Complex Conjugates in Quadratic Equations: Solving for z

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Homework Help Overview

The discussion revolves around solving the equation \( z^{*2} = 4z \) for \( z = a + ib \), where \( z^* \) represents the complex conjugate. Participants explore various methods of approaching the problem, including algebraic manipulation and polar form representation.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss writing \( z \) and \( z^* \) in terms of real and imaginary components, leading to complications with quartic terms. There are inquiries about the validity of these approaches and whether clever substitutions or alternative methods, such as Euler's formula, could simplify the problem. Some participants suggest equating real and imaginary parts after substituting the expressions for \( z \) and \( z^* \).

Discussion Status

The discussion is active, with participants sharing their attempts and questioning each other's reasoning. Some have provided guidance on using polar form, while others are still exploring algebraic methods. There is a recognition of differing interpretations of the problem, and some participants have reported finding solutions, though consensus on the best approach has not been reached.

Contextual Notes

Participants express frustration with the algebraic complexity and the length of their attempts, indicating that the problem may have constraints or assumptions that complicate straightforward solutions.

cragar
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Homework Statement


Solve each equation for z=a+ib
z^{*2}=4z
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 doesn't look like it will boil down, It was like over 2 pages of algera, I don't 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 ?
 
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cragar said:

Homework Statement


Solve each equation for z=a+ib
z^{*2}=4z
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 doesn't look like it will boil down, It was like over 2 pages of algera, I don't 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.
 
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cragar said:

Homework Statement


Solve each equation for z=a+ib
z^{*2}=4z
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 doesn't look like it will boil down, It was like over 2 pages of algera, I don't 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.
 
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I would automatically use the polar form here.
 
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PeroK said:
I would automatically use the polar form here.
That's a possibility, but it isn't necessary here.
 
oh ok I think I got it
sub in x+iy and x-iy
then equate real and imaginary
Real = x^2+y^2=4x
I am = -2xy=4y
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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