How to easily see that the complex conjugate of $\dfrac{1}{1-a e^{-j 2 \pi f}}$ is $\dfrac{1}{1-a e^{j 2 \pi f}}$.

Question

Given the function of $f$ with $j = \sqrt{-1}$ $$g(f) = \dfrac{1}{1-a e^{-j 2 \pi f}}$$ My textbook on signal processing quickly (no steps) shows that the complex conjugate of this function is $$h(f) = \dfrac{1}{1-a e^{j 2 \pi f}}$$

However, the complex conjugate of a function $g(f) = \text{Re}(g(f)) + i \text{Im}(g(f))$ is defined as $g^*(f) = \text{Re}(g(f)) - i \text{Im}(g(f))$

Clearly, the $g(f)$ as given is NOT of the above form and requires further manipulation. We are talking about a few more steps here.

But I wonder if the book author is using a quicker way of showing that the complex conjugate of $g(f)$ should be given as $h(f)$ (i.e., without converting it into rectangular form first). Is changing the $+j$ to $-j$ enough?

Answer 1

The conjugate of numbers of the form

$$ z = e^{j\theta}$$

is

$$ z^* = e^{-j\theta} $$

As a consequence of Euler's formula

$$ e^{j\theta} = \cos \theta + j\sin \theta$$

An easier way to think about the conjugate is just going through the expression and replace every instance of $j$ by $-j$

---

Here's a more rigorous derivation. From above, we know that

$$ (1-ae^{-j2\pi f})^* = 1 -ae^{j2\pi f} $$

Thus we can rationalize the denominator to get

$$ \begin{align} g^*(f) &= \left(\frac{1-ae^{j2\pi f}}{(1-ae^{-j2\pi f})(1-ae^{j2\pi f})}\right)^* \\ &= \frac{(1-ae^{j2\pi f})^*}{(1-ae^{-j2\pi f})(1-ae^{j2\pi f})} \\ &= \frac{1-ae^{-j2\pi f}}{(1-ae^{-j2\pi f})(1-ae^{j2\pi f})} \\ &= \frac{1}{1-ae^{j2\pi f}} \end{align} $$

Answer 2

HINT: you need to know some facts about complex numbers:

$$z^{-1}=\bar z/|z|^2\\ z=|z|e^{i\phi}\iff \bar z=|z|e^{-i\phi}$$

for $z\in\Bbb C\setminus\{0\}$. Also we have that for $w,z\in\Bbb C$ $$\overline{z+w}=\bar z+\bar w\quad\text{and}\quad \overline{z\cdot w}=\bar z\cdot\bar w$$

By last its easy to check that $\overline{z^n}=(\bar z)^n$ for $n\in\Bbb Z$ and $z\in\Bbb C$. Take a look here for more identities related to the conjugation of complex numbers.

Answer 3

This is quite easy to do.

Start with $a\in\mathbb{R}$ and $z=x+iy\in \mathbb{C}$ and consider $$ Z=\frac{1}{a+z}=\frac{a+z^*}{(a+z)(a+z^*)} = \frac{a+z^*}{(a^2+a(z+z^*)+zz^*)} $$ Note that the denominator is certainly real. Now consider \begin{align} Z^*=\left(\frac{1}{a+z}\right)^*&=\frac{a+z}{a^2+a(z+z^*)+zz^*}\, ,\\ &=\frac{a+z}{(a+z)(a+z^*)}\, ,\\ &=\frac{1}{a+z^*} \end{align} so it’s enough to take the complex conjugate $z$ in $1/(a+z)$ to obtain the conjugate of the whole expression.