Number of solutions to complex eq.

[kasse]

1. The problem statement, all variables and given/known data

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?

[dextercioby]

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

[kasse]

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

I learnt that the number of solutions equals the power of the eq., that is 2...

[dextercioby]

Under what conditions ?

[kasse]

Under what conditions ?

w^n=z

[dextercioby]

Okay, but did you solve the equation ?

[kasse]

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

[dextercioby]

Okay then. What follows next ?

[kasse]

Okay then. What follows next ?

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

But then...no idea.

[dextercioby]

Well, if 2xy=x, then i see 2 solutions, either x=0, or y=1/2.