1. ## Prooving DeMoivre's Theorem

I am having so much trouble prooving this theorem. This is what I have to prove:

PROVE: If z = r(cos u + i sin u), then z^n = r^n(cos nu + i sin nu) when n = 2

I am so frustrated at this point. I have to list all my steps and no matter how I try to fix it, it never comes out correctly.

2. Originally Posted by missyd819
I am having so much trouble prooving this theorem. This is what I have to prove:

PROVE: If z = r(cos u + i sin u), then z^n = r^n(cos nu + i sin nu) when n = 2

I am so frustrated at this point. I have to list all my steps and no matter how I try to fix it, it never comes out correctly.

What you are being asked to prove is:

$z^2=r^2(\cos(2u)+i\sin(2u))$

Well:

$z^2=r^2(\cos(u)+i\sin(u))^2=r^2((\cos(u))^2-(\sin(u))^2+2i\sin(u)\cos(u))$

but we have identities:

$\cos(2u)=(\cos(u))^2-(\sin(u))^2$

and

$
\sin(2u)=2\sin(u)\cos(u)
$

Hence:

$z^2=r^2(\cos(2u)+i\sin(2u))$

as required

RonL

3. In general, it is true for any integer $n$.

note that you can also write $z = r(\cos\theta + i\sin\theta)$ as $z=re^{i\theta}$

thus, if $n \in \mathbb{Z}$ we have

$z^n=\left(re^{i\theta}\right)^n = r^ne^{i(n\theta)}$

then transform the RHS to trigonometric form, you will get

$z^n = r^n (\cos (n\theta) + i\sin (n\theta))$

4. Originally Posted by kalagota
In general, it is true for any integer $n$.

note that you can also write $z = r(\cos\theta i\sin\theta)$ as $z=re^{i\theta}$

thus, if $n \in \mathbb{Z}$ we have

$z^n=\left(re^{i\theta}\right)^n = r^ne^{i(n\theta)}$

then transform the RHS to trigonometric form, you will get

$z^n = r^n (\cos (n\theta) + i\sin (n\theta))$
Once you can write a complex number in exponential form there is nothing left to prove!

RonL