trigonometry - Something strange regarding Euler's Identity

By Euler's identity,
$$e^{i\theta}=\cos\theta+i\sin\theta$$
And so if we let
$$\phi=e^{i\theta}=\cos\theta+i\sin\theta$$
Then we have
$$\phi=e^{i\theta}$$

$$\theta=\frac{\ln\phi}{i}$$
$$\theta=-i\ln\phi$$
and
$$\phi=\cos\theta+i\sin\theta$$
and, using an inversion formula that I explained in my question here,
$$\theta=\arcsin\frac{\phi}{\sqrt{1^2+i^2}}-\arcsin\frac{1}{\sqrt{1^2+i^2}}$$
$$\theta=\arcsin\frac{\phi}{\sqrt{1-1}}-\arcsin\frac{1}{\sqrt{1-1}}$$
$$\theta=\arcsin\frac{\phi}{0}-\arcsin\frac{1}{0}$$
and so it seems that
$$-i\ln\phi=\arcsin\frac{\phi}{0}-\arcsin\frac{1}{0}$$

Is there any meaning in this whatsoever? Why does this happen when I try to manipulate Euler's formula?

Answer

Your problem resides in taking the $\ln$ of a complex exponential. You assert that $e^{i \theta} = \phi$, but notice that:

$$e^{i \tau} = e^{2 i \tau} = e^{3 i \tau} = ... = 1$$

(I use $\tau = 2 \pi$). This shows that the complex exponential is not a one-to-one function, and so you can't just take its inverse with $\ln$. Thus your assertion that:

$$\theta = \frac{\ln \phi}{i}$$

May not be true.

[EDIT:] In regards to why you obtain a division by zero inside your $\arcsin$, the issue lies in the derivation of your formula for inverting $a \cos x + b \sin x$.

In your question that you referenced, you multiply and divide your equation by $\sqrt{a^2 + b^2}$. In this case, since $a^2 + b^2 = 0$, you're dividing by $0$, which isn't allowed. You may have to find a different way to get an inverse while avoiding dividing by zero.