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Basic questions of complex number's geometry?

  • Thread starter vkash
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  • #1
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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?
 

Answers and Replies

  • #2
fzero
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For (1), you are missing one solution. You have two factors and the equation is true when either one is zero. [itex]z=-4[/itex] 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

[itex] | 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),[/itex]

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 [itex]k=0,\pm 3[/itex] also helps.
 
  • #3
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For (1), you are missing one solution. You have two factors and the equation is true when either one is zero. [itex]z=-4[/itex] 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

[itex] | 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),[/itex]

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 [itex]k=0,\pm 3[/itex] 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 [itex]k<|\frac{1}{\sqrt{3}}|[/itex]
 
  • #4
HallsofIvy
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For the third one, [itex]z= x+ iy= k+3+ (3- k^2)i[/itex].

x= k+ 3, [itex]y= \sqrt{3- k^2}[/itex]. Solve the first equation for k, the replace k in the second equation with that.
 
  • #5
ehild
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And what is the locus of z when 3-k2<0? You need to give the points for all real k values.

ehild
 
  • #6
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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
 
  • #7
ehild
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I do not really understand your pictures. In case 3-k2<0 z is real. (The imaginary part is zero.)

ehild
 

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