Complex Conjugates in Quadratic Equations: Solving for z

[cragar]

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 ?

 

[Ray Vickson]

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.

 

[Mark44]

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.

 

[PeroK]

I would automatically use the polar form here.

 

[Mark44]

I would automatically use the polar form here.

That's a possibility, but it isn't necessary here.

 

[cragar]

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