Exponential conjugate equals to reciprocal?

Question

$$\Im[e^{-i x}]=- \sin x $$ 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?

Answer 1

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: $$ \Im\left[\frac{1}{e^{-ix}}\right] = \Im\left[e^{ix}\right] = \sin x $$

Answer 2

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.