Imaginary and Complex Numbers with Exponents

We can perform any mathematical operation with imaginary and complex numbers. Similar to how we can add, subtract, multiply and divide these numbers, we can also raise them to powers.

Here, we will learn what is the result of raising the imaginary unit to several powers. In addition, we will learn a formula that will allow us to raise any complex number to different powers.

ALGEBRA
imaginary and complex numbers with exponents

Relevant for

Exploring imaginary and complex numbers with exponents.

See examples

ALGEBRA
imaginary and complex numbers with exponents

Relevant for

Exploring imaginary and complex numbers with exponents.

See examples

Imaginary numbers with exponents

By raising imaginary numbers to incremental powers, we get the following:

{{i}^0}=1

{{i}^1}=i

{{i}^2}=-1

{{i}^3}=-i

{{i}^4}=1

We can see that even powers result in real numbers and odd powers result in imaginary numbers. There is a pattern of 1, i, -1, -i that is repeated when we take the powers of i, starting from {{i}^0}.

If we want to simplify large powers of i, we can decompose the powers to form smaller parts. Remembering that {{i}^4}=1, we can factor {{i}^4}, which is equal to 1. We can do this repeatedly until we have small powers of i less than or equal to 3.


Complex numbers with exponents

To solve problems of powers of complex numbers easily, we have to use the exponential form of a complex number.

Remember that the exponential form of a complex number is z=r{{e}^{i \theta}, where r represents the distance from the origin to the complex number and \theta represents the angle of the complex number.

If we have a complex number z = a + bi, we can find its radius with the formula:

r=\sqrt{{{a}^2}+{{b}^2}}

And we find its angle with the formula:

 \tan(\theta)=\frac{b}{a}

Then we apply any power n to the complex number in its polar form:

{{z}^n}={{r}^n}{{e}^{in\theta}}

The expression can be simplified and written in the form a+bi by using Euler’s formula:

{{e}^{i\theta}}=\cos(\theta)+i~\sin(\theta)


Complex numbers with exponents – Examples with answers

Practice what you have learned about the powers of imaginary and complex numbers with the following examples. Each example has its respective answer, but it is recommended that you try to solve the exercises yourself before looking at the solution.

EXAMPLE 1

Simplify the expression {{i}^{58}}.

We can simplify this by remembering that {{i}^4}=1. Therefore, we obtain factors that have exponents divisible by 4 and simplify:

{{i}^{58}}={{i}^{56}}\times {{i}^2}

={{({{i}^{4}})}^{14}}\times {{i}^2}

={{1}^{14}}\times {{i}^2}

={{i}^2}

 {{i}^{58}}=-1

EXAMPLE 2

Simplify the expression {{i}^{91}}.

To simplify we use the fact that {{i}^4}=1. We can extract the factor {{i}^3} to obtain a factor divisible by 4:

{{i}^{91}}={{i}^{88}}\times {{i}^3}

={{({{i}^{4}})}^{22}}\times {{i}^3}

={{1}^{22}}\times {{i}^3}

={{i}^3}

 {{i}^{91}}=-i

EXAMPLE 3

Find the result of {{(2-3i)}^3}.

Since this is a small power, we can solve this by multiplying the complex number by itself three times. Therefore, we have:

{{(2-3i)}^3}=(2-3i)(2-3i)(2-3i)

=(4-6i-6i+9{{i}^2})(2-3i)

=(4-12i-9)(2-3i)

=(-5-12i)(2-3i)

=-10+15i-24i+36{{i}^2}

=-10-9i-36

=-46-9i

EXAMPLE 4

What is the result of {{(4+4i)}^5}?

We have to start by writing this number in its exponential form. Therefore, we start by finding the radius:

r= \sqrt{{{4}^2}+{{4}^2}}

= \sqrt{16+16}

 = \sqrt{32}

=4\sqrt{2}

Now, we find the angle:

 \tan(\theta)=\frac{3}{3}

 \theta={{\tan}^{-1}(1)

 \theta=\frac{\pi}{4}

Thus, the number in exponential form is z=4\sqrt{2}{{e}^{i\frac{\pi}{4}}}. Therefore, raising to the power, we have:

{{(4+4i)}^5}={{(4\sqrt{2})}^5}{{e}^{i\frac{5\pi}{4}}}

We can use Euler’s formula to simplify the expression obtained:

=4096\sqrt{2}(\cos(\frac{5\pi}{4})+i~\sin(\frac{5\pi}{4})

=4096\sqrt{2}(-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i)

=-4096-4096i

EXAMPLE 5

Solve the expression {{(1- \sqrt{3}i)}^6}.

We rewrite the complex number in its exponential form, so we start with its radius:

r= \sqrt{{{1}^2}+{{(-\sqrt{3})}^2}}

= \sqrt{1+3}

 = \sqrt{4}

=2

Now, we find the angle:

 \tan(\theta)=\frac{\sqrt{3}}{1}

 \theta={{\tan}^{-1}(\sqrt{3})

 \theta=\frac{\pi}{3}

Therefore, the number in exponential form is z=2{{e}^{i\frac{\pi}{3}}}. Then,
when raising to the power, we have:

{{(1+\sqrt{3}i)}^6}={{2}^6}{{e}^{i\frac{6\pi}{3}}}

We can use Euler’s formula to simplify the expression obtained:

=64(\cos(2\pi)+i~\sin(2\pi)

=64(1+0i)

=64

EXAMPLE 6

Solve the expression {{( \sqrt{3}-i)}^6}.

We find the radius of the complex number to write it in its exponential form:

r= \sqrt{{{(\sqrt{3})}^2}+{{(-1)}^2}}

= \sqrt{3+1}

 = \sqrt{4}

=2

Now, we find the angle:

 \tan(\theta)=\frac{-1}{\sqrt{3}}

 \theta={{\tan}^{-1}}(\frac{-1}{\sqrt{3}})

 \theta=-\frac{\pi}{6}

Thus, the number in exponential form is z=2{{e}^{i\frac{-\pi}{6}}}. Therefore, raising to the power, we have:

{{(\sqrt{3}-i)}^6}={{2}^6}{{e}^{-i\frac{6\pi}{6}}}

We can use Euler’s formula to simplify the expression obtained:

=64(\cos(-\pi)+i~\sin(-\pi)

=64(-1+0i)

=-64


Complex numbers with exponents – Practice problems

Test your knowledge of powers of complex numbers to solve the following problems. Select an answer and click “Check” to verify that you selected the correct one.

Simplify the expression {{i}^{33}}.

Choose an answer






What is the result of {{(3+3i)}^5}?

Choose an answer






What is equal to {{(1-i)}^8}?

Choose an answer






Solve the expression {{(2+2i)}^9}.

Choose an answer







See also

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