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

[Saitama]

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
$$|z_2|≠1$$

 

[vela]

It should be $|z_2| \ne 1$ and
$$\left\lvert \frac{z_1-2z_2}{2-z_1\overline{z}_2} \right\rvert = 1$$Square both sides and use the fact that $|z|^2 = z\bar z$.

EDIT: Also |w/z| = 1 means |w|=|z|.

 

[Saitama]

It should be $|z_2| \ne 1$ and
$$\left\lvert \frac{z_1-2z_2}{2-z_1\overline{z}_2} \right\rvert = 1$$Square both sides and use the fact that $|z|^2 = z\bar z$.

Thanks for the reply vela!

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. :(

 

[Simon Bridge]

Thanks for the reply vela!

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?

 

[Saitama]

:) 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? :)

 

[Saitama]

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. :)

 

[Simon Bridge]

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).

 

[Saitama]

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. :)

 

[Simon Bridge]

Sweet dessert: That one is usually nasty because everyone tries to brute-force it. :)