How in this world can I simplify this $\sqrt 2\cdot(1/(\sqrt2)-1/(\sqrt2)\cdot i)^{31}$ ????

Question

I have a problem, obviously. I am doing some maths and now I have to simplify this: $\sqrt 2\cdot(1/(\sqrt2)-1/(\sqrt2)\cdot i)^{31}$ ????. But I just don´t know how ???? I´ve started simplifying by multiplying numerator and denominator with $\sqrt 2$ but that´s it I just can´t go any further because I have really no idea how to continue. So please is there anybody who can help me ? Because i really tried and also google won´t help :D

Answer 1

HINT:

$$(x+iy)^n=\left(\sqrt{x^2+y^2}e^{i\arctan2(y,x)}\right)^n \tag 1$$

where $\arctan2(y,x)$ is given HERE.

SPOILER ALERT: Scroll over the highlighted area to reveal the solution.

Using $(1)$ with $x=\frac{1}{\sqrt 2}$, $y=-\frac{1}{\sqrt 2}$, and $n=31$, we find that $$\begin{align}\sqrt 2\left(x+iy\right)^{n}&=\sqrt 2\left(\sqrt{\frac12+\frac12}e^{-i\pi/4}\right)^{31}\\\\&=\sqrt {2}e^{-i31\pi/4}\\\\&=1+i\end{align}$$

Answer 2

$$\sqrt{2}\left(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right)^{31}=$$

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• $$\left|\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right|=\sqrt{\left(\frac{1}{\sqrt{2}}\right)^2+\left(\frac{1}{\sqrt{2}}\right)^2}=1$$
• $$\arg\left[\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right]=-\arctan\left(\frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}\right)=-\frac{\pi}{4}$$

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$$\sqrt{2}\left(e^{-\frac{\pi i}{4}}\right)^{31}=\sqrt{2}e^{-\frac{31\pi i}{4}}=\sqrt{2}e^{\frac{\pi i}{4}}=1+i$$

Answer 3

Have you thought of using Euler identities? How about this: $$\sqrt{2}(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i)^{31}=\sqrt{2}[\cos(-\frac{\pi}{4})+i\sin(-\frac{\pi}{4})]^{31}=\sqrt{2}(e^{-i\frac{\pi}{4}})^{31}=\sqrt{2}e^{-i\frac{31\pi}{4}}$$

Answer 4

\begin{align} & \left( \frac 1 {\sqrt 2} - \frac 1 {\sqrt 2} i \right)^{31} \\[10pt] = {} & \underbrace{\left( \cos(-45)^\circ + i\sin(-45^\circ) \right)^{31} = \cos(-31\cdot 45^\circ) + i \sin(-31\cdot45^\circ)}_\text{de Moivre's theorem} \\[10pt] = {} & \cos(+45^\circ) + i \sin(+45^\circ) = \cdots\cdots \end{align} (and then multiply both sides by $\sqrt 2$).