[SOLVED] Argument of complex number

**Moo** · June 18th 2008, 12:52 PM

Hi !

Sorry if it isn't the correct subforum...

The question is :
Find the argument of $\frac{(z_1+z_2)^2}{z_1z_2}$ , given that $|z_1|=|z_2|=1$

I found it to be $2arg(z_1+z_2)-arg(z_1)-arg(z_2)$ , but couldn't go any further.. Did I miss something ?

Thanks in advance

**flyingsquirrel** · June 18th 2008, 01:00 PM

Hi

Originally Posted by Moo

[...] given that $|z_1|=|z_2|=1$

This gives us $z_1=\exp(i\theta)$ and $z_2=\exp(i\varphi)$ .

**Moo** · June 18th 2008, 01:05 PM

Originally Posted by flyingsquirrel

Hi

This gives us $z_1=\exp(i\theta)$ and $z_2=\exp(i\varphi)$ .

I know, but well, the z_1+z_2 is disturbing...

I got to expressions like $\cos \frac{\theta_1-\theta_2}{2}$ but couldn't simplify :/

**flyingsquirrel** · June 18th 2008, 01:25 PM

Originally Posted by Moo

I know, but well, the z_1+z_2 is disturbing...

I got to expressions like $\cos \frac{\theta_1-\theta_2}{2}$ but couldn't simplify :/

$\frac{(z_1+z_2)^2}{z_1z_2}=\frac{\mathrm{e}^{i2\th eta_1}+2\mathrm{e}^{i(\theta_1+\theta2)}+\mathrm{e }^{i2\theta_2}}{\mathrm{e}^{ i(\theta_1+\theta_2)}}=\mathrm{e}^{i(\theta_1-\theta_2)}+2+\mathrm{e}^{i(\theta_2-\theta_1)}$

If this is the expression you got to, you have the answer since this is a real number.

**Moo** · June 18th 2008, 02:00 PM

Originally Posted by flyingsquirrel

$\frac{(z_1+z_2)^2}{z_1z_2}=\frac{\mathrm{e}^{i2\th eta_1}+2\mathrm{e}^{i(\theta_1+\theta2)}+\mathrm{e }^{i2\theta_2}}{\mathrm{e}^{ i(\theta_1+\theta_2)}}=\mathrm{e}^{i(\theta_1-\theta_2)}+2+\mathrm{e}^{i(\theta_2-\theta_1)}$

If this is the expression you got to, you have the answer since this is a real number.

I got it, but what I need is the argument... ?

I'm not sure I'm getting what you're saying :/

**flyingsquirrel** · June 18th 2008, 02:16 PM

Originally Posted by Moo

I got it, but what I need is the argument... ?

I'm not sure I'm getting what you're saying :/

I should have added that $\frac{(z_1+z_2)^2}{z_1z_2}=\mathrm{e}^{i(\theta_1-\theta_2)}+2+\mathrm{e}^{i(\theta_2-\theta_1)}=2\cos(\theta_1-\theta_2)+2\geq0$

This tells us that the argument is $0\mod 2\pi$ except when $z_1=-z_2$ .

**Moo** · June 18th 2008, 02:19 PM

Originally Posted by flyingsquirrel

I should have added that $\frac{(z_1+z_2)^2}{z_1z_2}=\mathrm{e}^{i(\theta_1-\theta_2)}+2+\mathrm{e}^{i(\theta_2-\theta_1)}=2\cos(\theta_1-\theta_2)+2\geq0$

This tells us that the argument is $0\mod 2\pi$ except when $\theta_1=-\theta_2$ .

Oh yeah, that's better

Hmm $\theta_1=-\theta_2$ ? ^^

I'd say "arg=0 mod 2pi and theta_1-theta_2=-pi mod 2pi"

Edit : it's ok now

**bobak** · June 18th 2008, 02:22 PM

Hey Moo

I just have to post the geometric approach to this. hope you like my handwriting!

Bobak

Edit: I wrote equilateral when I meant isosceles, Thanks for spotting that Moo.