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 $\displaystyle z = a + ib$, then $\displaystyle z^2 = (a^2 - b^2) + 2iab$, so
$\displaystyle (a^2 - b^2) + 2iab + (1 + 3i)(a + ib) + \frac{3}{4}i = 0$
$\displaystyle (a^2 - b^2) + 2iab + (a + ib + 3ia - 3b) + \frac{3}{4}i = 0$
Resolving this into real and complex components:
$\displaystyle a^2 - b^2 + a - 3b = 0$
and
$\displaystyle 2ab + b + 3a + \frac{3}{4} = 0$
Now solve the system of equations.
-Dan