Math Help - Complex quadratic equation

Find the complex solutions of z^2 + (1 + 3i)z − 1 + (3/4) i = 0

Im not sure how to solve this, any help would be much appreciated

Thanks

2. Originally Posted by callumh167
Find the complex solutions of z^2 + (1 + 3i)z − 1 + (3/4) i = 0

Im not sure how to solve this, any help would be much appreciated

Thanks
One way:
Let $z = a + ib$, then $z^2 = (a^2 - b^2) + 2iab$, so
$(a^2 - b^2) + 2iab + (1 + 3i)(a + ib) + \frac{3}{4}i = 0$

$(a^2 - b^2) + 2iab + (a + ib + 3ia - 3b) + \frac{3}{4}i = 0$

Resolving this into real and complex components:
$a^2 - b^2 + a - 3b = 0$
and
$2ab + b + 3a + \frac{3}{4} = 0$

Now solve the system of equations.

-Dan