| Author | Complex Numbers Question |
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GIVEN:|z-2|=2|z-1| To Prove:|z|^2=4/3Re(z)
.............................Mod z square =3/4 multiplied by real part of z (To prove in english)
z is any complex number.
| signifies mod
Re(z) signifies the real part of the complex number
eg. if x+iy is our complex number , x=real part
No solutions on the internet .
Unable to solve this question , tried many ways =((
If someone could help will really appreciate it.
Thanks Alot! |
| Ask at mathoverflow.com |
Square both sides of given equation.
Simplify and you will get it. |
Use these results:
Let z be complex number.
z* Conjugate of z=Mod(z)^2
z + Conjugate of z=2 * Re(z) |
This is easier than it looks. You almost solved it in the question lol, just set z = x + iy and the algebra that follows is fairly direct.
First of all, |z|^2 = x^2 + y^2 = 4/3*x is what you have to prove.
Given: |z-2|^2 = 4|z-1|^2
And to clarify:
z-2 = (x-2) + iy
z-1 = (x-1) + iy
Therefore,
(x-2)^2 + y^2 = 4[(x-1)^2 + y^2]
x^2 -4x +4 +y^2 = 4x^2 -8x +4 +4y^2
0 = 3x^2 -4x +3y^2
x^2 + y^2 = 4/3*x, as required. |
| Resort to solving in Euler's notation as last resort,sometimes it is lengthy. |
I'm so dumb :X
0 = 3x^2 -4x +3y^2
I solved till here , after it I thought I'm doing wrong :X
THank you guys ^^
You are the best :P |
| GOsh I feel so embarrassed :\ |
| closed by Bewear (2018-07-18 09:55:15) |
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