Find the principal value of $z=\ln\left(i\tanh\left(\frac{\pi}{2}\right)\right)$

Question

Steps that I have taken: Substitute $\tanh(π/2)$ with $C$. Then we have $\ln(i\cdot C)=\ln|c| +i(\arg z)$. I am desperately stuck here. I also tried expressing $\tanh$ by definition using the powers of $e$, but yielding no results.

Answer 1

$$z=\ln\left[i\tanh\left(\frac\pi2\right)\right]$$ $$z=\ln\left[i\left(\frac{e^{\pi/2}-e^{-\pi/2}}{e^{\pi/2}+e^{-\pi/2}}\right)\right]$$ $$z=\ln\left[i\left(\frac{e^{\pi}-1}{e^{\pi}+1}\right)\right]$$

$$\ln(x+iy)=\ln{\sqrt{x^2+y^2}}+i\arctan\left(\frac yx\right)$$

See Complex Logarithm

since quantity inside $\ln$ is imaginary $\arg(z)=\pi/2$ $$z=\ln \left(\frac{e^{\pi}-1}{e^{\pi}+1}\right)+i\left(\frac\pi2\right)$$

Note: General value can be acquired by replacing principle value of $\arg=\theta$ by it's general value $2n\pi+\theta$ .