Exponential conjugate equals to reciprocal? $$\Im[e^{-i x}]=- \sin x $$
Is this true too?  $$\frac{1}{\sin x}= \Im[e^{-ix}]$$
If is not true, how can I express the above sine conjugate in terms of exponential?
 A: You have Euler's Theorem:
$$
e^{ix} = \cos x + i \sin x
$$
So the first claim is correct. You can see what happens to the second claim as well...
UPDATE
One way to remedy your claim is to apply inversion in a different place:
$$
\Im\left[\frac{1}{e^{-ix}}\right] = \Im\left[e^{ix}\right] = \sin x
$$
A: It is not true that the imaginary part of a reciprocal of a function $f(z)$ is the reciprocal of the imaginary part $f(z)$.  That is, 
$$\text{Im}\left(\frac{1}{f(z)}\right)\ne \frac{1}{\text{Im}\left(f(z)\right)} \tag 1$$
To see this, let $f(z)=u(x,y)+iv(x,y)$ where $z=x+iy$ and both $u(x,y)$ and $v(x,y)$ are real functions of the real variables $x$ and $y$.  
Then, for $f(z)\ne 0$
$$\text{Im}\left(\frac{1}{f(z)}\right)=\frac{-v(x,y)}{u^2(x,y)+v^2(x,y)} \tag 2$$
and 
$$ \frac{1}{\text{Im}\left(f(z)\right)}=\frac{1}{v(x,y)} 
\tag 3$$
Setting the right-hand side of $(2)$ equal to the right-hand side of $(3)$ reveals
$$u^2(x,y)=-2v^2(x,y) \tag 4$$
Inasmuch as the left-hand side of $(4)$ is non-negative while the right-hand side in non-positive, we conclude that $u(x,y)=v(x,y)=0$.  But, $f(z)\ne 0$ inasmuch as division by $0$ is undefined.  Therefore, the statement given by $(1)$ is true.
