# Complex Conjugate

• Apr 30th 2009, 09:16 PM
jzellt
Complex Conjugate
Let (z) be the complex conjugate of z.

Here's the problem:

Let c e C be non-zero(c is a non-zero element of the complex numbers)
Show (z)^-1 = (z^-1)
• Apr 30th 2009, 10:41 PM
red_dog
$\displaystyle (\overline{z})^{-1}=\frac{1}{\overline{z}}$

$\displaystyle \overline{z^{-1}}=\overline{\left(\frac{1}{z}\right)}=\frac{\ove rline{1}}{\overline{z}}=\frac{1}{\overline{z}}$
• Apr 30th 2009, 11:42 PM
Prove It
Alternatively, using DeMoivre's Theorem...

$\displaystyle z = r(\cos{\theta} + i\sin{\theta}), \overline{z} = r(\cos{\theta} - i\sin{\theta})$.

$\displaystyle z^{-1} = r^{-1}[\cos{(-\theta)} + i\sin{(-\theta)}]$

$\displaystyle = \frac{1}{r}(\cos{\theta} - i\sin{\theta})$.

So $\displaystyle \overline{z^{-1}} = \frac{1}{r}(\cos{\theta} + i\sin{\theta})$.

$\displaystyle (\overline{z})^{-1} = r^{-1}[\cos{(-\theta)} - i\sin{(-\theta)}]$

$\displaystyle = \frac{1}{r}(\cos{\theta} + i\sin{\theta})$

$\displaystyle = \overline{z^{-1}}$.

Or using Cartesians:

$\displaystyle z = a+ib, \overline{z} = a-ib$

$\displaystyle z^{-1} = \frac{1}{z}$

$\displaystyle = \frac{1}{a + ib}$

$\displaystyle = \frac{a - ib}{a^2 + b^2}$

$\displaystyle = \frac{a}{a^2 + b^2} -i\left(\frac{b}{a^2 + b^2}\right)$

$\displaystyle \overline{z^{-1}} = \frac{a}{a^2 + b^2} + i\left(\frac{b}{a^2 + b^2}\right)$.

$\displaystyle (\overline{z})^{-1} = \frac{1}{\overline{z}}$

$\displaystyle = \frac{1}{a - ib}$

$\displaystyle = \frac{a + ib}{a^2 + b^2}$

$\displaystyle = \frac{a}{a^2 + b^2} + i\left(\frac{b}{a^2 + b^2}\right)$

$\displaystyle = \overline{z^{-1}}$.