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Question: Find the values of $n$ for which

$$(1+i)^{2n}=(1-i)^{2n}$$

wolfram alpha tells me that the answer should be :

$$n=\frac{2i\pi m}{\log(1-i)-\log(1+i)}$$

$$n=-\frac{i(2\pi m+\pi)}{\log(1-i)-\log(1+i)}$$

PS:I have no idea about the answer. I know de Moivre's theorem and complex logs and I am sure this one could be solved by combinig the two.

My miserable attempt:

$2^{n} e^{\frac{i\pi n}{2}}=2^{n} e^{\frac{-i\pi n}{2}}$

Now assuming $n$ is real, cancelling $2^n$ on both sides

$ e^{\frac{i\pi n}{2}}= e^{\frac{-i\pi n}{2}}$

$\frac{n \pi}{2}=\frac{-n \pi}{2}+ 2 \pi k$

$n\pi =2 \pi k$

$n=2k$ which is far from being correct

Hashir Omer
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  • It will help to realize that Wolfram Alpha gave you an overly-complicated form of the answer. $\log(1-i)-\log(1+i)$ simplifies quite a bit. (Here again, writing $1+i$ and $1-i$ in exponential form is key.) – Blue Jul 21 '14 at 10:57
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    $n=2k$ where $k\in\mathbb{Z}$ is the correct answer. – Scientifica Jul 21 '14 at 11:01

5 Answers5

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

$$\implies(-1)^n=1$$

lab bhattacharjee
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Hint: Write $(1+i)$ and $(1-i)$ in the exponential form.

Scientifica
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Multiply both sides with $(1+i)^{2n}$ to arrive at $$ (1+i)^{4n}=\bigl((1+i)(1-i)\bigr)^{2n}=2^{2n}\in\mathbb R_{>0}$$

Hagen von Eitzen
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Hint

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

Claude Leibovici
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Let $w=\frac{1+i}{1-i}$ therefore we have $ w^{2n}=1$ and so $$w=\cos\frac{2k\pi}{2n}+i\sin\frac{2k\pi}{2n} \ \ \text{for}k=0,1,\cdots, 2n-1$$

user62498
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