Interesting mathematical artifact: Equality sign wrong for exponential function. I found an interesting case where it seems like an equality sign works wrong.
Let's consider the following construction:
$\frac{1+\Lambda}{2} e^{i\Lambda \phi}$ where $\Lambda = \pm1$, so $\Lambda^2=1$.
Then I apply Euler formula:
$\frac{1+\Lambda}{2} e^{i\Lambda \phi} = \frac{1+\Lambda}{2} (\cos \phi + i\Lambda \sin \phi)= \frac{1}{2} \cos \phi + \frac{\Lambda}{2} \cos \phi + \frac{i\Lambda}{2} \sin \phi + \frac{i}{2} \sin \phi = \frac{1+\Lambda}{2} e^{i\phi}$
where I have used $\sin(\Lambda \phi) = \Lambda \sin \phi$ and $\cos (\Lambda \phi) = \cos \phi$.
However, this is just wrong! $e^{i\Lambda \phi}\neq e^{i \phi}$ even though the equality sign was not broken anywhere in between (at least it doesn't seem to be broken to me). What am I doing wrong?

The problem is the division by zero:
The identity $\tfrac{1+\Lambda}{2}e^{i\Lambda\phi}=\tfrac{1+\Lambda}{2}e^{i\phi}$ holds as you have shown.
For $\Lambda=1$, this implies $e^{i\Lambda\phi}=e^{i\phi}$, which is obviously correct. However, for $\Lambda=-1$, this means that $\tfrac{1+\Lambda}{2}=0$ and therefore, the last implication $\tfrac{1+\Lambda}{2}e^{i\Lambda\phi}=\tfrac{1+\Lambda}{2}e^{i\phi}\implies e^{i\Lambda\phi}=e^{i\phi}$ is not correct in this case.
