I have attempted to solve it but arrive at either no solution or a solution much different from the textbook's.

1. If w = $\displaystyle \frac{z + 1}{z - 1}$ and |z| = 1, find Re(w).

I plugged in a + bi for z and attempted to simplify.

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- Jan 3rd 2012, 08:19 AMLemons123A few complex number problems
I have attempted to solve it but arrive at either no solution or a solution much different from the textbook's.

1. If w = $\displaystyle \frac{z + 1}{z - 1}$ and |z| = 1, find Re(w).

I plugged in a + bi for z and attempted to simplify. - Jan 3rd 2012, 09:08 AMFernandoRevillaRe: A few complex number problems
Use that $\displaystyle \textrm{Re}(w)=\frac{1}{2}(w+\bar{w})$ and $\displaystyle |z|^2=z\bar{z}=1$ .

- Jan 3rd 2012, 12:50 PMOpalgRe: A few complex number problems
If If $\displaystyle w=\frac{z + 1}{z - 1}$ then $\displaystyle w(z-1) = z+1.$ Solve that for z to get $\displaystyle z = \frac{w+1}{w-1}.$ If $\displaystyle |z|=1$ it follows that $\displaystyle |w+1| = |w-1|.$ Thus w is equidistant from 1 and –1, and so Re(w) = 0.

- Jan 3rd 2012, 02:04 PMFernandoRevillaRe: A few complex number problems
Or (using $\displaystyle z\bar{z}=1$) :

$\displaystyle \textrm{Re}(w)=\frac{1}{2}(w+\bar{w})=\frac{1}{2} \left(\frac{z+1}{z-1}+\frac{\bar{z}+1}{\bar{z}-1} \right)=\frac{1}{2}\dfrac {0}{(z-1)(\bar{z}-1)}=0$