Question

Express the complex number $sin\frac{\pi }{5}+i(1-cos\frac{\pi }{5})$ in polar form.

Answer 1

Let $z=sin\frac{\pi }{5}+i(1-cos\frac{\pi }{5})$.

Let its polar form be $z=r(cos\theta +isin\theta )$.

Now, $r^2=|z{|}^2=si{n}^2\frac{\pi }{5}+{(1-cos\frac{\pi }{5})}^2=(si{n}^2\frac{\pi }{5}+co{s}^2\frac{\pi }{5})+1-2cos\frac{\pi }{5}$

$\Rightarrow r^2=2(1-cos\frac{\pi }{5})=4si{n}^2\frac{\pi }{10}\Rightarrow r=2sin\frac{\pi }{10}$.

Let $\alpha$ be the acute angle, given by

$tan\alpha =\left|\frac{Im\left(z\right)}{Re\left(z\right)}\right|=\left|\frac{1-cos\frac{\pi }{5}}{sin\frac{\pi }{5}}\right|=\frac{2si{n}^2\frac{\pi }{10}}{2.sin\frac{\pi }{10}.cos\frac{\pi }{10}}=\tan \frac{\pi }{10}\Rightarrow \alpha =\frac{\pi }{10}$.

Clearly, the point representing z lies in the first quadrant as x>0 and y>0.

$\therefore arg\left(z\right)=\theta =\alpha =\frac{\pi }{10}$.
Thus, $r=2sin\frac{\pi }{10}$ and $\theta =\frac{\pi }{10}$.

Hence, the required polar form is $2sin\frac{\pi }{10}(cos\frac{\pi }{10}+isin\frac{\pi }{10})$.