Question

(cosA+isinA)⁴/(sinB+icosB)⁵​

Answer 1

Answer:

Euler's formula

e^{i\theta}=cos\theta+i\:sin\thetae

iθ

=cosθ+isinθ

Now,

\frac{(cos\:\theta+i\:sin\:\theta)^4}{(sin\:\theta+i\:cos\:\theta)^5}

(sinθ+icosθ)

5

(cosθ+isinθ)

4

=\frac{(cos\:\theta+i\:sin\:\theta)^4}{i^5(cos\:\theta-i\:sin\:\theta)^5}=

i

5

(cosθ−isinθ)

5

(cosθ+isinθ)

4

\begin{gathered}=\frac{(e^{i\theta})^4}{i(e^{-i\theta})^5}\\\\=(-i)\frac{e^{i4\theta}}{e^{-i5\theta}}\\\\=(-i)e^{i9\theta}\end{gathered}

=

i(e

−iθ

)

5

(e

iθ

)

4

=(−i)

e

−i5θ

e

i4θ

=(−i)e

i9θ

\begin{gathered}=-i[cos9\theta+i\:sin9\theta]\\\\=-icos9\theta-i^2sin9\theta\\\\=sin9\theta-icos9\theta\end{gathered}

=−i[cos9θ+isin9θ]

=−icos9θ−i

2

sin9θ

=sin9θ−icos9θ