# Thread: kernal and image with complex numbers

1. ## Re: kernal and image with complex numbers

For the kernel I get so far

Ker = { z in C* : z / (the complex conjugate) z = 1}

= { z in C* : z = (the complex conjugate) z }

Then get stuck. Sorry I have tried looking at #2 and #5 to help but just confueses me a little more.

2. ## Re: kernal and image with complex numbers

Let's use Plato's hint. We need to find all z such that $\displaystyle \frac{z^2}{|z|^2}=1$. This equality implies that $\displaystyle z^2$ is a real number, so either $\displaystyle z\in\mathbb{R}$ or z = ir for some $\displaystyle r\in\mathbb{R}$. In the second case, |z| = r and $\displaystyle z^2=-r^2$, so $\displaystyle \frac{z^2}{|z|^2}=-1$. What about the first case?

If we use the polar form, $\displaystyle z=re^{i\varphi}$ for some $\displaystyle r, \varphi\in\mathbb{R}$. Then $\displaystyle \bar{z}=re^{-i\varphi}$, so $\displaystyle \frac{z}{\bar{z}}=?$ What restriction does it put on $\displaystyle \varphi$?

3. ## Re: kernal and image with complex numbers

suppose f(z) = z/z* is our homomorphism.

if z is in ker(f), then z/z* = 1, so

z = z*
z-z* = 0
(1/(2i))(z-z*) = Im(z) = 0.

thus all elements of the kernel are real.

if however, z is real, then z = z*, so z/z* = 1, thus z is in ker(f).

so ker(f) = R

note that |f(z)| = |z/z*| = |z|/|z*| = |z|/|z| = 1. thus im(f) is a subset of the unit circle.

on the other hand, suppose that w is any complex number on the unit circle (that is: |w| = 1).

there exists at least one complex number z with z2 = w, and from |z|2 = |z2| = |w| = 1, we see that |z| = 1, as well.

so f(z) = z/z* = (z/z*)(1) = (z/z*)(z/z) = z2/(zz*) = z2/|z|2 = z2 = w, which means that im(f) is the entire unit circle.

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