# Seems easy but..

• Mar 15th 2010, 12:39 PM
Henryt999
Seems easy but..
I have this question I am suppose to solve and I think i misunderstood something. It is in a University algebra book and the questions is:
Show that if abs:z =1 then, abs: $\left[\frac{2z-1}{z-2}\right]=1$
I dont see what is so difficult here?? why, why put this into a University book??
If abs:z=1 then, abs:2z-1=1 and in the nominator abs:z-2 = -1 and absolute value of that is 1. So it is $\frac{1}{1}$?
Have I missunderstood something?
• Mar 15th 2010, 12:47 PM
icemanfan
If $|z| = 1$, then $z = -1$ or $z = 1$. You must show that $\left|\frac{2z - 1}{z - 2}\right| = 1$ for both values of z.
• Mar 15th 2010, 01:52 PM
Opalg
Quote:

Originally Posted by Henryt999
I have this question I am suppose to solve and I think i misunderstood something. It is in a University algebra book and the questions is:
Show that if abs:z =1 then, abs: $\left[\frac{2z-1}{z-2}\right]=1$
I dont see what is so difficult here?? why, why put this into a University book??
If abs:z=1 then, abs:2z-1=1 and in the nominator abs:z-2 = -1 and absolute value of that is 1. So it is $\frac{1}{1}$?
Have I missunderstood something?

This is surely a question about complex numbers. If |z| = 1 then z can be any point on the unit circle in the complex plane. The question is asking you to show that in that case $|z-2| = 2|z-\tfrac12|$.