# High powers of complex numbers [closed]

I have these two questions that I am trying to solve. I know that I am suppose to use De Movire's Theorem but I am getting stuck. Can you guys please help out? Thanks.

Compute the following powers and give your answer in the form $a+bi$. Use the square root symbol $\sqrt{}$ where needed to give an exact value for your answer. You may leave powers of real numbers in exponent form, e.g. $2^{11}$. \begin{align}&a)\quad \left(\frac 1{\sqrt{2}}+\frac 1{\sqrt{2}}i\right)^{95} \\ &b) \quad\left(-\sqrt{3}-i\right)^{13}\end{align}

## closed as off-topic by colormegone, Michael Albanese, user91500, Claude Leibovici, choco_addictedJun 23 '16 at 7:03

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• What is (1/rt(2)+i1/rt (2))^2? Then what is that squared? And that squared? – fleablood Jun 23 '16 at 0:35
• Note that $\cos(\frac{\pi}{4})=\sin(\frac{\pi}{4})=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$. What do you know about $\cos(\theta)+i\sin(\theta)$ in relation to the unit circle? What do you know about roots of unity? If not much, then look at $(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i)^n$ for $n$ a multiple of four or eight. – JMoravitz Jun 23 '16 at 0:35
• @fleablood yes it is – user3504306 Jun 23 '16 at 0:35
• Polar form tends to work a little nicer with powers. – user137731 Jun 23 '16 at 0:36
• For the other, what angle do the numbers $-\sqrt{3}$ and $-1$ remind you of? What about $2(-\frac{\sqrt{3}}{2})$ and $2(-\frac{1}{2})$? – JMoravitz Jun 23 '16 at 0:36

Big hint:

$\cos(\frac{\pi}{4})=\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$

$\sin(\frac{\pi}{4})=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$

$\cos(\frac{7\pi}{6})=-\frac{\sqrt{3}}{2}$

$\sin(\frac{7\pi}{6})=-\frac{1}{2}$

$r(\cos(\theta)+i\sin(\theta)) = re^{i\theta}$

$e^{2\pi i}=1$

So for the second one, we have: $(-\sqrt{3}-i)=2(-\frac{\sqrt{3}}{2}-\frac{1}{2}i)=2(\cos(\frac{7\pi}{6})+i\sin(\frac{7\pi}{6}))$. Now, raising that to a power...

Example worked out:

$(-\frac{1}{4}+\frac{\sqrt{3}}{4}i)^{32}$

We first convert to polar form by recognizing what angle gives us values like those above:

$(-\frac{1}{4}+\frac{\sqrt{3}}{4}i)^{32} = (\frac{1}{2}(-\frac{1}{2}+\frac{\sqrt{3}}{2}i))^{32} = (\frac{1}{2}(\cos(\frac{2\pi}{3})+i\sin(\frac{2\pi}{3})))^{32}$

$=(\frac{1}{2}e^{\frac{2\pi i}{3}})^{32}$

Apply properties of exponents: $(a^{b})^c = a^{bc}$

$=(\frac{1}{2})^{32}e^{\frac{64\pi i}{3}}$

Simplify by factoring out integer powers of $e^{2\pi i}$

$=\frac{1}{2^{32}}e^{\frac{60\pi i}{3}}\cdot e^{\frac{4\pi i}{3}} = \frac{1}{2^{32}}e^{\frac{4\pi i}{3}}$

Convert back to rectangular if you so desire.

$=\frac{1}{2^{32}}(-\frac{1}{2}-\frac{\sqrt{3}}{2}i) = -\frac{1}{2^{33}}-\frac{\sqrt{3}}{2^{33}}i$

$z = (\frac1{\sqrt 2} + \frac1{\sqrt 2}i)\\ z = \rho(\cos\theta + i \sin \theta)\\ z^{95} = \rho^{95}(\cos\theta + i \sin \theta)^{95}\\ z^{95} = \rho^{95}(\cos 95\theta + i \sin 95\theta) \text{ -- This is De Moivre's Theorem}\\ \cos (\theta + 2\pi) = \cos\theta,\sin (\theta + 2\pi) = \sin\theta$

Find the remainder of $\frac {95\theta}{2\pi}$. That is, $95\theta = k (2\pi) + r$

$z^{95} = \rho^{95}(\cos r + i \sin r)$

$(\frac1{\sqrt 2} + \frac1{\sqrt 2}i)^{95} = (\frac 1{\sqrt2} - \frac 1{\sqrt2}i)$

Now, I left out the calucations, but you have the process. See, if you can get an answer. Then try the second one on your own.

b) $z= - \sqrt{3} - i\\ \rho = \sqrt{x^2 + y^2} = \sqrt{3+1} = 2\\ z = 2(-\frac{\sqrt3}{2} - \frac 12 i)\\ z = 2(\cos \frac{4\pi}{3} + i\sin \frac{4\pi}{3}) \text{ -- 240 degrees if you prefer}\\ z^{13} = 2^{13}(\cos \frac{4\pi}{3} + i\sin \frac{4\pi}{3})^{13}$

$z^{12}$ cycles us around to a multiple of $2\pi$

$z^{13} = 2^{13}(-\frac{\sqrt 3}2 - \frac 12 i)\\ (-\sqrt 3 - i)^{13} = 2^{12}(-\sqrt 3 - i)$

• So I started working on the second one, thinking it might be easier but this is how far I got imgur.com/c4KrhTf. I am having a hard time going further – user3504306 Jun 23 '16 at 2:46
• looks like you got off to a bad start. $r = \sqrt{x^2 + y^2}$ you have a minus sign in there. And $\theta$ should be the same $\theta$ for both $\cos \theta$, and $\sin \theta$ – Doug M Jun 23 '16 at 3:46