Real part and argument of $e^{i\ z}$ Question and my attempt at it
I got the correct answer to this question from two different methods but on writing $e^{i\theta}$ in the form of $cis(\theta)$  in the first method , why do I get a different answer to this problem?
Please don't give a solution to this question, just tell me where am I wrong?
 A: Note that $$e^{iz}=e^{-\sin\theta}e^{i\cos \theta}=e^{-\sin\theta}(\cos(\cos \theta)+i\sin(\cos\theta)).$$ Thus the argument is $$\arctan\dfrac{\sin(\cos\theta)}{\cos(\cos\theta)}=\arctan\tan(\cos\theta)=\cos\theta.$$
(Note that you forgot to write $\sin\cos\theta$.)
A: Consider the more general problem with $z=x+iy$, so $iz=-y+ix$ and
$$
e^{iz}=e^{-y}e^{ix}=e^{-y}(\cos x+i\sin x)
$$
Then $|e^{iz}|=e^{-y}$ and $\arg(e^{iz})=x$ (with reduction to the required interval for the argument).
If now $z=\cos\theta+i\sin\theta$, we have
$$
|e^{iz}|=e^{-{\sin\theta}},
\qquad
\arg(e^{iz})=\cos\theta
$$
(the latter with possible reduction needed).
If your arg function maps to $(-\pi,\pi]$, you have to do no reduction. If the arg function maps to $[0,2\pi)$,
$$
\arg(e^{iz})=\begin{cases}
\cos\theta & \text{if $\cos\theta\ge0$} \\
2\pi+\cos\theta & \text{if $\cos\theta<0$}
\end{cases}
$$
Note that the relation
$$
\arg(z)=\arctan\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}
$$
is not true in general. It is in this case because $-1\le \cos\theta\le 1$, so $\cos\theta\in(-\pi/2,\pi/2)$. However, with the $x+iy$ notation as before,
$$
\arctan\frac{\operatorname{Im}(e^{iz})}{\operatorname{Re}(e^{iz})}
=\arctan\frac{e^{-y}\sin x}{e^{-y}\cos x}=\arctan \tan x
$$
so you get nothing new than with the method above.
You just got confused: if $z=\cos\theta+i\sin\theta$ you'd get
$$
\arctan\frac{\operatorname{Im}(e^{iz})}{\operatorname{Re}(e^{iz})}
=\arctan\frac{e^{-{\sin\theta}}\sin\cos\theta}{e^{-{\sin\theta}}\cos \cos\theta}=\arctan \tan \cos\theta=\cos\theta
$$
Also you seem to be confusing $|w|$ with $\operatorname{Re}(w)$.
