Given-that-e-i-npi-n-N-show-that-1-2-1-2i-sin-2-1-n-2-n-cos-n-1-2-e-1-2-in-

Question Number 124595 by physicstutes last updated on 04/Dec/20

Given that ω = e^{i θ}, θ \neq n π, n \in N

show that

(1) \frac{ω^{2} - 1}{ω} = 2 i \sin θ

(2) {(1 + ω)}^{n} = 2^{n} \cos^{n} (\frac{1}{2} θ) e^{\frac{1}{2} (i n θ)}

Answered by Ar Brandon last updated on 04/Dec/20

f (ω) = \frac{ω^{2} - 1}{ω} = \frac{e^{2 i θ} - 1}{e^{i θ}} = e^{i θ} - e^{- i θ}

= (\cos θ + isin θ) - (\cos (- θ) + isin (- θ))

= (\cos θ + isin θ) - (\cos θ - isin θ)

= 2 isin θ

Commented by physicstutes last updated on 04/Dec/20

thats great

Answered by mnjuly1970 last updated on 04/Dec/20

1 : ω^{2} - 1 = (c o s (2 θ) + i s i n (2 θ)) - 1 = 2 i s i n (θ) c o s (θ) - 2 {s i n}^{2} (θ)

= 2 i s i n (θ) [c o s (θ) + i s i n (θ)]

\Rightarrow \frac{ω^{2} - 1}{ω} = 2 i s i n (θ)

c o r a l l a r y : Ω = \int_{0}^{\frac{π}{2}} l n (I m (\frac{ω^{2} - 1}{2 w})) d θ = - \frac{π}{2} l n (2) ✓

Answered by Ar Brandon last updated on 04/Dec/20

f (ω) = {(1 + ω)}^{n} = {(1 + \cos θ + isin θ)}^{n}

= {({2 \cos}^{2} \frac{θ}{2} + 2 isin \frac{θ}{2} \cos \frac{θ}{2})}^{n}

= {[2 \cos \frac{θ}{2} (\cos \frac{θ}{2} + isin \frac{θ}{2})]}^{n}

= 2^{n} \cos^{n} \frac{θ}{2} e^{i \frac{n θ}{2}}

Commented by Ar Brandon last updated on 04/Dec/20

1 = \cos^{2} \frac{θ}{2} + \sin^{2} \frac{θ}{2}

\cos θ = \cos^{2} \frac{θ}{2} - \sin^{2} \frac{θ}{2}

\sin θ = 2 \sin \frac{θ}{2} \cos \frac{θ}{2}

\cos \frac{θ}{2} + isin \frac{θ}{2} = e^{i \frac{θ}{2}}

Answered by mnjuly1970 last updated on 04/Dec/20

2 . {(1 + ω)}^{n} = {(1 + c o s (θ) + i s i n (θ))}^{n}

= {(2 {c o s}^{2} (\frac{θ}{2}) + 2 i s i n (\frac{θ}{2}) c o s (\frac{θ}{2}))}^{n}

= 2^{n} ({c o s}^{n} (\frac{θ}{2})) {(c o s (\frac{θ}{2}) + i s i n (\frac{θ}{2}))}^{n}

= 2^{n} {c o s}^{n} (\frac{θ}{2}) e^{i (\frac{n θ}{2})} ✓