# Solve quadratic into complex numbers

• Feb 16th 2011, 11:00 AM
eggy524
Hi

I have this equation:
$\displaystyle s^2 + \sqrt{2}s + 4$

and it somehow factorises to this:

$\displaystyle (s + 2e^{+j(\pi/4)})(s + 2e^{-j(\pi/4)})$

I've tried using the quadratic formula but cant get it into the form with $\displaystyle e^{x}$

Would anyone be able to shed a little light on this?

Thanks
• Feb 16th 2011, 11:14 AM
TheEmptySet
Quote:

Originally Posted by eggy524
Hi

I have this equation:
$\displaystyle s^2 + \sqrt{2}s + 4$

and it somehow factorises to this:

$\displaystyle (s + 2e^{+j(\pi/4)})(s + 2e^{-j(\pi/4)})$

I've tried using the quadratic formula but cant get it into the form with $\displaystyle e^{x}$

Would anyone be able to shed a little light on this?

Thanks

Well it doesn't if you multiply it out you get

$\displaystyle (s + 2e^{+j(\pi/4)})(s + 2e^{-j(\pi/4)})=$
$\displaystyle \displaystyle s^2+2se^{-j\frac{\pi}{4}}+2se^{j\frac{\pi}{4}}+4$
$\displaystyle \displaystyle s^2+2s\left(e^{j\frac{\pi}{4}}+e^{-j\frac{\pi}{4}} \right)+4$
$\displaystyle \displaystyle s^2+2s(2\cos\left(\frac{\pi}{4} \right))+4$
$\displaystyle s^2+2\sqrt{2}s+4$
• Feb 16th 2011, 11:36 AM
HallsofIvy
Quote:

Originally Posted by eggy524
Hi

I have this equation:
$\displaystyle s^2 + \sqrt{2}s + 4$

and it somehow factorises to this:

$\displaystyle (s + 2e^{+j(\pi/4)})(s + 2e^{-j(\pi/4)})$

I've tried using the quadratic formula but cant get it into the form with $\displaystyle e^{x}$

Would anyone be able to shed a little light on this?

Thanks

When you use the quadratic formula to solve the equation you get $\displaystyle s=\frac{-\sqrt{2}\pm\sqrt{2- 4(1)(4)}}{2}= \frac{-\sqrt{2}\pm\sqrt{-14}}{2}$
so that $\displaystyle x^2+ \sqrt{2}x+ 4= \left(s+ \frac{\sqrt{2}}{2}+i\frac{\sqrt{14}}{2}\right)\lef t(s+ \frac{\sqrt{2}{2}}- i\frac{\sqrt{14}}{2}\right)$.

You can change those numbers to "polar" form but they are not anything like "$\displaystyle 2e^{i\pi/4}$" or "$\displaystyle e^{-i\pi/4}$.

The modulus is correct: $\displaystyle \sqrt{\frac{1}{2}+ \frac{7}{2}}= \sqrt{4}= 2$
but the argument is $\displaystyle tan^{-1}\left(\frac{\sqrt{14}}{2}\frac{2}{\sqrt{2}}\righ t)= tan^{-1}\right(\sqrt{7})$ which is closer to 1.2 radians than to $\displaystyle \pi/4$.

Did you, as TheEmptySet suggests, drop a factor of "2" from the coefficient of x?
• Apr 8th 2011, 11:20 AM
eggy524
Thanks guys, makes more sense now!