Complex Solutions Homework - Find Zeros of Equation

[Firepanda]

Homework Statement

http://img238.imageshack.us/img238/1381/complexle5.jpg

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?

 

[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

 

[tiny-tim]

quadratic equation …

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.)

 

[HallsofIvy]

Homework Statement

http://img238.imageshack.us/img238/1381/complexle5.jpg

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.

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

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

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

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

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

But what to do from here?

Use the quadratic formula or complete the square.