complex number and images and kernel

**mathslearner** · May 7th 2012, 07:51 AM

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.

**FernandoRevilla** · May 7th 2012, 08:59 AM

You have correctly found the kernel i.e. $\ker \phi=(\mathbb{R}-\{0\})i.$ Now, if $w\in\textrm{Im}\;\phi$ then, there exists $z\in\mathbb{C}-\{0\}$ such that $w=z/\bar{z}$ . This implies $|w|=|z/\bar{z}|=|z|/|\bar{z}|=1$ . Could you continue?

**mathslearner** · May 7th 2012, 09:29 AM

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

**mathslearner** · May 7th 2012, 09:30 AM

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

**FernandoRevilla** · May 7th 2012, 11:29 AM

Originally Posted by mathslearner

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

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

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

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