Proving that: $-\frac{2 \;i\log(i^2)}{2} = \pi$

Question

I'm trying to prove that: $$-\frac{2 \;i\log(i^2)}{2} = \pi$$

This is what I've tried:

$$-\frac{2 \;i\log(i^2)}{2} = -i \log(i^2) = -i (i \pi)\implies$$ $$-x\;(x y)=-x\;y\;x\implies$$ $$-i\;(i \pi) = -i\;i\;\pi = \pi \implies$$ $$-i\;i = (-i^2) = (-(-1)) = 1$$

Is this considered to be sufficient proof?

Is there an alternative approach?

Answer 1

Assuming the non-negative reals branch cut for the logarithm function (i.e., $\;\Bbb C-\Bbb R_+\;$) :

$$-\frac{2i\log i^2}2=-\frac{2i\log(-1)}2=-\frac{2i\left(\log 1+i\pi\right)}2=-\frac{-2\pi}{2}=\pi$$