Question

If z= 1/(1-cosθ-isinθ) then Re(z)=

Answer 1

u want to solve it ?

Answer 2

yes

Answer 3

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\(Hmm...\)
(1 - cosθ - isinθ) = 2cos^2(θ/2) - 2isin(θ / 2)cos(θ / 2) =2cos(θ / 2)[cos(θ / 2)-isin(θ / 2)]
(1-cosθ+isinθ= 2sin^2(θ / 2)+2isin(θ / 2)cos(θ / 2)=2sin(θ / 2)[sin(θ / 2)+icos(θ / 2)]

so z=cot(θ /2) [cos(θ / 2)-isin(θ / 2)]/[sin(θ / 2)+icos(θ / 2)]

\(Multiply~top~and~bottom~by~'i~' ~\)

\(iC\)(c-is)/(-c+is)
\(where....\)
C=cot(θ / 2), c=cos(θ / 2) and s=sin(θ / 2)

This gives \(-iC\), Since c-is=-(-c+is)

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Answer 4

\(Note:~It's~just~A~Sample,~Showing~how~to~solve~and~its~not~exact~Answer~!!\)

Answer 5

\[\large \begin{align*}z&=\frac{1}{1-\cos\theta-i\sin\theta}\\\\
&=\frac{1}{1-\cos\theta-i\sin\theta}\cdot\frac{1-\cos\theta+i\sin\theta}{1-\cos\theta+i\sin\theta}\\\\
&=\frac{1-\cos\theta+i\sin\theta}{(1-\cos\theta)^2+\sin^2\theta}\\\\
&=\frac{1-\cos\theta+i\sin\theta}{1-2\cos\theta+\cos^2\theta+\sin^2\theta}\\\\
&=\frac{1-\cos\theta+i\sin\theta}{2-2\cos\theta}\\\\
&=\frac{1-\cos\theta}{2(1-\cos\theta)}+i\frac{\sin\theta}{2-2\cos\theta}\\\\
&=\frac{1}{2}+i\frac{\sin\theta}{2-2\cos\theta}
\end{align*}\]