# Thread: complex number and images and kernel

1. ## 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.

2. ## Re: complex number and images and kernel

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?

3. ## 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

4. ## Re: complex number and images and kernel

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

5. ## Re: complex number and images and kernel

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

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