The following is a trial proof in which I try to establish a link between $\pi$ and $\ln(i)$. I want your hints if it is false.
We can use Euler's formula to relate $e^{i\pi}$ to $\cos(\pi)$ and $\sin(\pi)$, and the definition of the natural logarithm to relate $\ln(i)$ to its real and imaginary parts. Then we can combine these to obtain a formula that relates $e^{i\pi}$ and $\ln(i)$.
Euler's formula states that $e^{i\theta} = \cos(\theta) + i\sin(\theta)$ for any real number theta. So, if we let $\theta = \pi$, we have:
$e^{i\pi} = \cos(\pi) + i\sin(\pi)$
Now, $\cos(\pi) = -1$ and $\sin(\pi) = 0$, so we have:
$e^{i\pi} = -1$
Next, we use the definition of the natural logarithm to write $\ln(i)$ in terms of its real and imaginary parts. Recall that for any complex number $z = x + iy$ (where $x$ and $y$ are real numbers), we have:
$\ln(z) = \ln(|z|) + i\arg(z)$
where $|z|$ is the modulus (or absolute value) of $z$, and $\arg(z)$ is the argument (or angle) of $z$.
For $i$, we have $|i| = 1$ and $\arg(i) = \frac{\pi}{2}$ (since $i$ is the complex number that lies on the positive imaginary axis). So we have:
$\ln(i) = \ln(1) + i\frac{\pi}{2}$
Simplifying, we have:
$$\ln(i) = i\frac{\pi}{2}$$