# Number of solutions to complex eq.

1. Sep 1, 2007

### 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?

2. Sep 1, 2007

### dextercioby

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

3. Sep 1, 2007

### kasse

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

4. Sep 1, 2007

### dextercioby

Under what conditions ?

5. Sep 1, 2007

w^n=z

6. Sep 1, 2007

### dextercioby

Okay, but did you solve the equation ?

7. Sep 1, 2007

### kasse

Can't make it further than this:

(x^2-y^2-y-(1/4)+i(2xy-x)=0

Last edited: Sep 1, 2007
8. Sep 1, 2007

### dextercioby

Okay then. What follows next ?

9. Sep 1, 2007

### kasse

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

But then...no idea.

10. Sep 1, 2007

### dextercioby

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