View Full Version : Complex Numbers
Pranav-Arora
Oct28-11, 10:20 PM
1. The problem statement, all variables and given/known data
If \frac{z_1-2z_2}{2-z_1\overline{z}_2} is unimodulus and z2 is not unimodulus, then find |z1|.
2. Relevant equations
3. 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
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|.
Pranav-Arora
Oct28-11, 11:37 PM
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. :(
Simon Bridge
Oct29-11, 12:13 AM
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?
Pranav-Arora
Oct29-11, 12:15 AM
:) Follow your nose: if it was
x + 4y = 4 + xy
and you had to find x, what would you do?
I still dont understand. :(
Can you give me one more hint? :)
Pranav-Arora
Oct29-11, 12:52 AM
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
Oct29-11, 07:00 AM
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).
Pranav-Arora
Oct29-11, 07:19 AM
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
Oct29-11, 07:41 AM
Sweet dessert: That one is usually nasty because everyone tries to brute-force it. :)
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