# Complex numbers problem.

• January 23rd 2010, 12:57 AM
Complex numbers problem.
• January 23rd 2010, 01:43 AM
Prove It
Quote:

Originally Posted by .::MadMaX::.

Why not just use the Quadratic Formula?

$Z^2 + (5 - 2i)Z + 6(1 - i) = 0$

$Z = \frac{-(5 - 2i) \pm \sqrt{(5 - 2i)^2 - 4(1)[6(1 - i)]}}{2(1)}$

$= \frac{-5 + 2i \pm \sqrt{25 - 20i + 4i^2 - 24(1 - i)}}{2}$

$= \frac{-5 + 2i \pm \sqrt{25 - 20i - 4 - 24 + 24i}}{2}$

$= \frac{-5 + 2i \pm \sqrt{-3 + 4i}}{2}$.

Now to find $\sqrt{-3 + 4i}$, convert to polars and use DeMoivre's Theorem.
• January 23rd 2010, 01:50 AM
tonio
Quote:

Originally Posted by .::MadMaX::.

I really can't understand what you're trying to do: why not the good-old and well-known formula to find the roots of a quadratic equation?!!?

We have $z^2+(5-2i)z+6(1-i)=0\Longrightarrow a=1\,,\,b=5-2i\,,\,c=6(1-i)\Longrightarrow$ $Discriminant=\Delta=b^2-4ac=21-20i-24(1-i)=-3+4i$ , so:

$z_{1,2}=\frac{-b\pm \sqrt{\Delta}}{2a}=\frac{-(5-2i)\pm \sqrt{-3+4i}}{2}=\frac{(-5+2i)\pm(1+2i)}{2}=\left\{\begin{array}{l}-2+2i\\-3\end{array}\right.$

Tonio
• January 23rd 2010, 03:38 AM
HallsofIvy
Quote:

Originally Posted by .::MadMaX::.

In both both methods you are really assuming that $Z^2= 0$. That's why you got the wrong answer!
• January 24th 2010, 03:00 PM