# Evaluate $(\frac{1}{-\sqrt{2}+\sqrt{2}i})^{2011}$

Evaluate $$(\frac{1}{-\sqrt{2}+\sqrt{2}i})^{2011}$$

So ...

$$(\frac{1}{-\sqrt{2}+\sqrt{2}i})^{2011} = (-\sqrt{2}+\sqrt{2}i)^{-2011}$$

$$\theta=\pi - \arctan(\frac{\sqrt{2}}{\sqrt{2}}) = \frac{3\pi}{4}$$

$$-\sqrt{2}+\sqrt{2}i=\cos{\theta} + i \sin{\theta}$$

$$(-\sqrt{2}+\sqrt{2}i)^{-2011} =\cos{(-2011\theta)} + i \sin{(-2011\theta)} = e^{i(-2011)\theta}$$

Is it correct?

...

$$\arg{z} = \frac{3\pi}{4}, \qquad z=2e^{i \frac{3\pi}{4}}$$

$$LHS = (\frac{1}{z})^{2011}=2^{-2011}e^{-2011(\frac{3\pi}{4})i} = 2^{-2011} e^{-(\color{red}{1508\pi i} + \frac{\pi i }{4})} = 2^{-2011}e^{-\frac {\pi i}{4}} = 2^{-2011} \color{blue}{\frac{1 - i}{\sqrt{2}}} = ...$$

Why is the red $1508\pi i$ removed in the following step?

How do I get the blue $\frac{1 - i}{\sqrt{2}}$ from the prev step?

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$\frac{1}{-\sqrt{2}+i\sqrt{2}}=\frac{-\sqrt{2}-i\sqrt{2}}{4}=\frac{1}{2}\left( -\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2} \right)$ This explains the term $2^{-2011}$, but the argument is $\frac{5}{4}\pi$. –  Mark Apr 14 '12 at 6:22
How is it possible for $\cos\theta =-\sqrt{2}$ and $\sin\theta=\sqrt{2}$? You have to convert $z$ fully to polar form $re^{i\theta}$. Here the modulus $r$ is not $1$. –  anon Apr 14 '12 at 6:31
Updated question fixing signs –  Jiew Meng Apr 14 '12 at 8:05

$$\frac{1}{-\sqrt{2}+i\sqrt{2}}\frac{-\sqrt{2}-i\sqrt{2}}{-\sqrt{2}-i\sqrt{2}}=\frac{-\sqrt{2}-i\sqrt{2}}{4}=\frac12\frac{-1-i}{\sqrt{2}}$$ which has absolute value $\dfrac12$ and argument $\dfrac{5\pi}{4}$, so $\left(\frac{1}{-\sqrt{2}+i\sqrt{2}}\right)^{2011}$ has absolute value $1/2^{2011}$ and argument $\frac{5\pi}{4}\cdot2011\equiv-\frac{\pi}{4}\pmod{2\pi}$. Therefore, $$\left(\frac{1}{-\sqrt{2}+i\sqrt{2}}\right)^{2011}=\frac{1}{2^{2011}}\frac{1-i}{\sqrt{2}}$$ If, as the title suggests, the exponent is $-2011$, we simply need to take the reciprocal: $$\left(\frac{1}{-\sqrt{2}+i\sqrt{2}}\right)^{-2011}=2^{2011}\frac{1+i}{\sqrt{2}}$$

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Updated my question, fixing the signs. Anyways, how do I get from $2^{-2011}e^{-\frac {\pi i}{4}}$ to $2^{-2011} \frac{1 - i}{\sqrt{2}}$ –  Jiew Meng Apr 14 '12 at 8:15
Oh I simply "reverse" the polar form? –  Jiew Meng Apr 14 '12 at 8:18
Yes, polar and cartesian form: $$e^{-\frac{\pi i}{4}}=\frac{1-i}{\sqrt{2}}$$ –  robjohn Apr 14 '12 at 12:35

Ok, lots of bits here and there.

With respect to your answer, it is not true that $-\sqrt{2}+\sqrt{2}i=\cos{\theta} + i \sin{\theta}$, as in particular the length of the element on the left is $2$ and the length of the element on the right is $1$.

With respect to your question about the red bit, recall that $e^{2 i \pi} = 1$. So one can simply ignore the part of the exponent properly divisible by 2.

Similarly, they get the blue part by simply evaluating $e^{-i\pi/4 }$.

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First off, let's get the signs correct:

$$\left(\frac{1}{-\sqrt{2}+\sqrt{2}~i}\right)^{\color{Red}+2011}=(-\sqrt{2}+\sqrt{2}~i)^{-2011}.$$

Second, if $z=-\sqrt{2}+\sqrt{2}i=\sqrt{2}(-1+i)$, then we do have a magnitude of two:

$$|z|=\sqrt{2}\cdot |-1+i|=\sqrt{2}\cdot\sqrt{(-1)^2+1}=2.$$

The angle $\theta=\frac{3}{4}\pi$ of $z$ is correct so finally

$$z^{-2011}=(2e^{\frac{3}{4}\pi i})^{-2011}=2^{-2011}\exp\left(-\frac{6033}{4}\pi i\right)=2^{2011}e^{-\pi i/4}$$

because $6033\equiv1\bmod8$. We could also evaluate $e^{-\pi i/4}$ as $(1-i)/\sqrt{2}$ and get $2^{-(2011+1/2)}(1-i)$.

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\begin{eqnarray*} 2^{-2011} e^{-(\color{red}{1508\pi i} + \frac{\pi i }{4})} &=& 2^{-2011}e^{-(\color{red}{1508\pi i})} \cdot e^{\left( \frac{\pi i }{4}\right)} \\ &=& 2^{-2011}{\left(e^{(\color{red}{2\pi i})}\right)}^{-\frac{1508}{2}} \cdot e^{\left( \frac{\pi i }{4}\right)} \\ &=& 2^{-2011}{(\color{red}{1})}^{-\frac{1508}{2}} \cdot e^{\left( \frac{\pi i }{4}\right)} \hspace{5mm} \text{(Using the fact that $e^{2 \pi i } = 1$)} \\ &=& 2^{-2011} e^{\left( \frac{\pi i }{4}\right)}. \end{eqnarray*}