# Thread: Complex Numbers De Moirves Theorm

1. ## Complex Numbers De Moirves Theorm

Use De Moivres Theorm to express sin 3 theta as a polynomial in sin theta.

Where do I start?

2. Well if

$\displaystyle z = \cos{\theta} + i\sin{\theta}$

then

$\displaystyle z^3 = (\cos{\theta} + i\sin{\theta})^3$

$\displaystyle = \cos^3{\theta} + 3i\cos^2{\theta}\sin{\theta} - 3\cos{\theta}\sin^2{\theta} - i\sin^3{\theta}$

$\displaystyle = \cos^3{\theta} - 3\cos{\theta}\sin^2{\theta} + i(3\cos^2{\theta}\sin{\theta} - \sin^3{\theta})$.

But $\displaystyle z^3 = \cos{3\theta} + i\sin{3\theta}$ by DeMoivre's Theorem.

So $\displaystyle \cos^3{\theta} - 3\cos{\theta}\sin^2{\theta} + i(3\cos^2{\theta}\sin{\theta} - \sin^3{\theta}) = \cos{3\theta} + i\sin{3\theta}$.

Now equate real and imaginary parts.

3. Excellent ,Thankyou Prove It!

4. One more thing...

$\displaystyle \sin{3\theta}$ is usually written only in terms of $\displaystyle \sin{\theta}$

and

$\displaystyle \cos{3\theta}$ is usually written only in terms of $\displaystyle \cos{\theta}$

so use the Pythagorean Identity

$\displaystyle \cos^2{\theta} + \sin^2{\theta} = 1$

to clean it up.

5. Thanks,

Iam having troubles with this one too...

Determine all solutions for: z^4 + 3z^2 -4=0
The ans is +-1,+-2i

I got too,

r^4 (cos 4theta + i sin 4theta) + 3r^2(cos 2theta + i sin 2 theta) = 4 (cos 0 + 2pi k + i sin 0 +2pi k).

but don't know what to do from here,solving for R ,I get R=1,R=-4 ? and theta = 1/3 pi k ?

6. $\displaystyle z^4 + 3z^2 - 4 = 0$.

Let $\displaystyle Z = z^2$ to turn the polynomial into

$\displaystyle Z^2 + 3Z - 4 = 0$

$\displaystyle (Z + 4)(Z - 1) = 0$

$\displaystyle Z = -4$ or $\displaystyle Z = 1$

$\displaystyle z^2 = -4$ or $\displaystyle z^2 = 1$

$\displaystyle z = \pm 2i$ or $\displaystyle z = \pm 1$.

So the solutions are $\displaystyle z = -1, z = 1, z = -2i, z = 2i$.

For the other one, notice that if $\displaystyle z = r(\cos{\theta} + i\sin{\theta})$

then $\displaystyle z^2 = r^2(\cos{2\theta} + i\sin{2\theta}$ and $\displaystyle z^4 = r^4(\cos{4\theta} + i\sin{4\theta})$.

Also note that $\displaystyle \cos{(0 + 2\pi k)} = 1$ and $\displaystyle \sin{(0 + 2\pi k)} = 0$.

So that means you can write the equation

$\displaystyle r^4(\cos{4\theta} + i\sin{4\theta}) + 3r^2(\cos{2\theta} + i\sin{2\theta}) = 4[\cos{(0 + 2\pi k) + i\sin{(0 + 2\pi k)}}]$

as

$\displaystyle z^4 + 3z^2 = 4$

$\displaystyle z^4 + 3z^2 - 4 = 0$.

Do you see something in common here?

7. Thanks for explaining all that, very helpful.