1. ## Complex numbers question

Question:
if the complex number $\displaystyle x + iy$ denoted by z then the complex conjugate number $\displaystyle x-iy$denoted by $\displaystyle a*$.
a) Express $\displaystyle \left |\ z^* \right |$ and $\displaystyle arg(z^*)$ in terms of $\displaystyle \left |\ z \right |$ and $\displaystyle arg(z)$

b)if a, b, and c are real numbers prove that if $\displaystyle az^2+bz+c=0$
then $\displaystyle a(z^*)^2 + b{z^*}+ c=0$

c)if p and q are complex numbers and $\displaystyle q \neq 0$, prove that $\displaystyle \left ( \frac{p}{q} \right )^{*}=\frac{p^*}{q^*}$

My attempt

a) $\displaystyle \left |\ z^* \right | = \left |\ x-iy \right | = \sqrt{x^2+(-y)^2} = \sqrt{x^2+y^2}$
therefore $\displaystyle \left |\ z^* \right | = \left |\ z \right | = \sqrt{x^2+y^2}$
$\displaystyle arg(z) = \Theta$ and $\displaystyle arg(z^*) = - \Theta$
therefore $\displaystyle arg(z^*) = - arg(z)$
[/FONT][/COLOR]
) $\displaystyle a(z^*)^2 + b{z^*}+ c=0$[
sub $\displaystyle z^* = x-iy$ into the equation
$\displaystyle a(x-iy)^2+b(x-iy) +c=0$
$\displaystyle a(x^2-y^2)+bx+c -iy(2ax-b)=0$
I am stuck here
plus I dont know how to approach part c please can some one help me thank you

2. ## Re: Complex numbers question

b) $a z^2 + b z + c = 0$

$z = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$

suppose $Im(z) \neq 0$

$z = \dfrac{-b \pm i \sqrt{4ac-b^2}}{2a} = -\dfrac{b}{2a} \pm i \dfrac{\sqrt{4ac-b^2}}{2a}$

It's seen by inspection that both $z$ and $z^*$ are roots

to cheat a bit less

$az^2 + bz+c = 0 \Rightarrow a(x^2-y^2)+bx + c = 0,~2axy+b y = 0$ (note $c \in \mathbb{R}$)

$a (z^*)^2 + b z^* + c = a (x^2 - y^2 )+bx + c + i(-2axy - by) = 0 -i(0) = 0$

3. ## Re: Complex numbers question

c) use $z = |z|e^{i \arg(z)}$ for $p$ and $q$

Spoiler:

let

$p = |p|e^{i \arg(p)},~q = |q|e^{i \arg(q)}$

$\left(\dfrac p q \right)^* = \\ \left(\dfrac{|p|}{|q|}e^{i( \arg(p)-\arg(q))}\right)^* = \\ \dfrac{|p|}{|q|} e^{i(-\arg(p) - (-\arg(q)))} = \\ \dfrac{|p|e^{-i\arg(p)}}{|q|e^{-i\arg(q)}} =\\ \dfrac{p^*}{q^*}$