The complex number power formula is used to compute the value of a complex number which is raised to the power of “n”. To recall, a complex number is the form of x + iy, where x and y are the real numbers and “i” is an imaginary number. The “i” satisfies i^{2} = -1.

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## Formula to Calculate the Power of a Complex Number

The complex number power formula is given below.

**z ^{n} = (re^{Iθ })^{n} = r^{n} e^{inθ}**

**Example Question**

**Question 1:Compute: (3+3i) ^{5}**

**Solution:**

Here is the exponential form of 3+3i

r = √ (9+9) = 3 √2

tan θ = (3/3)

⇒arg z = (π/4)

3 + 3i = 3√2e^{i(π/4)}

Now, (3 + 3i)^{5} = (3 √2)^{5} e^{i(5π/4)}

= 972 √2 (cos(5π/4) + isin (5π/4))

= 972 √2 [(−√2/2)−(√2/2)i]

= − 972 − 972i

**Question 2: Compute: (1 – √3i) ^{6}**

Solution:

Given complex number is (1 – √3i)

^{6}

The exponential form of 1 – √3i is:

r = √(1+3) = 2

tan θ = (√3/1)

⇒arg z = (π/3)

1 – √3i = 2e

^{i(π/3)}

Now, (1 – √3i)

^{6}= (2)6e

^{i(6π/3)}

= 64 e

^{iπ}

= 64 [cos π + i sin π] = 64[1 + i(0)] = 64

**Question 3: Write the square root of 5 + 12i in the polar form.**

Solution:

Given complex number is: 5 + 12i

Square root of the given complex number = √(5 + 12i) = (5 + 12i)^{½}

r = √(25 + 144) = √169 = 13

tan θ = (12/5)

θ = tan-1(12/5) = 67.38

⇒arg z = 67.38

5 + 12i = 13^{ei67.38}

(5 + 12i)^{½} = (13)^{1/2}e^{(i67.38)/2}

√(5 + 12i) = √13 e^{i33.69}

= √13 (cos 33.69 + i sin 33.69)

## Powers and Roots of Complex Numbers

## Learning Objectives

- Use De Moivre’s Equation to find powers of complex numbers
- Prove De Moivre’s Theorem
- Use De Moivre’s theorem to find all solutions to an equation

## Introduction

Whether it is adding, subtracting, multiplying, dividing or some other mathematical operation that is being done on two or more complex numbers, there will be more than one method- using rectangular form or polar form

## De Moivre’s Theorem

How do we raise a complex number to a power? Let’s start with an example

(-4 – 4*i*)^{3} = (-4 – 4*i*) x (-4 – 4*i*) x (-4 – 4*i*) =

In rectangular form, this can get very complex. What about in r cis θ form?

In words: Raise the *r*-value to the same degree as the complex number is raised and then multiply that by *cis* of the angle multiplied by the number of the degree.

Based on the example above, it appears that we can generalize

(*a* + *bi*)* ^{n}* =

*r cis*(

*nθ*) =

*r*(cos(

*nθ*) +

*i*sin(

*nθ*))

If this is correct, then the polar form provides a much faster result for raising a complex number to a power than doing the problem in rectangular form.

**Example:** Find the value of (1+3–√i)^{4}

=16((−0.5)−0.866i)

Finally we have

*z*^{4} = -8 – 13.856*i*

Check by complex rectangular multiplication. First we will find *z*^{2}, then square that result:

(1+3–√i)2=1+23–√i−3

=−2+23–√i

Now by squaring the square, the fourth power is found or (−2+23–√i)2

4−83–√i−12=−8−83–√i

=−8−13.856i

The general rule for raising a complex number to any power is stated by **De Moivre’s**

**Equation:**

**Let ***z*** = ***r***(cos ***θ*** + ***i*** sin ***θ***) be a complex number in rcisθ form. If ***n*** is a positive integer, ***z*^{n}** is***z** ^{n}* =

*r*

*(cos(*

^{n}*nθ*) +

*i*sin(

*nθ*))

Proof:

The proof of De Moivre’s equation uses **mathematical induction**. Mathematical induction is a method of proving something is true for an infinite number of cases. In this example, we want to show that De Moivre’s equation works for all integer values of *n*.

The way we do this (and any induction proof) is first we will check that the theorem is true for *n* = 1. Then we assume it is true for *n* = *k* (called the induction step). Finally, we use the case *n* = *k* to show that if the theorem is true for *n* = *k* then it is also true for *n* = *k* + 1. Once we have done this, we have proven the theorem always works for any integer *n*.

If n = 1, then the theorem is

*z*^{1} = *r*^{1} (cos(1 ⋅ *θ*) + *i* sin(1 ⋅ *θ*))

Which is the rcis form of *z*

Now we assume that

*z** ^{k}* =

*r*

*(cos(*

^{k}*kθ*) +

*i*sin(

*kθ*))

Then we check the case of *z*^{k}^{+ 1}.

*z*^{k}^{+ 1} = *z*^{k}*z*

By the induction step,

zk+1=rk(cos(kθ)+i sin(kθ))r(cos(θ)+i sin(θ))

=rk+1(cos(kθ)+i sin(kθ))(cos(θ)+i sin(θ))

=rk+1(cos(kθ)cos(θ)+i sin(kθ)cos(θ)+i cos(kθ)sin(θ)+i2 sin(kθ)sin(θ))

Gathering like terms and using *i*^{2} = -1

=rk+1(cos(kθ)cos(θ)−sin(kθ)sin(θ)+i (sin(kθ)cos(θ)+cos(kθ)sin(θ)))

Finally, we use the angle addition identities, and we get

=rk+1(cos(kθ+θ)+i sin(kθ+θ))

=rk+1(cos((k+1)θ)+i sin((k+1)θ))

And now we are done. We have shown for any positive integer *n*,

zn=rn(cos(nθ)+i sin(nθ)).

## nth Root Theorem

Often in mathematics whenever an operation is presented, the inverse operation follows. This will be the case for the current topic. The inverse operation of finding a power for a number is to find a root of the same number.

The problem, for example is find the square root of (1 + *i*).

Previously you have learned that the square root of a number, *x*, was shown to equal *x*^{1/2}.

Any root, say the n^{th} root can be written as *x*^{1/}^{n}

The formula for De Moivre’s theorem also works for fractional powers (though we won’t prove that here). That means that this same formula can be used for finding roots:

In decimal form, we get

=1.189( 0.924 + 0.383*i*)

=1.099 + 0.455i

To check, we will multiply the result by itself in rectangular form:

(1.099+0.455i) ⋅ (1.099+0.455i)=1.0992+1.099(0.455i)+1.099(0.455i) + (0.455i)2

=1.208+0.500i+0.500i+0.208i2

=1.208+i−0.208 or

=1+i

**The Fundamental Theorem of Algebra** states that the number of solutions for an equation is equal to the degree of an equation. Finding

That would mean that there are two values for *x*, and in the solution above, but we only found one value. What is the other value? How is it found?

If the square root of a real number, *n*, is *b*, then both *b*^{2} and (-*b*)^{2} will equal *n*. Can this also be true for complex numbers?

or in approximate decimal form: 1.099 + 0.455*i*.

The opposite of this number is (-1.099 – 0.455*i*).

Squaring the opposite,

(−1.099−0.455i)2=(−1.099)2+(−1.099)(−0.455i)+(−1.099)(−0.455i) + (−0.455i)2

=1.208+0.500i+0.500i+0.208i2

=1.208+i−0.208

=1+i

Graphing both of these complex solutions on the same polar axis reveals an important feature of the roots of complex numbers:

What this reveals is that both solutions are the same distance from the origin, and but they are rotated by π radians when graphed. This suggests is that the roots of complex numbers, when graphed, are spaced apart in such a manner, that the difference between their angles is equal. In this case, both solutions are separated by π radians. Finding the cube root of a

between the roots. Add this angle “*n* – 1” times to the first angle, until a complete circle has been formed.

The procedures, when combined, are known as De Moivre’s Theorm.

## Solve Equations using De Moivre’s Theorem

Check any one of these solutions, for example the second solution, to see if the results are confirmed. To check the first and third solution will be given as an exercise at the back of the chapter.

## Applications, Technological Tools

After utilizing De Moivre’s Theorem, answers can be checked using a graphing calculator. Convert the given values into *a* + *bi* form and then put the calculator into “pol” mode. Perform the operation.

## Lesson Summary

There is only one practical method for finding the power or roots of a complex number. Before raising a complex number to a power or finding a root or roots of a complex number, the complex number must be in polar or cis form, and then De Moivre’s Theorem can be utilized.

## Points to Consider

Having a number in polar form can yield results that the rectangular form could NOT! Recall the *Correspondence Principle*. Sometimes a method learned at an earlier level in mathematics will work up to a point. When that method will no longer work, newer, sometimes more complicated, or seemingly more complicated methods are introduced, often to the protestations of students. Often these new methods are helpful for solving a wider range of problems.

## Review Questions

- Perform indicated operation on these complex numbers:

## Powers And Roots

Powers of complex numbers are just special cases of products when the power is a positive whole number. We have already studied the powers of the imaginary unit *i* and found they cycle in a period of length 4.

and so forth. The reasons were that (1) the absolute value |*i*| of *i* was one, so all its powers also have absolute value 1 and, therefore, lie on the unit circle, and (2) the argument arg(*i*) of *i* was 90°, so its *n*th power will have argument *n*90°, and those angles will repeat in a period of length 4 since 4*·*90° = 360°, a full circle.

More generally, you can find *z*^{n} as the complex number (1) whose absolute value is |*z*|^{n}, the *n*^{th} power of the absolute value of *z*, and (2) whose argument is *n* times the argument of *z.*

In the figure you see a complex number *z* whose absolute value is about the sixth root of 1/2, that is, |*z*| = 0.89, and whose argument is 30°. Here, the unit circle is shaded black while outside the unit circle is gray, so *z* is in the black region. Since |*z*| is less than one, it’s square is at 60° and closer to 0. Each higher power is 30° further along and even closer to 0. The first six powers are displayed, as you can see, as points on a spiral. This spiral is called a *geometric* or *exponential* sprial.

### Roots.

Note that in the last example, *z*^{6} is on the negative real axis at about -1/2. That means that *z* is just about equal to one of the sixth roots of -1/2.

There are, in fact, six sixth roots of any complex number. Let *w* be a complex number, and *z* any of its sixth roots. Since *z*^{6} = *w,* it follows that

- the absolute value of
*w*, |*w*| is |*z*|^{6}, so |*z*| = |*w*|^{1/6}, and - 6 arg(
*z*) is arg(*w*), so arg(*z*)=arg(*w*)/6.

Actually, the second statement isn’t quite right since 6 arg(*z*) could be any multiple of 360° more than arg(*w*), so you can add multiples of 60° to arg(*w*) to get the other five roots.

For example, take *w* to be -1/2, the green dot in the figure to the right. Then |*w*| is 1/2, and arg(*w*) is 180°. Let *z* be a sixth root of *w.* Then (1) |*z*| is |*w*|^{1/6} which is about 0.89. Also, (2) the argument of *w* is arg(*w*) = 180°. But the same angle could be named by any of180°, 540°, 900°, 1260°, 1610°, or 1970°.

If we take 1/6 of each of these angles, then we’ll have the possible arguments for *z*:30°, 90°, 150°, 210°, 270°, or 330°.

Since each of the angles for *z* differs by 360°, therefore each of the possible angles for *z* will differ by 60°. These six sixth roots of -1/2 are displayed in the figure as blue dots.

### More roots of unity.

Recall that an “*n*th root of unity” is just another name for an *n*th root of one. The fourth roots are ±1, ±*i,* as noted earlier in the section on absolute value. We also saw that the eight 8th roots of unity when we looked at multiplication were ±1, ±*i,* and ±√2/2 ± *i*√2/2.

Let’s consider now the sixth roots of unity. They will be placed around the circle at 60° intervals. Two of them, of course, are ±1. Let *w* be the one with argument 60°. The triangle with vertices at 0, 1, and *w* is an equilateral triangle, so it is easy to determine the coordinates of *w.* The *x*-coordinate is 1/2, and the *y*-coordinate is √3/2. Therefore, *w* is (1 + *i*√3)/2. The remaining sixth roots are reflections of *w* in the real and imaginary axes. In summary, the six sixth roots of unity are ±1, and (±1 ± *i*√3)/2 (where + and – can be taken in any order).

Now some of these sixth roots are lower roots of unity as well. The number –1 is a square root of unity, (–1 ± *i*√3)/2 are cube roots of unity, and 1 itself counts as a cube root, a square root, and a “first” root (anything is a first root of itself). But the remaining two sixth roots, namely, (1 ± *i*√3)/2, are sixth roots, but not any lower roots of unity. Such roots are called *primitive,* so (1 ± *i*√3)/2 are the two primitive sixth roots of unity.

It’s fun to find roots of unity, but we’ve found most of the easy ones already.

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