Solving for m in $\left ( 1+i \right )^{m}= \left ( 1-i \right )^{m}, m\in \mathbb{Z}$

Question

Taking the log of both sides , I wound up with an equality that looks like this: $$m\cdot i\left ( \ln \left ( \frac{\pi }{4} +2k\pi\right )-\ln \left ( \frac{7\pi}{4} +2k\pi\right ) \right )=0$$ which to my knowledge is satisfied only when $m=0$. However, wolfram alpha gives multiple solutions: http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427edlc1a076ou

The two numbers called $k$ in your equation do not need to be equal — Julian Rosen, Oct 10 '15 at 01:09

score 4 · Accepted Answer · answered Oct 09 '15 at 20:40

4

$$\left(\frac{1+i}{1-i}\right)^m=1\iff i^m=1\iff m\equiv 0\pmod{4}$$

answered Oct 09 '15 at 20:40

user236182

13,324

But, what about wolfram-alpha's solution? – Emir Šemšić Oct 09 '15 at 20:49
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Suggestion: you should add the $\cdot \frac {1+i}{1+i}$ before $i^m=1$ for clarity. – YoTengoUnLCD Oct 09 '15 at 20:49
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@YoTengoUnLCD You don't need $\cdot\frac{1+i}{1+i}$ to prove $\frac{1+i}{1-i}=i$, which is equivalent to $1+i=i(1-i)$, which is true. – user236182 Oct 09 '15 at 20:52
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Wolfram-Alpha's solution doesn't suppose $m$ is an integer. – Bernard Oct 09 '15 at 21:51
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@Shemafied Try searching "solve in integers (1+i)^m=(1-i)^m". – user236182 Oct 09 '15 at 22:05
Oh, thanks for the hint. – Emir Šemšić Oct 10 '15 at 09:13

Solving for m in $\left ( 1+i \right )^{m}= \left ( 1-i \right )^{m}, m\in \mathbb{Z}$

1 Answers1