Complex Conjugate

**jzellt** · April 30th 2009, 09:16 PM

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)

**red_dog** · April 30th 2009, 10:41 PM

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

$\overline{z^{-1}}=\overline{\left(\frac{1}{z}\right)}=\frac{\ove rline{1}}{\overline{z}}=\frac{1}{\overline{z}}$

**Prove It** · April 30th 2009, 11:42 PM

Alternatively, using DeMoivre's Theorem...

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

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

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

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

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

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

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

Or using Cartesians:

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

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

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

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

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

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

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

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

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

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

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

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