# Complex Analysis- help needed with proof

• Dec 24th 2011, 07:24 AM
flaming
Complex Analysis- help needed with proof
Complex Argument: how do I show that

arg$\displaystyle \overline{z} = -$ arg$\displaystyle z$. If $\displaystyle z$ is not a real number
• Dec 24th 2011, 07:51 AM
Deveno
Re: Complex Analysis- help needed with proof
if we write $\displaystyle z = r(\cos\theta + i\sin\theta)$

then $\displaystyle \text{arg}(z) = \theta$

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

$\displaystyle =r(\cos\theta - i\sin\theta) = r(\cos(-\theta) + i\sin(-\theta))$

so $\displaystyle \text{arg}(\overline{z}) = -\theta = -\text{arg}(z)$
• Dec 24th 2011, 08:47 AM
FernandoRevilla
Re: Complex Analysis- help needed with proof
Another way (specially if the OP means the principal argument). Denoting $\displaystyle \arg z$ the principal argument of $\displaystyle z\neq 0$ we have $\displaystyle \arg(z_1z_2)=\arg z_1+\arg z_2$ if $\displaystyle z_1z_2\neq 0$ and $\displaystyle -\pi<\arg z_1+\arg z_2\leq \pi$ . These conditions are verified for $\displaystyle z_1=z$ and $\displaystyle z_2=\bar{z}$ and $\displaystyle 0\neq z\not\in\mathbb{R}$ . So $\displaystyle z\bar{z}=|z|^2$ which implies $\displaystyle \arg z+\arg\bar{z}=0$ i.e. $\displaystyle \arg\bar{z}=-\arg{z}$ .