Question

$\text{If}\pi <\theta <2\pi andz=1+cos\theta +isin\theta ,\text{then write the value of}|z|$

Answer 1

given $\pi <\theta <2\pi$

and $z=1+cos\theta +isin\theta$

$\Rightarrow |z|=\sqrt{(1+cos\theta {)}^2+si{n}^2\theta }=\sqrt{1+co{s}^2\theta +2cos\theta +si{n}^2\theta }=\sqrt{1+co{s}^2\theta +si{n}^2\theta +2cos\theta }=\sqrt{1+1+2cos\theta }=\sqrt{2+2cos\theta }=\sqrt{2(1+cos\theta )}=\sqrt{2\times 2co{s}^2\frac{\theta }{2}}(\because 1_{c}os\theta =2co{s}^2\frac{\theta }{2})\Rightarrow |z|=2cos\frac{\theta }{2}\therefore \pi <\theta <2\pi \Rightarrow \frac{\pi }{2}<\frac{\theta }{2}<\pi$

$i.e.\frac{\theta }{2}$ lies in second quadrant and in second quadrant cosine is negative

$\therefore |z|=-2cos\frac{\theta }{2}(\because |x|=-x\text{if}x<0)$