complex number and images and kernel

• May 7th 2012, 07:51 AM
mathslearner
complex number and images and kernel
I have attached a problem here and my attempt-
peobelm asks me to then identify quotentient group G/Ker (phi) up to isomorphism, I have found the kernel but not sure about Image.
• May 7th 2012, 08:59 AM
FernandoRevilla
Re: complex number and images and kernel
You have correctly found the kernel i.e. $\displaystyle \ker \phi=(\mathbb{R}-\{0\})i.$ Now, if $\displaystyle w\in\textrm{Im}\;\phi$ then, there exists $\displaystyle z\in\mathbb{C}-\{0\}$ such that $\displaystyle w=z/\bar{z}$. This implies $\displaystyle |w|=|z/\bar{z}|=|z|/|\bar{z}|=1$. Could you continue?
• May 7th 2012, 09:29 AM
mathslearner
Re: complex number and images and kernel
Not really as I have tried looking on net and my struggles are clear. If you can show me this time around I'll be comfortable with similar problems if you don't mind☺thanks
• May 7th 2012, 09:30 AM
mathslearner
Re: complex number and images and kernel
I think it's R+ but don't know how to mathematically get there
• May 7th 2012, 11:29 AM
FernandoRevilla
Re: complex number and images and kernel
Quote:

Originally Posted by mathslearner
I think it's R+ but don't know how to mathematically get there

It is the unit circle $\displaystyle S_1$. If $\displaystyle w\in S_1$ then, $\displaystyle w=\cos \theta +i\sin \theta$ for some $\displaystyle \theta \in [0,2\pi].$ i.e. we can express

$\displaystyle w=\dfrac{\cos (\theta/2)+i\sin (\theta/2)}{\cos (-\theta/2)+i\sin (-\theta/2)}=\dfrac{z}{\bar{z}}\in \textrm{Im}\;\phi.$