Complex quadratic equation

**callumh167** · March 3rd 2008, 09:55 AM

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

**topsquark** · March 3rd 2008, 10:40 AM

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

Thread: Complex quadratic equation

LinkBack

Thread Tools

Display

Complex quadratic equation

Similar Math Help Forum Discussions

Quadratic Equation for complex dataset

Complex roots of a quadratic equation

Complex Quadratic Equations

Complex quadratic equation problem

quadratic equation for complex numbers

Search Tags