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- Jan 23rd 2010, 12:57 AM.::MadMaX::.Complex numbers problem.
- Jan 23rd 2010, 01:43 AMProve It
Why not just use the Quadratic Formula?

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

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

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

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

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

Now to find $\displaystyle \sqrt{-3 + 4i}$, convert to polars and use DeMoivre's Theorem. - Jan 23rd 2010, 01:50 AMtonio

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

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

Tonio - Jan 23rd 2010, 03:38 AMHallsofIvy
- Jan 24th 2010, 03:00 PM.::MadMaX::.
Thank you all,but I just don't know how to handle this:

http://www.mathhelpforum.com/math-he...945312bf-1.gif

EDIT: Anyone?