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

Question

The question on our worksheet is: For which $z$ is the following true: $Log(iz^2) = \frac{i\pi}{2} + 2Log(z)$. We raised both sides by $e$ and concluded this equation holds for all $z\neq 0$, 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.

Answer 1

Substitute $z=re^{i\phi}$ with $-\pi < \phi \le \pi$.

Then we get: $$\text{Log}(ir^2e^{2i\phi})=\frac{i\pi}2 + 2\text{Log}(re^{i\phi})$$ $$2\log r + \text{Log}(e^{i(\pi/2 + 2\phi)})=\frac{i\pi}2 + 2\log r + 2i\phi$$ $$\text{Arg}(e^{i(\pi/2 + 2\phi)})=\frac{\pi}2 + 2\phi$$ $$-\pi < \frac\pi 2 + 2\phi \le \pi$$ $$-\frac{3\pi}4 < \phi \le \frac\pi 4$$

So the expression holds true for any $z \ne 0$ with $-\frac{3\pi}4 < \text{Arg } z \le \frac\pi 4$.