Complex conjugate properties

Here are some complex conjugate properties and identities that are useful to know for complex numbers \(z\) and \(w\).

Complex conjugation is distributive over addition, subtraction, multiplication and division. $$ \begin{align*}
\overline{z+w} &= \overline{z} + \overline{w} \\
\overline{z-w} &= \overline{z}-\overline{w} \\
\overline{zw} &= \overline{z} \, \overline{w} \\
\overline{\left( \frac{z}{w} \right)} &= \frac{\overline{z}}{\overline{w}}, \text{ if $w\ne0$.}
\end{align*} $$

We can relate conjugation with the real part and the imaginary part. $$\begin{align*} z + \overline{z} &= 2\, \text{Re}(z) \\ z-\overline{z} &= 2i \,\text{Im}(z). \end{align*}$$

We have an identity for powers of \(z\). $$\overline{z^n} = (\overline{z})^n \text{, for integer }n.$$

Lastly, the conjugate of a conjugate of a complex number \(z\) is \(z\). $$\overline{\overline{z}} = z.$$

Let’s prove a few of these.

Theorem. For complex numbers \(z\) and \(w\), we have \( z + \overline{z} = 2\, \text{Re}(z) \).
Proof. Let \(z = x+yi\), where \(x\) and \(y\) are real numbrs. Then $$
z + \overline{z} = x+yi + x-yi = 2x = 2\text{Re}(z).
$$

The next one is requires a bit more working, but relies on multiplying the top and the bottom of the fraction by the conjugate of the denominator.

Theorem. For complex numbers \(z\) and \(w\), we have $$\overline{\left( \frac{z}{w} \right)} = \frac{\overline{z}}{\overline{w}}.$$
Proof. Let \(z = a + bi\) and \(w = c+di\). Then $$\begin{align*}
\overline{\left( \frac{z}{w} \right)} &= \overline{\left( \frac{a+bi}{c+di} \right)} \\
&= \overline{\frac{a+bi}{c+di} \times \frac{c-di}{c-di}} \\
&= \overline{\left( \frac{ac + bd + (bc-ad)i}{c^2 + d^2} \right)} \\
&= \frac{ac + bd-(bc-ad)i}{c^2 + d^2}. \end{align*}$$

Theorem. For complex number \(z\), we have \(\overline{z^n} = (\overline{z})^n\).
Proof. Continually apply the identity \( \overline{zw} = \overline{z} \, \overline{w} \). That is, we have $$ \begin{align*} \overline{z^n} &= \overline{z^{n-1} \times z} \\
&= \overline{z^{n-1}} \times \overline{z} \\
&= \overline{z^{n-2} \times z} \times \overline{z} \\
&= \overline{z^{n-2}} \times \overline{z} \times \overline{z} \\
&= \dots = \overline{z}^n.\end{align*}$$

Problems

1. Prove the following for any complex \(z\) and \(w\).
   1. \(\overline{z+w} = \overline{z} + \overline{w}\)
   2. \(\overline{z-w} = \overline{z}-\overline{w}\)
   3. \(\overline{zw} = \overline{z} \, \overline{w}\)
   4. \(z-\overline{z} = 2i \,\text{Im}(z)\)
   5. \(\overline{\overline{z}} = z\)