If $\left( \frac{1 + \cos{\phi + i\sin\phi}}{1 + \cos{\phi - i\sin\phi}} \right) = u + iv$, where $u$ and $v$ all real numbers, then $u$ is

Question

Dr. Abubacker · Accepted Answer

(b)
$\left( \frac{1 + \cos{\phi + i\sin\phi}}{1 + \cos{\phi - i\sin\phi}} \right)^{n} = u + iv$
$\Rightarrow \ \ \ \ \ \left( \frac{2\cos^{2}{\ \frac{\phi}{2} + 2i\sin{\frac{\phi}{2}\cos\frac{\phi}{2}}}}{2\cos^{2}{\frac{\phi}{2} - 2\sin{\frac{\phi}{2}\cos\frac{\phi}{2}}}} \right)^{n} = u + iv$
$\Rightarrow \ \ \ \ \ \left( \frac{e^{i\ \frac{\phi}{2}}}{e^{- i\frac{\phi}{2}}} \right)^{n} = u + iv\ \ \ \ \  \Rightarrow \ \ \ \ \ \ \left( e^{in\phi} \right) = u + iv$
$\Rightarrow \ \ \cos{n\phi + i\sin{n\phi = u + iv}}$
$\Rightarrow \ \ \ \ u = \cos{n\phi,\ \ \ v = \sin{n\phi}}$

Q. If $\left( \frac{1 + \cos{\phi + i\sin\phi}}{1 + \cos{\phi - i\sin\phi}} \right) = u + iv$, where $u$ and $v$ all real numbers, then $u$ is

Explanation

Reviewed by Dr. Abubacker

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