prove through induction that [r(cosØ+isinØ)]^k = r^k (coskØ+isinkØ). n=k and n=k+1?

Question

Bonus Question

find or prove (incorrect or correct) or find the relation b/w i=1/i

Answer 1

The first poster used exponentials to prove it, which is perfectly valid. If you cannot use exponentials and have to use formal induction, here is the proof:

let us prove it for k=1:
[r(cosΘ+isinΘ)]^1 = r^1(cos(1*Θ) + isin(1*Θ)). Duh.

if it is true for k=n, let us prove it for k = n+1:
[r(cosΘ+isinΘ)]^(n+1) =r(cosΘ+isinΘ)*[r(cosΘ+isinΘ)]^n
since [r(cosΘ+isinΘ)]^n = r^n*(cosnΘ + isinnΘ),

[r(cosΘ+isinΘ)]^(n+1) = r(cosΘ+isinΘ)*r^n*(cosnΘ + isinnΘ)
[r(cosΘ+isinΘ)]^(n+1) = r^(n+1)*[cosΘcosnΘ + icosΘsinnΘ + isinΘcosnΘ - sinΘsinnΘ)
but cosΘcosnΘ - sinΘsinnΘ = cos(n+1)Θ and cosΘsinnΘ+ sinΘcosnΘ = sin(n+1)Θ, so
cosΘcosnΘ + icosΘsinnΘ + isinΘcosnΘ - sinΘsinnΘ = cos(n+1)Θ + isin(n+1)Θ.
finally [r(cosΘ+isinΘ)]^(n+1) = r^(n+1)*[cos(n+1)Θ + isin(n+1)Θ]
it is therefore true for any k > 0

Bonus question: i/(1/i) =i² = -1, so i = -1 * (1/i)

Answer 2

To prove this by induction, your inductive step is completed by using the addition formulas for sine and cosine.

Answer 3

r(cosØ+isinØ) = exp(irØ)
exp(irØ)^k = exp(ikrØ) = r^k(coskØ+isinkØ).