# Thread: Let w be a complex number with 0 < |w| < 1. Show that the set of all z...

1. ## Let w be a complex number with 0 < |w| < 1. Show that the set of all z...

Let w be a complex number with 0 < |w| < 1. Show that the set of all z with these properties is as described

1: Show that |z - w| < |1 - w*z| is the disc {z:|z| < 1}
2: Show that |z - w| = |1 - w*z| is the circle {z:|z| = 1}
3: Show that |z - w| > |1 - w*z| is the circle {z:|z| > 1}

I squared both sides and simplified... assuming that it would equal

a^2 + b^2 < 1

given z= a +bi and w= c + di

but i got

a^2 + b^2 + c^2 + d^2 - (a^2)(c^2) - (b^2)(c^2) - (b^2)(d^2) -(a^2)(d^2) < 1

which i simplified to...

(|z|^2)(1 - (|w|^2)) + |w|^2 < 1

anyone know what i am doing wrong?

2. ## So close!

Don't kick yourself, but you've already done all of the hard work! Assuming that your working is correct, the answer lies in a simple reaarangement to make $|z|$ the subject of

$|z|^{2}(1-|w|^{2})+|w|^{2}<1.$

Note that when rearranging you will be dividing by a possibly very small number. Can you state which condition allows you to do this without making a division by 0 error?

3. Originally Posted by nimon
Don't kick yourself, but you've already done all of the hard work! Assuming that your working is correct, the answer lies in a simple reaarangement to make $|z|$ the subject of

$|z|^{2}(1-|w|^{2})+|w|^{2}<1.$

Note that when rearranging you will be dividing by a possibly very small number. Can you state which condition allows you to do this without making a division by 0 error?
oook so i can jsut move |w|^2 over to the right and divide to get 1 on the right side?

4. Bingo!

5. Wow I am so thick in the head sometimes