$Z=\ln(i)/i$ , find $Z$.

Question

It feels like a trivial question, but i don't know if my answer is correct. My attempt : $$Z = \frac{\ln(i)}{i}$$ $$iZ= \ln(i)$$ $$e^{iZ} = i$$ $$\cos Z + i\sin Z = i$$ $$\cos Z = i(1-\sin Z)$$ Squaring both sides(i think doing this is incorrect): $${cos}^2Z = -(1 + {\sin}^2Z - 2\sin Z)$$ This gives : $$\sin Z= 0$$ $$or$$ $$\sin Z = 1 $$ So the solutions are : $$Z= n\pi + (-1)^n\pi/2$$ $$and$$ $$Z= n\pi $$ Is this correct? (I know it is not rigorous.)

Answer 1

In your derivation, when squaring at both sides, you forget that $i^2=-1$, and hence the signs are incorrect on the right-hand side.

I will present a different (and simpler) derivation. Note that $i=e^{i\frac{\pi}{2}+i2k\pi}=e^{i\pi\big(\frac{1}{2}+2k\big)}$ with $k\in\mathbb{Z}$, so

$$\begin{align} Z=\frac{\ln(i)}{i}=\frac{\ln(e^{i\pi\big(\frac{1}{2}+2k\big)})}{i}=\frac{i\pi\big(\frac{1}{2}+2k\big)}{i}=\frac{\pi}{2}+2k\pi\,\,,\,k\in\mathbb{Z}\,. \end{align}$$

If you assume that $\ln(i)$ is the principal logarithm, we dismiss the $2k\pi$ term.

Answer 2

Note that $\log(i)$ is the "principal" logarithm. The definition of $\log(z)$ is $\log|z|+i\arg(z),~z\neq0$. Thus, in your case, $z=\frac{\log(i)}{i}=\frac{\frac{\pi i}{2}}{i}=\frac{\pi }{2}$.