$Q. Z = \left(1+\cos \theta \right)+ i \sin \theta , \mathrm{\pi }<0<2\mathrm{\pi }, \mathrm{Find} \mathrm{out} : -$

1. Modulus
2. Argument
3. Convert in polar form.

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{z}=\left(1+\mathrm{cos\theta }\right)+\mathrm{isin\theta } \\ =\left(1+2{\cos }^2\frac{\mathrm{\theta }}{2}-1\right)+\mathrm{i}\left(2\sin \frac{\mathrm{\theta }}{2}\cos \frac{\mathrm{\theta }}{2}\right) \\ =2{\cos }^2\frac{\mathrm{\theta }}{2}+\mathrm{i}\left(2\sin \frac{\mathrm{\theta }}{2}\cos \frac{\mathrm{\theta }}{2}\right) \\ \mathrm{As} \mathrm{\pi }<\mathrm{\theta }<2\mathrm{\pi }, \mathrm{hence} \\ \frac{\mathrm{\pi }}{2}<\mathrm{\theta }<\mathrm{\pi } \mathrm{i}.\mathrm{e}. \mathrm{second} \mathrm{quadrant}, \mathrm{hence} \cos \frac{\mathrm{\theta }}{2} \mathrm{will} \mathrm{be} \mathrm{negative} \\ \mathrm{z}=-2\cos \frac{\mathrm{\theta }}{2}\left[-\cos \frac{\mathrm{\theta }}{2}-\mathrm{isin}\frac{\mathrm{\theta }}{2}\right] \\ =-2\cos \frac{\mathrm{\theta }}{2}\left[\cos \left(\mathrm{\pi }+\frac{\mathrm{\theta }}{2}\right)+\mathrm{isin}\left(\mathrm{\pi }+\frac{\mathrm{\theta }}{2}\right)\right] \mathrm{which} \mathrm{is} \mathrm{desired} \mathrm{polar} \mathrm{form} \\ \mathrm{Hence} \\ \mathrm{Modulus}=-2\cos \frac{\mathrm{\theta }}{2} \\ \mathrm{Argument}=\mathrm{\pi }+\frac{\mathrm{\theta }}{2} \end{gathered}$$

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