Question

pwease im begging for some help :3
will do what ever u want :P
\(e^{i\theta}=\cos\theta+i\sin\theta\)
prove that
\((e^{i\theta})^n=e^{i n \theta}\)
for all n in Z

Answer 1

Given :
\(\large e^{i\theta}=\cos\theta+i\sin\theta \)

you want to prove :
\((e^{i\theta})^n=e^{in\theta}\)

?

Answer 2

yes

Answer 3

its correct if i use exponential properties ,
but i should use the given equation

Answer 4

\(\color{blue}{\text{Originally Posted by}}\) @SithsAndGiggles
What you want to show:
\[\large e^{in\theta}=\cos n\theta+i\sin n\theta=(\cos \theta+i\sin\theta)^n=\left(e^{i\theta}\right)^n\]
Try it by induction, with the base case \(n=1\):
\[\large e^{i\theta}=\cos\theta+i\sin\theta=\left(e^{i\theta}\right)^1\]
This is true by the definition of \(e^{i\theta}\).

Assume equality holds for \(n=k\):
\[\large e^{ik\theta}=\cos k\theta+i\sin k\theta=(\cos\theta+i\sin\theta)^k=\left(e^{i\theta}\right)^k\]
Show it holds for \(n=k+1\):
\[\large e^{i(k+1)\theta}=e^{ik\theta}e^{i\theta}=\cdots=\left(e^{i\theta}\right)^{k+1}\]
Fill in the blanks.
\(\color{blue}{\text{End of Quote}}\)

Answer 5

good!it works for n>0
what about n<0 ?

Answer 6

oh well ty alote @SithsAndGiggles for ur help !
i got it now for \(n=0 \) and \(n<0\)