments 0, π, - π are real.
Collinearity criterion for points O, A, B:
In order for points A(a) and B(b) to be collinear with the initial point O, it is necessary and sufficient that the quotient a b be a real number, i.e. a b = ´ a ´ b or a ´ b = ´ a b (6 ) Now take points A(a), B(b), C(c), D(d). Vectors ⃗ BA and ⃗ DC collie are non-ary if and only if the points defined by complex numbers a-b and с-d, are collinear with the beginning O. Note: 1. Based on (6) we have: ⃗ AB ∨¿ ⃗ CD↔ (a − b) (´ c − ´ d) =(´ a − ´ b) (c − d) ; (8) 2. If points A, B, C, D belong to the unit circle z ´ z = 1, then ´ a = 1 a; ´ b = 1 b ; ´ c = 1 c ; ´ d = 1 d and therefore condition (8) takes the form: ⃗ AB ∨¿ ⃗ CD↔ ab = cd ; (9) 3. The collinearity of points A, B, C is characterized by the collinearity of the vectors ⃗AB and ⃗AC. Using (8), we obtain: (a − b) (´ a −´ c) =(´ a − ´ b) (a − c) (10) This is the criterion for the points A, B, C to belong to the same straight line. It can be represented in the symmetrical form a (´ b −´ c) + b (´ c −´ a) + c (´ a − ´ b) = 0 (11) 8
If points A and B belong to the unit circle z ´ z = 1, then ´ a = 1 a; ´ b = 1 b and therefore each of the relations (10) and (11) is transformed (after reduction by (a-b) into the following: c + ab ´ c = a + b (12) Points A and B are fixed, and the point We will consider C a variable, redesignating its coordinate as z. Then each of the obtained relations (10), (11), (12) will be an equation of the straight line AB: (´ a − ´ b) z + (b − a) ´ z + a ´ b − b ´ a = 0 , (10a) z + ab ´ z = a + b . (12a) In particular, the direct OA has the equation a ´ z = ´ a z . are purely imaginary. Therefore, OA ⊥ OB↔ a b = − ´ a ´ b or OA ⊥ OB↔a ´ b + ´ a b = 0 (13) The perpendicularity of the segments AB and CD is determined by the equality (a − b) (´ c − ´ d) + (´ a − ´ b) (c − d) = 0 (14) In particular, when points A, B, C, D belong to the unit circle z ´ z = 1, then dependence (14) is simplified: ab + cd = 0 (15) The scalar product of vectors is expressed. scalar product vectors ⃗ OA and ⃗ OB through the complex coordinates a and b of points A and B. Let a=x 1 +iy 1 , b=x 2 +iy 2 . Then a b + a b=(x 1 +iy 1)(x 2 −iy 2)+(x 1 −iy 1)(x 2 +iy 2)=2(x 1 x 2 +y 1 y 2)= 2 ⃗ OA∙⃗OB. So, ⃗ OA ∙ ⃗ OB = 1 2 (a b + ab) (16) 9
Let now four be given arbitrary points A(a), B(b), C(c), D(d) by their complex coordinates. Then 2 ⃗ AB ∙ ⃗ CB = 1 2 (a-b)(c - d)+(a - b)(c-d) (17) Angles Let us agree to denote by the symbol ∠ (AB ,CD) the positively oriented angle through which the vector ⃗ must be rotated AB so that it becomes co-directed with the vector ⃗ CD. Then, cos ∠ (AB, CD)= (d − c) (´ b − ´ a) +(´ d −´ c)(b − a) 2 | d − c || b − a | (18) sin ∠ (AB ,CD)= (d − c) (´ b −´ a) +(´ d −´ c)(b − a) 2 i | d − c || b − a | (19) Intersection point of secants to a circle If points A, B, C and D lie on the circle z ´ z = 1, then the complex coordinate of the intersection point is found by the formula ´ z = (a + b) − (c + d) ab − cd (20) If AB is perpendicular to CD, then z= 1 2 (a+b+c+d) (21) Intersection point of the tangents to the circle 10
The complex coordinate of the point of intersection of the tangents to the circle z ´ z =1 at its points A(a) and B(b) is found by the formula z= 2ab a + b (22) Orthogonal projection of a point onto a straight line Orthogonal projection of a point M(m) onto a straight line AB, where A(a) and B(b) is found by the formula In the case when A and B belong to the unit circle z= 1 2 (a + b + m − cb m) .
Chapter III.
Triangle geometry in complex numbers
On the plane of complex numbers, a triangle is defined by three complex numbers corresponding to its vertices. Centroid and orthocenter of a triangle. [ 2 ] It is known that for the centroid G (the intersection point of the medians) of triangle ABC and any point O the following equality is true: ⃗ OG = 1 3 (⃗ OA + ⃗ OB + ⃗ OC). Therefore, the complex coordinate g of the centroid G is calculated by the formula g = 1 3 (a + b + c) (23) Let us express h the complex coordinate of the orthocenter H of triangle ABC through the coordinates a, b, c of its vertices. Let the lines AH, BH, CH intersect the circumcircle of the triangle at points A1, B1, C1, respectively. Let this circle have the equation z ´ z =1, then according to (15) we have: a 1 = − bc a , b 1 = − ca b , c 1 = − ab c By formula (20) h = (a + a 1 ) −(b + b 1) a a 1 − bb 1 = ab + bc + ca abc = 1 a + 1 b + 1 c 11
Where h=a+b+c comes from. (24) The resulting expression includes the coordinates of the vertices of the triangle symmetrically, therefore the third altitude of the triangle passes through the intersection point of the first two. Similar triangles [2,1] Triangles ABC and A 1 B 1 C 1 are similar and identically oriented (similarity of the first kind), if B 1 =kAB, A 1 B 1 =kAC and angles B 1 A 1 C 1 and BAC are equal (angles are oriented). Using complex numbers, these equalities can be written as follows: |a 1 −b 1 |=k|a−b|, |a 1 −c 1 |=k|a−c|,arg c 1 − a 1 b 1 − a 1 =arg c − a b − a . The two equalities are equivalent to one with 1 − a 1 c − a = b 1 − a 1 b − a = σ , (25) where σ is a complex number, |σ|=k-similarity coefficient. If σ is real, then c 1 − a 1 c − a = ´ c 1 − ´ a 1 ´ c − ´ a , where AC║A 1 C 1. Consequently, triangles ABC and A 1 B 1 C 1 are homothetic. Relation (25) is necessary and sufficient condition so that triangles ABC and A 1 B 1 C 1 are similar and equally oriented. It can be given a symmetrical form ab 1 +bc 1 +ca 1 =ba 1 +cb 1 +ac 1 (25a) Equal triangles If | σ | = 1, then triangles ABC and A 1 B 1 C 1 are equal. Then relation (25) is a sign of equality of identically oriented triangles, and relation (26) is a sign of equality of oppositely oriented triangles. Regular triangles If you require that an oriented triangle ABC was similar to oriented triangle BCA, then triangle ABC will be regular. 12
Therefore, from (25) we obtain a necessary and sufficient condition for triangle ABC to be regular (a−b) 2 +(b−c) 2 +(c−a) 2 =0 (27) Area of the triangle (proved by the author) We derive formula for the area S of a positively oriented triangle ABC: S = 1 2 | AB || AC | sin ∠ (AB , AC)= 1 4i ((c − a) (´ b − ´ a) − (b − a) (´ c − ´ a)) = − 1 4i (a (´ b − ´ c) + b (´ c − ´ a) + c (´ a − ´ b)) or S = i 4 (a (´ b − ´ c) + b (´ c − ´ a) + c (´ a − ´ b )) (28) If triangle ABC inscribed in the circle z ´ z = 1, then formula (28) is transformed to the form: S = i 4 (a − b)(b − c)(c − a) abc (29) Theorem about the midline of a triangle (proved by the author)
Theorem
. middle line of the triangle is parallel to the base and equal to half of it. Proof. Let points M and N be the midpoints of sides AB and BC, then m = b 2 ; n = b + c 2 . Since z 2 =z ´ z, then MN 2 =(m-n)(´ m - ´ n)=(b 2 - b + c 2)(´ b 2 – ´ b + ´ c 2)= b ´ b 4 − b ´ b + b ´ c 4 − b ´ b + ´ b c 4 + b ´ b + b ´ c + ´ b c + c ´ c 4 = c ´ c 4 13
4MN 2 =c ´ c, AC 2 =(c-0)(c-0)=c ´ c, therefore 4MN 2 = AC 2 or 2MN=AC. Condition (8) of collinearity of vectors MN and AC is also satisfied, and therefore MN ║AC. Thales' theorem (proved by the author)
Theorem
. If on one side of an angle parallel lines cut off equal segments, then on the other side of the angle they cut off equal segments. Proof Let's assume that c=kb. Then if BD||CE, then we have (b-d)(´ c − 2 ´ d ¿= (´ b − ´ d) (c − 2d) Opening the brackets and bringing similar terms, we get the equation b ´ c − 2 b ´ d −´ c d = ´ b c − 2 ´ b d − c ´ d Replacing c with kb and ´ c with k ´ b , we get bk ´ b -2b ´ d -dk ´ b = ´ b kb-2 ´ b d-kb ´ d . Bringing similar terms again and moving everything to one side, we get 2b ´ d + dk ´ b − 2 ´ b d − kb ´ d =0. We'll take it out common multiplier and we get 2(b ´ d − ´ b d ¿+ k (´ b d − b ´ d) = 0. Hence k=2, i.e. c=2b. Similarly, it is proved that f=3b, etc. Pythagorean Theorem (proved by the author) B right triangle square of the hypotenuse equal to the sum square legs 14
Proof. The distance between points B and C is equal to BC=|b-c|=b, BC 2 =b ´ b. Since |z| 2 = z ´ z , then AC 2 =(a-c)(c ´ a − ´ ¿ ¿=(a − 0) (´ a - 0)=a ´ a . AB 2 =(a-b)(´ a − ´ b ¿= a ´ a − a ´ b - ´ a b+b ´ b. Since b is a real number, i.e. b= ´ b , then -a ´ b =− ab . on the Oy axis, then a = - ´ a, that is - ´ ab = ab. Thus, AB 2 = a ´ a -a ´ b - ´ ab +b ´ b = a ´ a +b ´ b = AC 2 +BC 2. The theorem is proven. Euler's straight line (proved by the author) Let us prove that the orthocenter, centroid and circumcenter of the triangle lie on the same straight line (this straight line is called the Euler straight line), and OG = 1/2GH 15.
Proof: Point G(g) is the centroid of triangle ABC, H(h) is the orthocenter, and O(o) is the center of the circumscribed circle of the triangle. In order for these points to be collinear, equality (10) must be satisfied: (g-о)(´ g - ´ h ¿ -(´ g − ´ o ¿ (g − h) =0 Let us take point O as the origin, then g(´ g - ´ h ¿ - ´ g (g − h) =g 2 -g ´ h −¿ (g 2 - h ´ g ¿ =-g ´ h + h ´ g (30) The complex coordinate of the orthocenter is calculated according to formula (24) h=a+b+c, (30a) and the centroid according to formula (23) g = 1 3 (a + b + c) (30c) Substitute into (30), we get 1 3 (a+b +c)(´ a + b + c)-(a+b+c)(´ a + b + c 1 3 ¿))=0. Equality (10) is satisfied, therefore, the centroid, orthocenter and center of the circumscribed triangle. the circles lie on the same line. OG=g= 1 3 (a+b+c) GH=h-g=a+b+c- 1 3 (a+b+c)= 2 3 (a+b+c) We got, that OG= 1 2 GH. The theorem is proven 16.
Euler's circle (nine point circle). Proved by the author Consider triangle ABC. Let's agree that | OA | = | OB | = | OC | =1, i.e. all vertices of the triangle belong to the unit circle z ´ z = 1 (the center of the circumcircle O is the origin, and the radius is the unit of length). Let us prove that the bases of the three altitudes of an arbitrary triangle, the midpoints of its three sides and the midpoints of the three segments connecting its vertices with the orthocenter lie on the same circle, and its center is the midpoint of the segment OH, where H, recall, is the orthocenter of the triangle ABC. Such a circle is called
Euler circle
. Let points K, L and M be the midpoints of the sides of triangle ABC, points Q, N, P the bases of its altitudes, points F, E, D the midpoints of three segments connecting its vertices with the orthocenter. Let's prove that points D, E, F, K, L, M, N, P, Q belong to the same circle. Assign the corresponding complex coordinates to the points: k = a + b 2 , l = b + c 2 ;m = a + c 2 ,o 1 = h 2 = a + b + c 2 d = 2a + b + c 2 ; e = 2 c + a + b 2 ; f = 2 b + a + c 2 n = 1 2 (a + b + c − ab c) , q = 1 2 (a + c + b − ac b) , p = 1 2 (c + b + a − cb a) O 1 K = | o 1 − k | = | c 2 | ,O 1 L = | o 1 − l | = | a 2 | , O 1 M = | o 1 − m | = | b 2 | O 1 D = | o 1 − d | = | a 2 | ,O 1 E = | o 1 − e | = | c 2 | ,O 1 F = | o 1 − f | = | b 2 | O 1 N= | o 1 − n | = 1 2 | ab c | = 1 2 | a || b | | c | , O 1 Q= 1 2 | a || c | | b | , O 1 F= 1 2 | b || c | | a | . 17
Because triangle ABC is inscribed in the circle z ´ z = 1, then | a | = | b | = | c | = 1,→ | a 2 | = | b 2 | = | c 2 | = 1 2 | a || b | | c | = 1 2 | a || c | | b | = 1 2 | b || c | | a | = 1 2 So, points D, E, F, K, L, M, N, Q, F belong to the same circle Gauss’s theorem If a line intersects the lines containing sides BC, CA, AB of triangle ABC, respectively, at points A 1, B 1 , C 1, then the midpoints of the segments AA 1, BB 1, СС 1 are collinear. Proof. Using (11), we write the conditions for the collinearity of triplets of points AB 1 C, CA 1 B, BC 1 A, A 1 B 1 C 1: 0,) b - a (c) a - c () c - b (a 0 ,) c - b a() b - a () a - c b(0,) a - c b() c - b () b - a c(0,) b - a (c) a - c () c - b a (1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 b c a b (31) If M, N, P are the midpoints of the segments AA 1, BB 1, CC 1 , then we have to show that 0) () () ( n m p m p n p n m (32) Since), (2 1), (2 1), (2 1 1 1 1 c c p b b n a a m then the equality being proved (31) is equivalent to this: 0))(())(())((1 1 1 1 1 1 1 1 1 b b a a c c a a c c b c b b a a or after multiplication: 0) () () () () () () () () () () () (1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 b a c b a with b a c b a c a c b a with b a c b a c b c b a c b a c b a c (33) Now it is easy to see that (33) is obtained by term-by-term addition of equalities (31). The proof is complete. . 18
Chapter IV.
Solving USE problems and various Olympiads using the complex number method.
Problem 1. Unified State Examination -2012, P-4 On a line containing the median AD of a right triangle ABC with right angle C, a point E is taken, distant from vertex A at a distance equal to 4. Find the area of triangle BCE if BC=6, AC= 4. First solution. According to the Pythagorean theorem AD=5. Then ED=1 Let point E lie on ray AD. The median AD is longer than AE, and point E lies inside triangle ABC (Fig. 1). Let us drop the perpendicular EF from point E to line BC and consider similar right triangles DEF and DAC. From the similarity of these triangles we find: EF = AC ∙ ED AD = 4 5 19
Therefore, S BCE = 1 2 ∙ 6 ∙ 4 5 = 2.4. Let now point A lie between E and D (Fig. 2). In this case ED=9 and EF = AC ∙ ED AD = 36 5 . Then S BCE = 1 2 ∙ 6 ∙ 36 5 = 21.6. Answer: 2.4; 21.6. Solving the problem using complex numbers. Case I: point E lies on ray AD. Since D is the middle of CB, then CD=3. And since CA=4, it is clear that AD=5, i.e. DE=1. Let's take point C as the initial point, and lines CA and CB as the real and imaginary axes. Then A(4), C(0), B(6i), D(3i), E(e). Points A, E and D are collinear, then e − 4 3i − e = 4 i.e. e= 12i + 4 5 . According to formula (25) S CBE =│ ´ i 4 (e6 ´ i +6i(− ´ e)│= e e − ´ ¿ 6 i 2 4 ¿ ¿ =2.4 Case II: point A lies between points D and E , then 4 − e 3i − 4 = 4 5 , i.e. e= 36 − 12 i 5 S CBE = | 3 i 2 2 (36 − 12 i 5 − − 36 − 12i 5) | : 2.4 and 21.6 To solve a problem in the first way, it is necessary to have a number of guesses, which may not appear immediately, but after quite a lot of reasoning. Although, if the student is well prepared, then the solution itself is formed instantly. When solving the problem in the second way, we. We use ready-made formulas, saving time on searching. However, we understand that without knowing the formulas, problems cannot be solved using the complex numbers method. As you can see, each method has its pros and cons.
Task 2 (MIOO, 2011):
“Point M lies on segment AB. On a circle with diameter AB, point C is taken, distant from points A, M and B at distances of 20, 14 and 15, respectively. Find the area of triangle BMC." 20
Solution: Since AB is the diameter of a circle, then ∆ ABC is rectangular, ∠ C = 90 ° Let’s take C as zero point plane, then A(20i), B(15), M(z). Since CM=14, the equality z ´ z = 196 is true, i.e. point M ∈ a circle with center at point C and r=14. Let's find the intersection points of this circle with line AB: Equation of line AB (10a): 20 i (15 −´ z) + 15 (´ z + 20 i) + z (− 20 i − 15) = 0 Replacing ´ z with 196 z and multiplying the entire equation by (4 i − 3) , we obtain a quadratic equation for z: 25 z 2 + 120 i (4 i − 3) z + 196 (4 i − 3) 2 = 0 z 1,2 = 2 (3 − 4 i) (6 i± √ 13) 5 Using formula (28), we find the area ∆ MBC: S = i 4 (z (´ b − ´ c) + b (´ c − ´ z) + c (´ z − ´ b)) Where c = 0, ´ c = 0, b = 15, ´ b = 15, ´ z = 196 ∗ 5 2 (3 − 4 i) (6 i ± √ 13) Having completed equivalent transformations, we get S = 54 ± 12 √ 13 sq. units Answer. 54 ± 12 √ 13 sq. units If you solve the problem geometric methods, then it is necessary to consider two different cases: 1st - point M lies between A and D; 2nd - between D and B. 21
When solving a problem using the method of complex numbers, the duality of the solution is obtained due to the presence of two points of intersection of a circle and a line. This circumstance allows us to avoid a common mistake.
Problem 3
The medians AA 1, BB 1 and CC 1 of triangle ABC intersect at point M. It is known that AB=6MC 1. Prove that triangle ABC is right triangle. Solution: Let C be the zero point of the plane, and assign a real unit to point A. The problem then reduces to proving that b is a purely imaginary number. AB 2 = (b − 1) (´ b − 1) . M is the centroid, its coordinate is 1 3 b + 1 3 MC 1 2 = (1 3 b + 1 3 − 1 2 b − 1 2)(1 3 ´ b + 1 3 − 1 2 ´ b − 1 2) = 1 3 b (b + 1) (´ b + 1) Since AB=6MC 1, then (b − 1) (´ b − 1) = (b + 1) (´ b + 1) . Having carried out the transformations, we obtain b =− ´ b, i.e. b is a purely imaginary number, i.e. the angle C is a straight line.
Task 4.
22
As a result of a 90° rotation around point O, segment AB turned into segment A "B". Prove that the median OM of the triangle OAB " is perpendicular to the line A " B . Solution: Let the coordinates O, A, B be equal to 0.1, b, respectively. Then points A " and B " will have coordinates a" = i and b" = bi, and the middle M of the segment AB " will have coordinates m = 1 2 (1 + bi). We find: a " − b m − 0 = i − b 1 2 (1 + bi) = 2 i (i − b) i − b = 2i number is purely imaginary. Based on the perpendicularity criterion (segments AB and CD are perpendicular if and only if the number a − b c − d is purely imaginary), the lines OM and A ’ B are perpendicular.
Problem 5
.
23
From the base of the altitude of the triangle, perpendiculars are dropped onto two sides that do not correspond to this altitude. Prove that the distance between the bases of these perpendiculars does not depend on the choice of the height of the triangle. Solution: Let triangle ABC be given, and the circle circumscribed around it has the equation z ´ z = 1. If CD is the height of the triangle, then d = 1 2 (a + b + c − ab c) The complex coordinates of the bases M and N of the perpendiculars dropped from point D to AC and BC, respectively, are equal to m = 1 2 (a + c + d − ac ´ d 2) n = 1 2 (b + c + d − bc ´ d 2) We find: m − n = 1 2 (a − b + c ´ d (b − a)) = 1 2 ( a − b) (1 − c ´ d) = (a − b) (a − c) (b − c) 4 ab Since | a | = | b | = 1,then | m − n | = | (a − b) × (b − c) (c − a) | 4 . This expression is symmetrical with respect to a, b, c, i.e. the distance MN does not depend on the choice of triangle height.
Conclusion
24
"Certainly! All problems can be solved without complex numbers. But the fact of the matter is that the algebra of complex numbers is another effective method solving planimetric problems. We can only talk about choosing a method that is more effective for a given task. Disputes about the advantages of a particular method are pointless if we consider these methods in general, without application to a specific problem” [2]. A large place in the study of the method is occupied by a set of formulas. This is
main disadvantage
method and at the same time
dignity
, since it allows you to solve enough complex tasks according to ready-made formulas with elementary calculations. In addition, I believe that when solving planimetry problems this method is universal.
Bibliography
1. Markushevich A.I. Complex numbers and conformal mappings - M.: State Publishing House of Technical and Theoretical Literature, 1954. - 52 p. 25
2. Ponarin Ya. P. Algebra of complex numbers in geometric problems: A book for students of mathematical classes of schools, teachers and students of pedagogical universities - M.: MTsNMO, 2004. - 160 p. 3. Shvetsov D. From Simson’s line to the Droz-Farny theorem, Kvant. - No. 6, 2009. – p. 44-48 4. Yaglom I. M. Geometric transformations. Linear and circular transformations. - State Publishing House of Technical and Theoretical Literature, 1956. – 612 p. 5. Yaglom I.M. Complex numbers and their application in geometry - M.: Fizmatgiz, 1963. - 192 p. 6. Morkovich A.G. and others, Algebra and the beginnings of mathematical analysis. 10th grade. In 2 hours. Part 1. Textbook for students of general education institutions (profile level) - M.: Mnemosyne, 2012. - 343 p. 7. Andronov I.K. Mathematics of real and complex numbers - M.: Prosveshchenie, 1975. - 158 p. 26
Application
Classical theorems elementary geometry
Newton's theorem.
In a quadrilateral circumscribed about a circle, the midpoints of the diagonals are collinear with the center of the circle. 27
Proof. Let us take the center of the circle as the origin, setting its radius equal to one. Let us denote the points of contact of the sides of this quadrilateral triangle A o B o C o D o by A, B, C, D (in a circular order) (Fig. 4). Let M and N be the midpoints of the diagonals A o C o and B o D o, respectively. Then, according to the formula for the intersection points of tangents to the circle z = 2ab a + b, points A o , B o , C o , D o will have complex coordinates, respectively: , 2 , 2 , 2 , 2 0 0 0 0 d c cd d c b bc c b a ab b d a ad a where a, b, c, d are the complex coordinates of points A, B, C, D. Therefore.) (2 1 ,) (2 1 0 0 0 0 d c cd b a ab d b n c b bc d a ad c a m Calculate.))(())((a d c b d c b a n m Since, 1 , 1 b b a a , 1 , 1 d d c c then directly it is clear that n m n m Based on (6), points O, M, N are collinear.
Pascal's theorem
.
The intersection points of lines containing opposite sides of an inscribed hexagon lie on the same line. 28
Proof. Let the hexagon ABCDEF and P FA CD N EF BC M DE AB ) () (,) () (,) () ( (Fig. 6) be inscribed in a circle (Fig. 6). Let us take the center of the circle as the zero point of the plane, and its radius is per unit length. Then, according to (17), we have: ,) (,) (,) (fa cd a f d c p ef bc f e c b n de ab e d b a m Calculate) )(())((ef bc de ab ab fa ef de cd bc e b n m and similarly.))(())((fa cd ef bc bc ab fa ef de cd f c p n Next we find: .))(())((de ab c f fa cd e b p n n m Since the numbers f e d c b a are equal, respectively, f e d c b a 1 , 1 , 1 , , 1, 1, then an oral check reveals that the found expression coincides with its conjugate, i.e., it is a real number. This means the points M, N, P are collinear.
Monge's theorem.
In a quadrilateral inscribed in a circle, the lines passing through the midpoints of the sides and. Each diagonal is perpendicular to opposite sides and, accordingly, the other diagonal intersects at one point. It is called the Monge point of a cyclic quadrilateral. Proof. The perpendicular bisectors to the sides of the quadrilateral ABCD intersect at the center of the circumscribed circle, which we take as the starting point. For each point M(z) of the perpendicular bisector to [AB] the number b a b a z ) (2 1 purely imaginary. 29
In particular, for z=0 it is equal to) (2) (b a b a . For each point N(z) of the line passing through the middle of the side CD perpendicular to (AB), the number b a d c z ) (2 1 will need to be purely imaginary and vice versa. But for z=) (2 1 d c b a it is equal) (2 b a b a i.e. purely imaginary. Therefore, point E with a complex coordinate) (2 1 d c b a lies on the indicated line And this expression is symmetrical with respect to the letters a, b, c, d. Therefore, the other five similarly constructed lines contain point E. 30
- We will be based on connections, not on mechanical formulas.
- Let's consider complex numbers as a complement to our number system, the same as zero, fractional or negative numbers.
- We visualize ideas in graphics to better understand the essence, and not just present them in dry text.
And ours secret weapon: learning by analogy. We'll get to complex numbers by starting with their ancestors, negative numbers. Here's a little guide for you:
For now, this table makes little sense, but let it be there. By the end of the article everything will fall into place.
Let's really understand what negative numbers are
Negative numbers are not so simple. Imagine that you are a European mathematician in the 18th century. You have 3 and 4, and you can write 4 – 3 = 1. It's simple.
But what is 3 – 4? What exactly does this mean? How can you take 4 cows away from 3? How can you have less than nothing?
Negative numbers were viewed as complete nonsense, something that “cast a shadow over the whole theory of equations” (Francis Maceres, 1759). Today it would be complete nonsense to think of negative numbers as something illogical and unhelpful. Ask your teacher if negative numbers violate basic math.
What happened? We invented a theoretical number that had useful properties. Negative numbers cannot be touched or felt, but they are good at describing certain relationships (like debt, for example). This is a very useful idea.
Instead of saying, “I owe you 30,” and reading the words to see if I'm in the black or in the black, I can just write down “-30” and know what that means. If I make money and pay off my debts (-30 + 100 = 70), I can easily write this transaction in a few characters. I'll be left with +70.
The plus and minus signs automatically capture the direction - you don't need a whole sentence to describe the changes after each transaction. Mathematics has become simpler, more elegant. It no longer mattered whether negative numbers were “tangible” - they had useful properties, and we used them until they became firmly established in our everyday life. If someone you know has not yet understood the essence of negative numbers, now you will help them.
But let's not belittle human suffering: Negative numbers were a real shift in consciousness. Even Euler, the genius who discovered the number e and much more, did not understand negative numbers as well as we do today. They were seen as "meaningless" results of calculations.
It is strange to expect children to calmly understand ideas that once confused even the best mathematicians.
Entering Imaginary Numbers
It's the same story with imaginary numbers. We can solve equations like this all day long:
The answers will be 3 and -3. But let’s imagine that some smart guy added a minus here:
Well well. This is the kind of question that makes people cringe when they see it for the first time. Do you want to calculate the square root of a number less than zero? This is unthinkable! (Historically there really were similar questions, but it’s more convenient for me to imagine some faceless wise guy, so as not to embarrass the scientists of the past).
It looks crazy, just like negative numbers, zero and irrational numbers (non-repeating numbers) looked back in the day. There's no "real" meaning to this question, right?
No it is not true. So-called “imaginary numbers” are as normal as any other (or just as abnormal): they are a tool for describing the world. In the same spirit that we imagine that -1, 0.3 and 0 "exist", let's suppose that there is some number i, where:
In other words, you multiply i by itself to get -1. What's happening now?
Well, at first we certainly have a headache. But by playing the game "Let's pretend that i exists" we actually make mathematics simpler and more elegant. New connections appear that we can easily describe.
You won't believe in i, just as those old grumpy mathematicians didn't believe in the existence of -1. All new concepts that twist the brain into a tube are difficult to perceive, and their meaning does not emerge immediately, even for the brilliant Euler. But as negative numbers have shown us, strange new ideas can be extremely useful.
I don't like the term "imaginary numbers" itself - it feels like it was chosen specifically to offend the feelings of i. The number i is as normal as the others, but the nickname “imaginary” has stuck to it, so we will also use it.
Visual understanding of negative and complex numbers
The equation x^2 = 9 actually means this:
Which transformation of x, applied twice, turns 1 into 9?
There are two answers: "x = 3" and "x = -3". That is, you can “scale by” 3 times or “scale by 3 and flip” (reversing or taking the reciprocal of the result are all interpretations of multiplying by negative one).
Now let's think about the equation x^2 = -1, which can be written like this:
Which transformation of x, applied twice, turns 1 into -1? Hm.
- We can't multiply twice positive number because the result will be positive.
- We cannot multiply a negative number twice because the result will again be positive.
What about... rotation! It sounds unusual, of course, but what if we think of x as a “90 degree rotation”, then by applying x twice we will make a 180 degree rotation by coordinate axis, and 1 will turn into -1!
Wow! And if we think about it a little more, we can make two revolutions in opposite direction, and also go from 1 to -1. This is a "negative" rotation or multiplication by -i:
If we multiply by -i twice, then on the first multiplication we get -i from 1, and on the second -1 from -i. So there are actually two square roots-1: i and -i.
This is pretty cool! We have something like a solution, but what does it mean?
- i is the "new imaginary dimension" for measuring number
- i (or -i) is what the numbers "become" when rotated
- Multiplying by i is rotating 90 degrees counterclockwise
- Multiplying by -i is a 90 degree clockwise rotation.
- Rotating twice in either direction gives -1: it takes us back to the "normal" dimension of positive and negative numbers (the x-axis).
All numbers are 2-dimensional. Yes, it's hard to accept, but it would have been just as hard for the ancient Romans to accept. decimals or long division. (How is it that there are more numbers between 1 and 2?). Looks weird like anyone new way think in mathematics.
We asked "How to turn 1 into -1 in two actions?" and found the answer: rotate 1 90 degrees twice. Quite a strange, new way of thinking in mathematics. But very useful. (By the way, this geometric interpretation of complex numbers appeared only decades after the discovery of the number i itself).
Also, do not forget that taking a revolution counterclockwise is positive result- this is a purely human convention, and everything could have been completely different.
Search for sets
Let's go a little deeper into the details. When you multiply negative numbers (like -1), you get a set:
- 1, -1, 1, -1, 1, -1, 1, -1
Since -1 does not change the size of the number, only the sign, you get the same number either with a “+” sign or with a “-” sign. For the number x you get:
- x, -x, x, -x, x, -x…
This is a very useful idea. The number "x" can represent good and bad weeks. Let's imagine that good week replaces the bad one; It's a good week; What will the 47th week be like?
X means it's going to be a bad week. See how negative numbers "follow the sign" - we can simply enter (-1)^47 into the calculator instead of counting ("Week 1 good, week 2 bad... week 3 good..."). Things that constantly alternate can be perfectly modeled using negative numbers.
Okay, what happens if we continue multiplying by i?
Very funny, let's simplify it all a little:
Here is the same thing presented graphically:
We repeat the cycle every 4th turn. That definitely makes sense, right? Any child will tell you that 4 turns to the left are the same as not turning at all. Now take a break from the imaginary numbers (i, i^2) and look at the total set:
- X, Y, -X, -Y, X, Y, -X, -Y…
Exactly how negative numbers are modeled mirror reflection numbers, imaginary numbers can model anything that rotates between two dimensions "X" and "Y". Or anything with a cyclical, circular dependence - do you have anything in mind?
Understanding Complex Numbers
There is one more detail to consider: can a number be both “real” and “imaginary”?
Don't even doubt it. Who said that we have to turn exactly 90 degrees? If we stand with one foot on the “real” dimension and the other on the “imaginary” one, it will look something like this:
We are at the 45 degree mark, where the real and imaginary parts are the same, and the number itself is “1 + i”. It's like a hot dog, where there is both ketchup and mustard - who said you have to choose one or the other?
Basically, we can choose any combination of real and imaginary parts and make a triangle out of it all. The angle becomes the "angle of rotation". A complex number is a fancy name for numbers that have a real and an imaginary part. They are written as “a + bi”, where:
- a - real part
- b - imaginary part
Not bad. But there's only one left last question: How “big” is a complex number? We cannot measure the real part or the imaginary part separately because we will miss the big picture.
Let's take a step back. Size negative number is the distance from zero:
This is another way to find absolute value. But how to measure both components at 90 degrees for complex numbers?
Is it a bird in the sky... or an airplane... Pythagoras is coming to the rescue!
This theorem pops up wherever possible, even in numbers invented 2000 years after the theorem itself. Yes, we are making a triangle, and its hypotenuse will be equal to the distance from zero:
Although measuring a complex number is not as simple as “just omitting the - sign,” complex numbers have very useful applications. Let's look at some of them.
Real Example: Rotations
We won't wait until college physics to practice complex numbers. We'll do this today. A lot can be said on the topic of multiplying complex numbers, but for now you need to understand the main thing:
- Multiplying by a complex number rotates by its angle
Let's see how it works. Imagine that I am on a boat, moving on a course of 3 units to the East every 4 units to the North. I want to change my course 45 degrees counterclockwise. What will my new course be?
Someone might say “It's easy! Calculate sine, cosine, google the tangent value... and then..." I think I broke my calculator...
Let's go over in a simple way: we are on a course of 3 + 4i (it doesn’t matter what the angle is, we don’t care for now) and we want to turn 45 degrees. Well, 45 degrees is 1 + i (ideal diagonal). So we can multiply our rate by this number!
Here's the gist:
- Initial heading: 3 units East, 4 units North = 3 + 4i
- Rotate counterclockwise 45 degrees = multiply by 1 + i
When multiplied we get:
Our new landmark- 1 unit to the West (-1 to the East) and 7 units to the North, you can draw the coordinates on the graph and follow them.
But! We found the answer in 10 seconds, without any sines and cosines. There were no vectors, no matrices, no tracking of which quadrant we were in. It was simple arithmetic and a little algebra to work out the equation. Imaginary numbers are great for rotation!
Moreover, the result of such a calculation is very useful. We have course (-1, 7) instead of angle (atan(7/-1) = 98.13, and it is immediately clear that we are in the second quadrant. How, exactly, did you plan to draw and follow the indicated angle? Using a protractor at hand?
No, you would convert the angle into cosine and sine (-0.14 and 0.99), find the approximate ratio between them (about 1 to 7) and sketch a triangle. And here complex numbers undoubtedly win - accurately, lightning fast, and without a calculator!
If you're like me, you'll find this discovery mind-blowing. If not, I'm afraid that mathematics doesn't excite you at all. Sorry!
Trigonometry is good, but complex numbers make calculations much easier (like finding cos(a + b)). This is just a small announcement; in the following articles I will provide you with the complete menu.
Lyrical digression: some people think something like this: “Hey, it’s not convenient to have a North/East course instead of simple angle for the passage of the ship!
Is it true? Okay, look at yours right hand. What is the angle between the base of your little finger and the tip index finger? Good luck with your calculation method.
Or you can simply answer, “Well, the tip is X inches to the right and Y inches up,” and you can do something about it.
Are complex numbers getting closer?
We went through my basic discoveries in the field of complex numbers like a tornado. Look at the very first illustration, it should now become more clear.
There is so much more to discover in these beautiful, wonderful numbers, but my brain is already tired. My goal was simple:
- Convince you that complex numbers were only seen as “crazy”, but in fact they can be very useful (just like negative numbers)
- Show how complex numbers can simplify some problems like rotation.
If I seem overly concerned about this topic, there is a reason for that. Imaginary numbers have been an obsession of mine for years - the lack of understanding irritated me.
But lighting a candle is better than wading through pitch darkness: these are my thoughts, and I am sure that the light will light in the minds of my readers.
Epilogue: But they're still pretty weird!
I know they still look weird to me too. I'm trying to think like the first person who discovered zero thought.
Zero is such a strange idea, “something” represents “nothing”, and this could not be understood in any way in Ancient Rome. It's the same with complex numbers - it's a new way of thinking. But both zero and complex numbers greatly simplify mathematics. If we had never introduced weird things like new number systems, we would still be counting everything on our fingers.
I repeat this analogy because it is so easy to start thinking that complex numbers are "not normal." Let's be open to innovation: in the future, people will only joke about how someone until the 21st century did not believe in complex numbers.
October 23, 2015
POSSIBILITY OF USING COMPLEX NUMBERS
IN THE COURSE OF MATHEMATICS IN GENERAL EDUCATION SCHOOL
Scientific adviser:
Municipal educational institution
Pervomaiskaya secondary school
With. Kichmengsky Town
St. Zarechnaya 38
The presented work is devoted to the study of complex numbers. Relevance: solving many problems in physics and technology leads to quadratic equations with negative discriminant. These equations have no solution in the region real numbers. But the solution of many such problems has a very definite physical meaning.
Practical significance: complex numbers and functions of complex variables are used in many questions of science and technology; they can be used in school to solve quadratic equations.
Object area: mathematics. Object of research: algebraic concepts and actions. Subject of research- complex numbers. Problem: complex numbers are not studied in a secondary school mathematics course, although they can be used to solve quadratic equations. The possibility of introducing complex numbers into Unified State Exam assignments in future. Hypothesis: You can use complex numbers to solve quadratic equations in secondary school. Target: to study the possibility of using complex numbers when studying mathematics in the 10th grade of a secondary school. Tasks: 1. Study the theory of complex numbers 2. Consider the possibility of using complex numbers in a 10th grade mathematics course. 3. Develop and test tasks with complex numbers.
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you just need to agree to act on such expressions according to the rules of ordinary algebra and assume that
In 1572, a book by the Italian algebraist R. Bombelli was published, in which the first rules for arithmetic operations on such numbers were established, up to the extraction from them cubic roots. The name “imaginary numbers” was introduced in 1637. French mathematician and philosopher R. Descartes, and in 1777 one of the largest mathematicians VIII century X..gif" width="58" height="19"> as an example of the use of complex numbers when studying mathematics in the 10th grade. Hence. The number x, the square of which is equal to –1, is called the imaginary unit and is denoted i. Thus, , from where ..gif" width="120" height="27 src=">.gif" width="100" height="27 src=">8th grade" href="/text/category/8_klass/" rel ="bookmark">8th grade in algebra.- M.: Education, 1994.-P.134-139.
2. encyclopedic Dictionary young mathematician / Comp. E-68. - M.: Pedagogy, 19с