# Basic questions of complex number's geometry?

## 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

|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
Homework Helper
Gold Member
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.

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.

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

ehild

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

ehild