Basic questions of complex number's geometry?

[vkash]

Homework Statement

(z' represent conjugate of complex number z,i is iota =sqrt(-1))
(1)find the locus of z.
|z|²+4z'=0

(2)|z₁|=1, |z₂|=2, then find the value of |z₁-z₂|²+|z₁+z₂|²

(3)z=(k+3)+i[sqrt(3-k²)] for all real k. find locus of z

Homework Equations

|z|²=z.z'

The Attempt at a Solution

(1) |z|²+4z'= |z|²+4z'*z/z
|z|²{1+4/z}=0
it is possible iff z= -4. so it has one solution.

(2) I take z₁= cos(a)+i sin(a) and z₂=2{cos(b)+i*sin(b)}
after putting these values in the required equation i got 10.

(3) let z=x+iy
x=k+3; y=sqrt(3-k²)
x-3=k; y²=3-k²
squaring x and adding it to y²,
(x-3)²+y²=k²+3-k²
that's circle.

unfortunately all of my answers are incorrect.
I want to know why?

 

[fzero]

For (1), you are missing one solution. You have two factors and the equation is true when either one is zero. z=-4 is the solution where the 2nd factor is zero, what is the solution when the 1st factor vanishes?

For (2), it would help to expand as

| z_1-z_2|^2 + | z_1+z_2|^2 = (z_1-z_2) (\bar{z}_1-\bar{z}_2)+ (z_1+z_2) (\bar{z}_1+\bar{z}_2),

then simplify the resulting expression.

For (3), you made a mistake to try to compute (x-3)²+y². (x-3)² does not appear in |z|². Instead, it might help to look at the real and imaginary parts independently as functions of k. Looking at some special points like k=0,\pm 3 also helps.

 

[vkash]

For (1), you are missing one solution. You have two factors and the equation is true when either one is zero. z=-4 is the solution where the 2nd factor is zero, what is the solution when the 1st factor vanishes?

For (2), it would help to expand as

| z_1-z_2|^2 + | z_1+z_2|^2 = (z_1-z_2) (\bar{z}_1-\bar{z}_2)+ (z_1+z_2) (\bar{z}_1+\bar{z}_2),

then simplify the resulting expression.

For (3), you made a mistake to try to compute (x-3)²+y². (x-3)² does not appear in |z|². Instead, it might help to look at the real and imaginary parts independently as functions of k. Looking at some special points like k=0,\pm 3 also helps.

(1) Oh it was y foolishness.

(2) even after opening the equation you place i got 2(|z₁|²+|z₂|²) =2(1+2²)=10

(3)I did not understand third answer.

THANKS FOR REPLY first 2 answer really helps me.
In third; 3 and-3 lies on the circle but not satisfy the complex number. so i think it is part of circle for all those for which k<|\frac{1}{\sqrt{3}}|

 

[HallsofIvy]

For the third one, z= x+ iy= k+3+ (3- k^2)i.

x= k+ 3, y= \sqrt{3- k^2}. Solve the first equation for k, the replace k in the second equation with that.

 

[ehild]

And what is the locus of z when 3-k²<0? You need to give the points for all real k values.

ehild

 

[vkash]

And what is the locus of z when 3-k²<0? You need to give the points for all real k values.

ehild

see this figure
this is circle(in real part)
but why it is different from this figure

 

[ehild]

I do not really understand your pictures. In case 3-k²<0 z is real. (The imaginary part is zero.)

ehild