Proving that $\log(1-ei) = 1-i(\frac{\pi}{2})$

Question

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.

Answer 1

I reckon that the identity is not true:

1. $\log(1-ei) = 1-i(\dfrac{\pi}{2}$)

2. $1-ei = e^{1-i(\dfrac{\pi}{2})}$

3. $1-ei = \dfrac{e}{e^\dfrac{i\pi}{2}} = \dfrac{e}{\cos\left(\dfrac{\pi}{2}\right)+i\sin\left(\dfrac{\pi}{2}\right)} = \dfrac{e}{i}$

4. $i+e = e$

5. $i = 0$

Which is obviously wrong.