# Thread: Roots of this auxiliary eqn?

1. ## Roots of this auxiliary eqn?

I'm trying to find the auxiliary roots of this DE

$y'''+iy=0$
So the auxiliary eqn is
$m^3+i=0$
and m=i ?

2. Use De Moivre's formula to find the three roots.

EDIT: $-i = e^{-i \pi /2} = \cos(-\pi/2) + i \sin(-\pi/2)$

$(-i)^{1/3} = e^{i(-\pi/6 + 2k \pi/3)} = \cos (-\pi/6 + 2k \pi /3) + i \sin(-\pi/6 + 2k \pi /3)$ for k = 0, 1, and 2

When k=0, $(-i)^{1/3} = \cos(-\pi/6) + i \sin(-\pi/6) = \frac{\sqrt{3}}{2} - \ \frac{i}{2}$

when k =1, $(-1)^{1/3} = \cos(\pi/2) + i \sin(\pi/2) = i$

and for the k=2, $(-1)^{1/3} = -\frac{\sqrt{3}}{2} - \ \frac{i}{2} \$

The complex roots are not conjugate pairs like in the other problem, so the general solution is $y(x) = C_{1}e^{(\frac{\sqrt{3}}{2} - \ \frac{i}{2})x} + C_{2}e^{ix} + C_{3}e^{(-\frac{\sqrt{3}}{2} - \ \frac{i}{2} )x}$

BTW, Wolfram Alpha (the web site associated with Mathematica) can solve differential equations like these.

3. Ah that formula is new to me. Just to be clear the $-i = e^{-\pi /2}$ Comes from Euler's formula and $\cos(-\pi/2) + i \sin(-\pi/2)$ comes from DeMoivre...is it $\frac{\pi}{2}$ because we're dealing with just $i$?

4. Originally Posted by mmattson07
Ah that formula is new to me. Just to be clear the $-i = e^{-\pi /2}$ Comes from Euler's formula and $\cos(-\pi/2) + i \sin(-\pi/2)$ comes from DeMoivre...is it $\frac{\pi}{2}$ because we're dealing with just $i$?

It should be $-i=e^{-i\pi/2} = \cos (-\pi/2) + i \sin (-\pi /2)$

Then $(-i)^{1/3} = e^{i (-\pi/6 + 2k \pi/3)}$

But the rest is correct.

I used De Moivre's formula to find the roots for the other problem, too.

$-1= e^{i \pi} = \cos \pi + i \sin \pi$

so $(-4)^{-1/4} = 4^{1/4}(-1)^{1/4} = \sqrt{2}e^{i(\pi /4 + 2k \pi /4)} = \sqrt{2} [(\cos(\pi /4 + 2k \pi /4) + i \sin (\pi /4 + 2k \pi /4)]$

Then let k =0,1,2,and 3 to find the four roots.

http://en.wikipedia.org/wiki/De_Moivre's_formula

5. Ok I see thanks for the link. Why is x $=\frac{\pi}{2}$ in this case?
$-i=e^{-i\pi/2} = \cos (-\pi/2) + i \sin (-\pi /2)$

6. Originally Posted by mmattson07
Ok I see thanks for the link. Why is x $=\frac{\pi}{2}$ in this case?
$-i=e^{-i\pi/2} = \cos (-\pi/2) + i \sin (-\pi /2)$
It's because of the location of -i on the complex plane.

I'll try to explain, but you should really study up a bit on complex numbers.

All real numbers (a+0i) are located along the x axis, all purely imaginary numbers (0+bi) are located along the y axis, and all complex numbers (a+bi) are located everywhere else.

$-i = 0-i$ is located on the negative y axis and can be represented as a vector of unit length (since $\sqrt{0^{2}+1^{2}} = 1$) with it's tail at the origin and it's head at y=-1 .

The angle (going counterclockwise) that the vector makes with the positive horizontal axis is $3 \pi/2$. But you usually want the angle to be between $\pi$and $-\pi$. The angle in that range that is equal to $3 \pi /2$ is $=-\pi/2$

You can write any number (real, imaginary, or complex) in the form $re^{i\theta} = r(cos \theta + i \sin \theta)$ where $r$ is the distance of the point from the origin in the complex plane and $\theta$ is the angle formed by the vector and the positive x axis.

So $-i= 1e^{-i \pi /2} = 1(\cos -\pi/2 + i \sin -\pi/2) = 1(0-i) = -i$

For the other problem the angle is $\pi$ because -1 is located on the negative x axis.

7. Should it be $i$ or $-i$? Because $(i)^3=-i$ satisfying $m^3+i=0$ and $(-i)^3=i$ which would give $2i=0$

8. Originally Posted by mmattson07
Should it be $i$ or $-i$? Because $(i)^3=-i$ satisfying $m^3+i=0$ and $(-i)^3=i$ which would give $2i=0$
You're correct. $-i$ is not a root. But I never said it was. We're looking for the three roots of $(-i)^{1/3}$. I need to first express $-i$ in complex polar form (which I did above) before I can apply De Moivre's formula to find the roots. One of those roots is $i$.

9. OH I see you get $(-i)^{1/3}$ from solving for m then apply DeMoivre. Ok thanks.