Proving that $\log(1-ei) = 1-i(\frac{\pi}{2})$
One thing that I know is $|ie|=e$ and $\log(z) = \ln|z| + i\arg(z)$, but I don't know how to calculate it because there is 1 there.
Solution 1:
I reckon that the identity is not true:
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$\log(1-ei) = 1-i(\dfrac{\pi}{2}$)
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$1-ei = e^{1-i(\dfrac{\pi}{2})}$
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$1-ei = \dfrac{e}{e^\dfrac{i\pi}{2}} = \dfrac{e}{\cos\left(\dfrac{\pi}{2}\right)+i\sin\left(\dfrac{\pi}{2}\right)} = \dfrac{e}{i}$
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$i+e = e$
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$i = 0$
Which is obviously wrong.