Exponentiation of a trigonometric number. Raising Complex Numbers to a Power

Let's start with a 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 of multiplying polynomials.

The second way is to use the well-known school formula for abbreviated multiplication:

For a complex number, it is easy to derive your own formula for abbreviated multiplication:

A similar formula can be derived for the square of the difference, as well as for the cube of the sum and the cube of the difference. But these formulas are more relevant for complex analysis tasks. What if a complex number needs to be raised to, say, the 5th, 10th, or 100th power? It is clear that in algebraic form it is almost impossible to do such a trick, really, 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 formula: If a complex number is presented in trigonometric form, then when it is raised to a natural power, the formula is correct:

Just outrageous.

Example 10

Given a complex number, find.

What should be done? First, you need to represent the given number in trigonometric form. Attentive readers Notice that in Example 8 we have already done this:

Then, according to the Moivre 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 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 subtract one revolution :. I hope everyone understands that they are the same angle.

Thus, the final answer will be written like this:

A separate kind of the problem of exponentiation is the exponentiation of purely imaginary numbers.

Example 12

Raise complex numbers to a power ,,

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”, getting an even power:

If there is a minus (or any valid coefficient), then it must first be separated:

Extracting roots from complex numbers. Quadratic equation with complex roots

Let's consider an example:

Can't extract the root? If we are talking about real numbers, then really it is impossible. In complex numbers, you can extract the root! Or rather, two root:

Are the found roots really a solution to the equation? Let's check:

Which is what was required to be verified.

An abbreviated notation is often used, both roots are written in one line under the "one comb":.

Such roots are also called conjugate complex roots .

How to extract square roots from negative numbers, I think everyone understands: ,,,, etc. In all cases it turns out two conjugate complex roots.

Let's start with a 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 of multiplying polynomials.

The second way is to use the well-known school formula for abbreviated multiplication:

For a complex number, it is easy to derive your own formula for abbreviated multiplication:

A similar formula can be derived for the square of the difference, as well as for the cube of the sum and the cube of the difference. But these formulas are more relevant for complex analysis tasks. What if a complex number needs to be raised to, say, the 5th, 10th, or 100th power? It is clear that in algebraic form it is almost impossible to do such a trick, really, 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 formula: If a complex number is presented in trigonometric form, then when it is raised to a natural power, the formula is correct:

Just outrageous.

Example 10

Given a complex number, find.

What should be done? First, you need to represent the given number in trigonometric form. Attentive readers will have noticed that in Example 8 we have already done this:

Then, according to the Moivre 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 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 subtract one revolution :. I hope everyone understands that they are the same angle.

Thus, the final answer will be written like this:

A separate kind of the problem of exponentiation is the exponentiation of purely imaginary numbers.

Example 12

Raise complex numbers to a power ,,

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”, getting an even power:

If there is a minus (or any valid coefficient), then it must first be separated:

Extracting roots from complex numbers. Quadratic equation with complex roots

Let's consider an example:

Can't extract the root? If we are talking about real numbers, then really it is impossible. In complex numbers, you can extract the root! Or rather, two root:

Are the found roots really a solution to the equation? Let's check:

Which is what was required to be verified.

An abbreviated notation is often used, both roots are written in one line under the "one comb":.

Such 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!

According to well-known school formulas, we get 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 "nth" degree has equal roots, some of which can be complex.

A simple example for a do-it-yourself solution:

Example 14

Find the roots of the equation and factor the quadratic binomial.

The factorization is carried out again according to the standard school formula.