Solve $\log(z^2-1)=\frac{i\pi}{2}$

Question

I was wondering if someone could help me solve the following complex logarithmic equation, such that $$ \text{Log}(z):=\log(z) \iff\arg z=\theta_p $$

$$ \forall z \in \mathbb{R} :\text{Log}(z^2-1)=i\pi/2 $$

So far, I have

$$ w=e^z \implies w^2+w+1=0. $$ Solving for $w$ using quadratic $$ w=-1/2+i\sqrt3/2 $$ or$$ w=-1/2-i\sqrt3/2 $$ From the text, I know a few things $$ \text{Log}(z)=\text{Log}(r)+i\theta_p : r=|z|>0, \theta_p=\arg(z), -\pi<\theta_p<=\pi $$ $$ \text{Arg}(z)=\theta_p+2\pi k : k=..., -3, -2, -1, 0, 1, 2, 3, ... $$ $$ \log(z)=\text{Log}(r)+i(\theta+2 \pi k) $$

Thank you in advance for your help!

Answer 1

Hint:

Eliminate the natural logarithm on both sides, i.e. $$\exp(\log (z^2 -1)) = \exp \left(\dfrac {i \pi}{2}\right)$$ Also notice that $$\exp \left(\dfrac {i \pi}{2}\right) \rightarrow \cos \dfrac {\pi}{2} +i \sin \dfrac {\pi}{2} = 0 + i (1) = i$$

Can you take it from here?