Something strange regarding Euler's Identity

Question

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 1

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.