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For my algebra assignment, I have to analyze the homomorphism $f: \mathbb C^* \rightarrow \mathbb C^*$ given by $$z \rightarrow\frac{z}{|z|}$$ I have to give the kernel, image, cosets and a plot of the homomorphism.

I know the kernel is the set of $z$ for which $f(z)=e$ (identity), so is that just all complex numbers with length $1$? Or the complex numbers $z = 1$?

I also know the image is given by $\{f(z): z \in \mathbb C^*\}$, so is this the set $\{\frac{z}{|z|}: z \in \mathbb C^*\}$?

It would be nice if someone could tell me if im on the right path :) Thanks in advance.

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The kernel is given by $ \{z\in\mathbb C^*~:~f(z)=1\}. $

Next, we check $$f(z)=1\Leftrightarrow \frac{z}{|z|}=1\Leftrightarrow z=|z|.$$ Since the RHS is real we conclude that $z$ has to be real too. And real numbers which are equal to their absolute value are ...

You are right that the image of $f$ is given by $$ img(f)=\left\{\frac{z}{|z|}~:~z\in\mathbb C^*\right\}. $$ What happens geometrically if you divide a vector by its length? (Hint: What is the length of the elements in $img(f)$?)

Using the hint, you might find a better description of the set, which has a name, because it is very special!

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  • $\begingroup$ Real numbers with are equal to their absolute value are the positives! For your second question, if you divide a vector by its length, you project it on a circle with radius 1, right? And also the length of the elements in img(f) is 1? $\endgroup$ – Katie Apr 4 '18 at 9:07
  • $\begingroup$ So is the set $img(f)$ = {$|z|=1: z \in C*$}? I'm not sure yet what the name is... $\endgroup$ – Katie Apr 4 '18 at 9:16
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    $\begingroup$ Yes, the kernel is the set of all positive real numbers. For the image, you should write $img(f)=\{z\in\mathbb C^*~:~|z|=1\}$ and this set is called the unit sphere of $\mathbb C$. $\endgroup$ – Mundron Schmidt Apr 4 '18 at 10:25
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Well, for the kernel, it's all complex numbers with $f(z) = e$, that is, $f(z)=1$.

Now, $f(z)=\frac z{|z|}$, so this means $\frac z{|z|}=1$, that is, $z=|z|$.

Which complex numbers have $z=|z|$?

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  • $\begingroup$ That must be all complex numbers with length one (so a circle with radius one), right? $\endgroup$ – Katie Apr 4 '18 at 8:55
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The identity element of $(\mathbb{C}\setminus\{0\},\times)$ is $1$. Therefore,$$\ker f=\left\{z\in\mathbb{C}\,\middle|\,f(z)=1\right\}.$$

And, yes, the image of $f$ is $\left\{\frac z{|z|}\,\middle|\,z\in\mathbb{C}\setminus\{0\}\right\}$, but that's not the best description of that set.

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