Complex numbers

Question

3 questions, not sure how to do them.

1. Let $z$ and $w$ be complex numbers such that $w=\frac{1}{1-z}$ and $|z|^2=1$. Find the real part of w.

2. If $z=e^{i\theta}$ prove that $\frac{z^2-1}{z^2+1}=i\tan\theta$.

3. $w=\frac{kz}{z^2+1}$ where $z^2$ is not equal to $-1$.

If $\Im w=\Im k=0$ and $\Im z$ is not equal to 0, prove that $|z|=1$.

Also, how do I get the not equals sign?

Answer 1

2) if $z=e^{i\theta}$, then \begin{align*} \frac{z^2-1}{z^2+1}=&\frac{e^{i2\theta}-1}{1}\frac1{e^{i2\theta}+1}\\ =& \frac{2i}{e^{-i\theta}}\frac{e^{i\theta}-e^{-i\theta}}{2i}\frac{2}{e^{i\theta}+e^{-i\theta}}\frac{e^{-i\theta}}{2}\\ =& \frac{2i}{e^{-i\theta}}\sin\theta\frac1{\cos\theta}\frac{e^{-i\theta}}{2}\\ =&i\tan\theta. \end{align*}

Answer 2

I'll start with part $1$. $|z|=1$, so let $z=e^{i\theta}$. Then $w=\frac{1}{1-z}=\frac{1}{1-e^{i\theta}}$, that is $w=(1-e^{-i\theta})\frac{1}{2-2\cos \theta}$. The denominator is real, so the real part of $w$ is the real part of $1-e^{-i\theta}$ multiplied by $\frac{1}{2-2\cos \theta}$. This is $(1-\cos \theta)\frac{1}{2-2\cos \theta}$. Hence the real part of $w$ is $1/2$.

For the next part another user has already posted the comment. For the last part, let $w=\frac{1}{1+z^2}kz$. If Im(w) is $0$, and (if $w=r$ where $r$ is real) then $rz^2-kz+r=0$. Note $k$ is real. Let the roots of $q(z)=rz^2-kz+r$ be $z_1$ and $z_2$.

As $q$ is a polynomial with real coefficients, $z_1$ and $z_2$ are forced to be complex conjugates, furthermore $z_1z_2=a_0/a_2$, if the polynomial $q$ is expressed as $q=a_2z^2+a_1z+a_0$. Hence $z_1z_2=\frac{r}{r}$, thus $1=z_1z_2$. Hence, writing $\bar{z_1}=z_2$, we see $z_1\bar{z_1}=|z_1|^2=1$.