Complex conjugate of an involved expression

Question

I understand the the complex conjugate of, say, $z:=\exp({a+ib})$ is $z:=\exp({a-ib})$.

However , I have a composite expression and I'm not sure how to attack taking it's complex conjugate.

Say $z:=i\exp({ib}) / ({a + ic})$

I would be tempted to say that the denominator becomes ${a - ic}$, that the denominator changes signs & the exponential as well, so:

$z*:=-i\exp({-ib}) / ({a - ic})$

I'm asking because I need to compute the norm of a complex expression (which structurally is similar to this exemple) and I feel I'm about to embark on a rather lengthy derivation, based in part on the computation of that norm... hence would like to know if my understanding of the complex conjugate is accurate in a more involved case.

Thanks

EDIT: wrt to comment: a, b, and c are real (e.g. I have explicited any imaginary part)

Answer 1

The things you state are true, and here are some basic facts about complex conjugates and norms, some of which you have used:$$ (zw)^* = z^*w^* \:\text{and}\: \left(\dfrac{z}{w}\right)^* = \frac{z^*}{w^*}$$

also

$$ |zw| = |z|\cdot |w| \:\text{and}\: \left| \dfrac{z}{w} \right| = \dfrac{|z|}{|w|}$$

(for any complex $z$ and $w$).

Normally these are proved (or merely stated) immediately after conjugates and norms are introduced. Unless you're just starting with complex numbers it's probably okay for you to use them.

Answer 2

If $z=e^{a+ib}$ then $z=e^a(\cos b+i\sin b)$, so $\bar z=e^a(\cos b-i\sin b)$ and then $\bar z=e^{a-ib}$.

Answer 3

In general, if $f$ has a power series with real coefficients then $\overline{f(z)} = f( \overline{z})$. Hence $\overline{e^z} = e^\overline{z}$.

Since $\overline{wz} = \overline{w} \ \overline{z}$ and $\overline{\left( w \over z \right)} = {\overline{w} \over \overline{z}}$, we have $\overline{ \left ( {i \over a+ic} e^{ib}\right ) } = \overline{ \left ( {i \over a+ic} \right ) }\ \overline{ \left ( e^{ib}\right ) } = \left ( {\overline{i} \over \overline{a+ic}} \right ) e^{-ib}= {-i \over a-ic}e^{-ib}$.