# Complex solutions

1. Mar 3, 2008

### Firepanda

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

http://img238.imageshack.us/img238/1381/complexle5.jpg [Broken]

3. The attempt at a solution

I figure if i just find 1 zero of the equation i can use its conjugate as the other zero, then I guess it's complete.

I know this:

where a is an element of C (Complex)

(z - a) (z - a(conj)) = z^2 - 2(Re a)z + |a|^2

so.. using this

2(Re a) = 1 + 3i
and
|a|^2 = -1 + 3i/4

But what to do from here?

Last edited by a moderator: May 3, 2017
2. Mar 3, 2008

### Firepanda

Hmm perhaps I have to find the discriminant:

which is -4 + 3i

so z = (-(1 + 3i) +- i(sqrt (4 - 3i)))/2

:/ doesn't look too correct

3. Mar 3, 2008

### tiny-tim

This is an ordinary quadratic equation.

Just solve it the ordinary way.

(The only difficulty is that you'll have to calculate the square root of a complex number.)

4. Mar 4, 2008

### HallsofIvy

Staff Emeritus
No, that's not true. The two roots of a quadratic equation must be conjugates only if the equation has all real coefficients.

but that's impossible: 2(Re a) must be a real number.
Again, the roots of this equation are not complex conjugates.

Once more, that's impossible. |a| is a real number, its square is a real number.

Use the quadratic formula or complete the square.

Last edited by a moderator: May 3, 2017