Question

$\frac{i-1}{i(1-cos\frac{2\pi }{5})+\sin \frac{2\pi }{5}}$

• A. $\theta =\frac{13\pi }{20},r=\frac{cosec{18}^0}{\sqrt{2}}$
• B. $\theta =\frac{13\pi }{20},r=\frac{cosec{36}^0}{\sqrt{2}}$
• C. $\theta =\frac{11\pi }{20},r=\frac{cosec{18}^0}{\sqrt{2}}$
• D. $\theta =\frac{11\pi }{20},r=\frac{cosec{36}^0}{\sqrt{2}}$

Answer 1

The correct option is D $\theta =\frac{11\pi }{20},r=\frac{cosec{36}^0}{\sqrt{2}}$

Consider the denominator

$i(1-cos\frac{2\pi }{5})+sin\left(\frac{2\pi }{5}\right)$

$=i2si{n}^2\left(\frac{\pi }{5}\right)+2sin\left(\frac{\pi }{5}\right).cos\left(\frac{\pi }{5}\right)$

$=2sin\left(\frac{\pi }{5}\right)\left[cos\right(\frac{\pi }{5})+isin(\frac{\pi }{5}\left)\right]$

$=2sin\left(\frac{\pi }{5}\right).e^{i\frac{\pi }{5}}$

Hence by substituting in the equation, we get

$\frac{i-1}{2sin\left(\frac{\pi }{5}\right).e^{i\frac{\pi }{5}}}$

$=\left[\frac{cosec\frac{\pi }{5}}{2}\right](-1+i).e^{-i\frac{\pi }{5}}$
$=\left[\frac{cosec\frac{\pi }{5}}{2}\right]\sqrt{2}.e^{i\frac{3\pi }{4}}.e^{-i\frac{\pi }{5}}$
$=\frac{cosec\frac{\pi }{5}}{\sqrt{2}}.e^{i(\frac{3\pi }{4}-\frac{\pi }{5})}$
$=\frac{cosec{36}^0}{\sqrt{2}}.e^{i\frac{11\pi }{20}}$

Hence

$|z|=r=\frac{cosec{36}^0}{\sqrt{2}}$

$Arg\left(z\right)=\theta =\frac{11\pi }{20}$