# For which z is the following true: Log(iz^2)=frac(i\pi)(2)+2Log(z)

Makaila Simon 2022-09-17 Answered
For which $z$ is the following true: $Log\left(i{z}^{2}\right)=\frac{i\pi }{2}+2Log\left(z\right)$. We raised both sides by $e$ and concluded this equation holds for all $z\ne 0$, but are not sure if we have all the work for it. We are mostly concerned with how the Arg part comes into play here and if that will limit what $z$ can be.
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## Answers (1)

Rachael Conner
Answered 2022-09-18 Author has 8 answers
Substitute $z=r{e}^{i\varphi }$ with $-\pi <\varphi \le \pi$
Then we get:
$\text{Log}\left(i{r}^{2}{e}^{2i\varphi }\right)=\frac{i\pi }{2}+2\text{Log}\left(r{e}^{i\varphi }\right)$
$2\mathrm{log}r+\text{Log}\left({e}^{i\left(\pi /2+2\varphi \right)}\right)=\frac{i\pi }{2}+2\mathrm{log}r+2i\varphi$
$\text{Arg}\left({e}^{i\left(\pi /2+2\varphi \right)}\right)=\frac{\pi }{2}+2\varphi$
$-\pi <\frac{\pi }{2}+2\varphi \le \pi$
$-\frac{3\pi }{4}<\varphi \le \frac{\pi }{4}$
So the expression holds true for any $z\ne 0$ with

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