Number of solutions to complex eq.

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Homework Statement



How many solutions does

z^2-iz(conjugate)=1/4

have?

2. The attempt at a solution

z^2=(1/4)+y+ix

Since the RHS is a complex number, the eq. has two solutions.

Correct?
 

Discussion

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dextercioby said:
I found 3 solutions. Why don't you solve the equation ?

I learned that the number of solutions equals the power of the eq., that is 2...
 
dextercioby said:
Under what conditions ?

w^n=z
 
dextercioby said:
Okay, but did you solve the equation ?

Can't make it further than this:

(x^2-y^2-y-(1/4)+i(2xy-x)=0
 
Last edited:
dextercioby said:
Okay then. What follows next ?

Ah, y=2x since 2xy-x has got to be 0?

But then...no idea.
 

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