Question

The argument of the complex number $\left(-1+i\sqrt{3}\right)\left(1+i\right)\left(\cos \theta +i\sin \theta \right)$ is ____________.

Answer 1

$$\begin{gathered} \mathrm{For}z=\left(-1+i\sqrt{3}\right)\left(1+i\right)\left(\cos \theta +i\sin \theta \right) \\ \arg z=\arg \left(-1+i\sqrt{3}\right)+\arg \left(1+i\right)+\arg \left(\cos \theta +i\sin \theta \right) \\ \arg \left(-1+i\sqrt{3}\right):-z_1=-1+i\sqrt{3} \\ {\theta }_1={\tan }^{-1}\left|\left(\frac{\sqrt{3}}{1}\right)\right| \\ \theta =\frac{\mathrm{\pi }}{3} \end{gathered}$$
Since z₁ lies in II quadrant
$$\begin{gathered} \Rightarrow \arg z_1=\mathrm{\pi }-\frac{\mathrm{\pi }}{3}=\frac{2\mathrm{\pi }}{3} \\ \arg \left(1+i\right):-z_2=1+i \\ {\theta }_2={\tan }^{-1}\left|\frac{1}{1}\right|={\tan }^{-1}1 \\ \mathrm{i}.\mathrm{e}{\theta }_2=\frac{\mathrm{\pi }}{4} \end{gathered}$$
Since z₂ lies in I quadrant
$$\begin{gathered} \Rightarrow \arg z_2=\frac{\mathrm{\pi }}{4} \\ \arg z_3=\arg \left(\cos \theta +i\sin \theta \right)=\theta \\ \Rightarrow \arg z=\arg z_1+\arg z_2+\arg z_3 \\ \arg z=\frac{2\mathrm{\pi }}{3}+\frac{\mathrm{\pi }}{4}+\theta =\left|\frac{11\mathrm{\pi }}{12}+\theta \right| \end{gathered}$$