Finding |z1| when |z1-2z2 / 2-z1z2*| = 1 and |z2| ≠ 1

  • Thread starter Thread starter Saitama
  • Start date Start date
AI Thread Summary
The discussion revolves around solving the equation |(z1 - 2z2) / (2 - z1z2*)| = 1, given that |z2| ≠ 1. The user initially struggles with complex numbers but eventually derives the equation |z1|^2 + 4|z2|^2 = 4 + |z1|^2|z2|^2. After further hints and guidance, they simplify the equation to find that |z1| = 2. The conversation highlights the step-by-step problem-solving process in complex analysis. The final solution is confirmed as |z1| = 2.
Saitama
Messages
4,244
Reaction score
93

Homework Statement


If \frac{z_1-2z_2}{2-z_1\overline{z}_2} is unimodulus and z2 is not unimodulus, then find |z1|.

Homework Equations


The Attempt at a Solution


I am a complete dumb at Complex numbers, please someone guide me in the right direction.
In this question, what i understand is this, and nothing else.
|\frac{z_1-2z_2}{2-z_1\overline{z}_2}|=1
and
|z_2|≠1
 
Last edited:
Physics news on Phys.org
It should be |z_2| \ne 1 and
\left\lvert \frac{z_1-2z_2}{2-z_1\overline{z}_2} \right\rvert = 1Square both sides and use the fact that |z|^2 = z\bar z.

EDIT: Also |w/z| = 1 means |w|=|z|.
 
vela said:
It should be |z_2| \ne 1 and
\left\lvert \frac{z_1-2z_2}{2-z_1\overline{z}_2} \right\rvert = 1Square both sides and use the fact that |z|^2 = z\bar z.

Thanks for the reply vela! :smile:

I squared both the sides and using the fact |z|^2 = z\bar z, i get:-
|z_1|^2+4|z_2|^2=4+|z_1|^2|z_2|^2

But now i am stuck here. :(
 
Pranav-Arora said:
Thanks for the reply vela! :smile:

I squared both the sides and using the fact |z|^2 = z\bar z, i get:-
|z_1|^2+4|z_2|^2=4+|z_1|^2|z_2|^2

But now i am stuck here. :(

:) Follow your nose: if it was

x + 4y = 4 + xy

and you had to find x, what would you do?
 
Simon Bridge said:
:) Follow your nose: if it was

x + 4y = 4 + xy

and you had to find x, what would you do?

I still don't understand. :(

Can you give me one more hint? :)
 
Thank you both for the help. I have figured it out. :)

x+4y=4+xy
or x-xy=4-4y
or x(1-y)=4(1-y)
or x=4
or |z1|=2.

Thanks again. :)
 
Well done!

For dessert:

z = \left ( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i \right )^{12}
...rewrite z in it's simplest form (it will be exact).
 
Simon Bridge said:
Well done!

For dessert:

z = \left ( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i \right )^{12}
...rewrite z in it's simplest form (it will be exact).

z=-1. :)
 
Sweet dessert: That one is usually nasty because everyone tries to brute-force it. :)
 
Back
Top