Suppose $z=re^{iθ}$, prove : $|e^{iz}|=e^{-r\sinθ}$.

Question

Suppose $z=re^{iθ}$, prove $|e^{iz}|=e^{-r\sin θ}$.

I tried but the result was not as expected: $$e^{iz}=\cosh z+i(\sinh z)\\|e^{iz}|=(\cosh z)^2+[i(\sinh z)]^2=\cosh(2z)$$

Answer 1

We have $$e^{iz}=e^{ire^{i\theta}}=e^{ir(\cos\theta+i\sin\theta)}=e^{-r\sin\theta}e^{ir\cos\theta}$$

Now, taking the norm, note that $|e^{ir\cos\theta}|=1$, and since the norm is multiplicative, we get $$|e^{iz}|=|e^{-r\sin\theta}e^{ir\cos\theta}|=|e^{-r\sin\theta}|\cdot|e^{ir\cos\theta}|=e^{-r\sin\theta}$$

Edit: $|e^{ir\cos\theta}|=1$ because $$e^{ir\cos\theta}=\cos(r\cos\theta)+i\sin(r\cos\theta)$$ so $$|e^{ir\cos\theta}|^2=\cos^2(r\cos\theta)+\sin^2(r\cos\theta)=1$$