Powers of Complex Numbers Introduction

We can find powers of Complex numbers, like (2+i)^{4}, by either performing the multiplication by hand or by using the Binomial Theorem for expansion of a binomial x+y. This can be somewhat of a laborious task. Fortunately, there is a nifty shortcut that we can apply to shorten the process and it involves the Polar form of Complex numbers.

Example

Find (2+i)^{4}

(2+i)^{4}=((2+i)(2+i))((2+i)(2+i))=((4-1)+(2+2)i))((4-1)+(2+2)i))=(3+4i)(3+4i)=(9-16)+(12+12)i=-7+24i

with a modulus |(2+i)^{4}|=|2+i||2+i||2+i||2+i|=(\sqrt{5})(\sqrt{5})(\sqrt{5})(\sqrt{5})=(5)(5)=25

which equals

|-7+24i|=\sqrt{(-7)^{2}+(24)^{2}}=\sqrt{576+49}=\sqrt{625}=25

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Let's go

Powers of the Polar Form

We saw in the Polar Representation section the proof that

z^{2}=r^{2}(\cos{(\theta+\theta)}+i\sin{(\theta+\theta)})=r^{2}(\cos{2\theta}+i\sin{2\theta})

and we're going to extend this definition to show that the power of any Complex number z^{n}=(r(\cos{\theta}+isin{\theta}))^{n} also has a very special and useful result.

Example

We'll extend the result for z^{2}=r^{2}(\cos{2\theta}+i\sin{2\theta}) and apply it to z^{3}

If z=1+i=\sqrt{2}(\cos{\frac{\pi}{4}}+i\sin{\frac{\pi}{4}}) with 2\theta=\frac{2\pi}{4}=\frac{\pi}{2} then

z^{2}=(\sqrt{2})^{2}(\cos{\frac{\pi}{2}}+i\sin{\frac{\pi}{2}})=2(0+i)=0+2i=2i

then

z^{3}=z^{2}z=2(\cos{\frac{\pi}{2}}+i\sin{\frac{\pi}{2}})\sqrt{2}(\cos{\frac{\pi}{4}}+i\sin{\frac{\pi}{4}})=2\sqrt{2}(\cos({\frac{\pi}{2}+\frac{\pi}{4})}+i\sin{(\frac{\pi}{2}+\frac{\pi}{4})})=2\sqrt{2}(\cos{\frac{3\pi}{4}}+i\sin{\frac{3\pi}{4}})=2\sqrt{2}(-\frac{\sqrt2}{2}+\frac{i\sqrt{2}}{2}

Example

If z=1+\sqrt{3}i=2(\cos{\frac{\pi}{3}}+i\sin{\frac{\pi}{3}}) with 3\theta=\frac{3\pi}{3}=\pi then

z^{3}=(2)^{3}(\cos{\pi}+i\sin{\pi})=8(-1+0i)=-8

We can now infer that

z^{n}=r^{n}(\cos{n\theta}+i\sin{n\theta})

De Moivre's Theorem and The Unit Circle

By setting r=1 and using the Unit Circle, we obtain De Moivre's Theorem

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

This is an extremely useful theorem for finding powers and roots of Complex numbers.

We can express the sine or cosine function of a multiple of  an angle \theta, n\theta, by powers of the sine and cosine of the original angle \theta. We do this by applying the Binomial Theorem for a power n to the product (\cos{\theta}+i\sin{\theta})^{n}.

Example

If \cos{5\theta}+i\sin{5\theta}=(\cos{\theta}+i\sin{\theta})^{5} then we can expand the right side in powers of the sine and cosine of \theta by using the binomial expansion for n=5

Binomial Expansion for(x+y)^{n} n=5:

1x^{5}y^{0}  5x^{4}y^{1}  10x^{3}y^{2}  10x^{2}y^{3}  5x^{1}y^{4}  1x^{0}y^{5}

for decreasing powers of cosine from n=5 and increasing powers of sine from n=0

then

\cos{5\theta}+i\sin{5\theta}=\cos^{5}{\theta}+5i\sin{\theta}\cos^{4}{\theta}+10i^{2}\sin^{2}{\theta}\cos^{3}{\theta}+10i^{3}\sin^{3}{\theta}\cos^{2}{\theta}+5i^{4}\sin^{4}{\theta}\cos{\theta}+i^{5}\sin^{5}{\theta}

The Real part \cos{5\theta} of the expansion is

\cos^{5}{\theta}-10i\sin^{2}{\theta}\cos^{3}{\theta}+5i^{4}\sin^{4}{\theta}\cos{\theta}

and the Imaginary part \sin{5\theta} of the expansion is

5i\sin{\theta}\cos^{4}{\theta}-10i\sin^{3}{\theta}\cos^{2}{\theta}+i\sin^{5}{\theta}

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Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.