Question

Express in the form A + iB

$\left(\frac{1+sin\alpha +icos\alpha }{1+sin\alpha -icos\alpha }\right)$

• A. $cos(\frac{n\pi }{2}-n\alpha )-isin(\frac{n\pi }{2}-n\alpha )$
• B. $cos(n\pi -n\alpha )-isin(n\pi -n\alpha )$
• C. $cos(\frac{n\pi }{2}-n\alpha )+isin(\frac{n\pi }{2}-n\alpha )$
• D. $cos(n\pi -n\alpha )+isin(n\pi -n\alpha )$

Answer 1

The correct option is C $cos(\frac{n\pi }{2}-n\alpha )+isin(\frac{n\pi }{2}-n\alpha )$

Simplifying, we get

$\frac{1+cos(\frac{\pi }{2}-\theta )+isin(\frac{\pi }{2}-\theta )}{1+cos(\frac{\pi }{2}-\theta )-isin(\frac{\pi }{2}-\theta )}$

$=\frac{2co{s}^2(\frac{\pi }{4}-\frac{\theta }{2})+i2cos(\frac{\pi }{4}-\frac{\theta }{2})sin\left(\right(\frac{\pi }{4}-\frac{\theta }{2})}{2co{s}^2(\frac{\pi }{4}-\frac{\theta }{2})-i2cos(\frac{\pi }{4}-\frac{\theta }{2})sin\left(\right(\frac{\pi }{4}-\frac{\theta }{2})}$

$=\frac{cos(\frac{\pi }{4}-\frac{\theta }{2})+isin\left(\right(\frac{\pi }{4}-\frac{\theta }{2})}{cos(\frac{\pi }{4}-\frac{\theta }{2})-isin\left(\right(\frac{\pi }{4}-\frac{\theta }{2})}$

$=\frac{e^{i(\frac{\pi }{4}-\frac{\theta }{2})}}{e^{-i(\frac{\pi }{4}-\frac{\theta }{2})}}$

$=e^{i\left(2\right(\frac{\pi }{4}-\frac{\theta }{2}\left)\right)}$

$=e^{i(\frac{\pi }{2}-\theta )}$

Or

$e^{i(\frac{n\pi }{2}-n\theta )}$