1. ## theorem

let z=r(cosθ+isinθ) be any complex number & n be any positive integer. Then zn =rn (cos(nθ)+isin(nθ)) or equivelently, [rn , nθ].

Prove it in special cases n=2,3,4

2. ## Re: theorem

Originally Posted by franios
Prove it in special cases n=2
So, why don't you do this?

3. ## Re: theorem

I'm just confused, do I have to just plug in n? What about theta?

4. ## Re: theorem

Do you know how to prove that for n = 2 it is the case that for all x and y, (x + y)^n = x^n + 2xy + y^n?

Now prove that for n = 2 it is the case that for all r and θ, (r(cosθ+isinθ))^n = rn(cos(nθ)+isin(nθ)).

5. ## Re: theorem

Originally Posted by franios
I'm just confused, do I have to just plug in n? What about theta?
\begin{align*}z\cdot z &=r(cos(t)+i\sin(t))r(cos(t)+i\sin(t))\\&=r^2(\cos ^2(t)-\sin^2(t))+i(\cos(t)\sin(t))\\&=r^2(cos(2t)+i\sin( 2t))\end{align*}

Now $z^3=z^2\cdot z$