Let's start with our favorite square.
Example 9
Square a complex number
Here you can go in two ways, the first way is to rewrite the degree as a product of factors and multiply the numbers according to the rule for multiplying polynomials.
The second method is to use the well-known school formula for abbreviated multiplication:
For a complex number it is easy to derive your own abbreviated multiplication formula:
A similar formula can be derived for the square of the difference, as well as for the cube of the sum and cube of the difference. But these formulas are more relevant for complex analysis problems. What if you need to raise a complex number to, say, the 5th, 10th or 100th power? It is clear that it is almost impossible to perform such a trick in algebraic form; indeed, think about how you will solve an example like?
And here the trigonometric form of a complex number comes to the rescue and the so-called Moivre's formula: If a complex number is represented in trigonometric form, then when it is raised to a natural power, the following formula is valid:
It's just outrageous.
Example 10
Given a complex number, find.
What should be done? First you need to represent this number in trigonometric form. Attentive readers will have noticed that in Example 8 we have already done this:
Then, according to Moivre's formula:
God forbid, you don’t need to count on a calculator, but in most cases the angle should be simplified. How to simplify? Figuratively speaking, you need to get rid of unnecessary turns. One revolution is a radian or 360 degrees. Let's find out how many turns we have in the argument. For convenience, we make the fraction correct:, after which it becomes clearly visible that you can reduce one revolution:. I hope everyone understands that this is the same angle.
Thus, the final answer will be written like this:
A separate variation of the exponentiation problem is the exponentiation of purely imaginary numbers.
Example 12
Raise complex numbers to powers
Here, too, everything is simple, the main thing is to remember the famous equality.
If the imaginary unit is raised to an even power, then the solution technique is as follows:
If the imaginary unit is raised to an odd power, then we “pinch off” one “and”, obtaining an even power:
If there is a minus (or any real coefficient), then it must first be separated:
Extracting roots from complex numbers. Quadratic equation with complex roots
Let's look at an example:
Can't extract the root? If we are talking about real numbers, then it really is impossible. It is possible to extract the root of complex numbers! More precisely, two root:
Are the roots found really a solution to the equation? Let's check:
Which is what needed to be checked.
An abbreviated notation is often used; both roots are written on one line under the “same comb”: .
These roots are also called conjugate complex roots.
I think everyone understands how to extract square roots from negative numbers: ,,,, etc. In all cases it turns out two conjugate complex roots.
Let's start with our favorite square.
Example 9
Square a complex number
Here you can go in two ways, the first way is to rewrite the degree as a product of factors and multiply the numbers according to the rule for multiplying polynomials.
The second method is to use the well-known school formula for abbreviated multiplication:
For a complex number it is easy to derive your own abbreviated multiplication formula:
A similar formula can be derived for the square of the difference, as well as for the cube of the sum and cube of the difference. But these formulas are more relevant for complex analysis problems. What if you need to raise a complex number to, say, the 5th, 10th or 100th power? It is clear that it is almost impossible to perform such a trick in algebraic form; indeed, think about how you will solve an example like?
And here the trigonometric form of a complex number comes to the rescue and the so-called Moivre's formula: If a complex number is represented in trigonometric form, then when it is raised to a natural power, the following formula is valid:
It's just outrageous.
Example 10
Given a complex number, find.
What should be done? First you need to represent this number in trigonometric form. Attentive readers will have noticed that in Example 8 we have already done this:
Then, according to Moivre's formula:
God forbid, you don’t need to count on a calculator, but in most cases the angle should be simplified. How to simplify? Figuratively speaking, you need to get rid of unnecessary turns. One revolution is a radian or 360 degrees. Let's find out how many turns we have in the argument. For convenience, we make the fraction correct:, after which it becomes clearly visible that you can reduce one revolution:. I hope everyone understands that this is the same angle.
Thus, the final answer will be written like this:
A separate variation of the exponentiation problem is the exponentiation of purely imaginary numbers.
Example 12
Raise complex numbers to powers
Here, too, everything is simple, the main thing is to remember the famous equality.
If the imaginary unit is raised to an even power, then the solution technique is as follows:
If the imaginary unit is raised to an odd power, then we “pinch off” one “and”, obtaining an even power:
If there is a minus (or any real coefficient), then it must first be separated:
Extracting roots from complex numbers. Quadratic equation with complex roots
Let's look at an example:
Can't extract the root? If we are talking about real numbers, then it really is impossible. It is possible to extract the root of complex numbers! More precisely, two root:
Are the roots found really a solution to the equation? Let's check:
Which is what needed to be checked.
An abbreviated notation is often used; both roots are written on one line under the “same comb”: .
These roots are also called conjugate complex roots.
I think everyone understands how to extract square roots from negative numbers: ,,,, etc. In all cases it turns out two conjugate complex roots.
Example 13
Solve quadratic equation
Let's calculate the discriminant:
The discriminant is negative, and the equation has no solution in real numbers. But the root can be extracted in complex numbers!
Using well-known school formulas, we obtain two roots: – conjugate complex roots
Thus, the equation has two conjugate complex roots:,
Now you can solve any quadratic equation!
And in general, any equation with a polynomial of the “nth” degree has equal roots, some of which may be complex.
A simple example to solve on your own:
Example 14
Find the roots of the equation and factor the quadratic binomial.
Factorization is again carried out according to the standard school formula.