Algebra in the complex plane

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  • #1
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Homework Statement


Solve each equation for z=a+ib
[itex] z^{*2}=4z [/itex]
where z* is the complex conjugate

The Attempt at a Solution


I wrote z and z* in terms of x and iy , and tried solving for x and y, but I get quartic terms for y, it doesnt look like it will boil down, It was like over 2 pages of algera, I dont think that is how it is supposed to be done, I tired some alternative forms of the modulus and it didnt go anywhere, Are my approaches the right way, or do I need some clever substitution, or can it be solved with Eulers formula ?
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement


Solve each equation for z=a+ib
[itex] z^{*2}=4z [/itex]
where z* is the complex conjugate

The Attempt at a Solution


I wrote z and z* in terms of x and iy , and tried solving for x and y, but I get quartic terms for y, it doesnt look like it will boil down, It was like over 2 pages of algera, I dont think that is how it is supposed to be done, I tired some alternative forms of the modulus and it didnt go anywhere, Are my approaches the right way, or do I need some clever substitution, or can it be solved with Eulers formula ?

Show us the equations in x and y that you actually get. What you are claiming sounds wrong to me.
 
  • #3
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Homework Statement


Solve each equation for z=a+ib
[itex] z^{*2}=4z [/itex]
where z* is the complex conjugate

The Attempt at a Solution


I wrote z and z* in terms of x and iy , and tried solving for x and y, but I get quartic terms for y, it doesnt look like it will boil down, It was like over 2 pages of algera, I dont think that is how it is supposed to be done, I tired some alternative forms of the modulus and it didnt go anywhere, Are my approaches the right way, or do I need some clever substitution, or can it be solved with Eulers formula ?
If z = a + bi, then ##\bar{z} = a - bi##
If I substitute these into the given equation, I don't get a quartic, but I do get a solution fairly quickly. Please show what you did.
 
  • #4
PeroK
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I would automatically use the polar form here.
 
  • #6
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oh ok I think I got it
sub in x+iy and x-iy
then equate real and imaginary
Real = [itex] x^2+y^2=4x [/itex]
Im = [itex] -2xy=4y [/itex]
I solved for x then put it in the other equation, then I solved for y,
and I got the answer in the back of the book, thanks for your posts
 

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