Question

Write the argument of $\left(1+i\sqrt{3}\right)\left(1+i\right)\left(\cos \theta +i\sin \theta \right)$.

Disclaimer: There is a misprinting in the question. It should be $\left(1+i\sqrt{3}\right)$ instead of $\left(1+\sqrt{3}\right)$.

Answer 1

Let the argument of $\left(1+i\sqrt{3}\right)$ be α. Then,
$$\begin{gathered} \tan \alpha =\frac{\sqrt{3}}{1}=\tan \frac{\mathrm{\pi }}{3} \\ \Rightarrow \alpha =\frac{\mathrm{\pi }}{3} \end{gathered}$$

Let the argument of $\left(1+i\right)$ be β. Then,
$$\begin{gathered} \mathrm{tan\beta }=\frac{1}{1}=\tan \frac{\mathrm{\pi }}{4} \\ \Rightarrow \mathrm{\beta }=\frac{\mathrm{\pi }}{4} \end{gathered}$$

Let the argument of $\left(\mathrm{cos\theta }+i\mathrm{sin\theta }\right)$ be γ. Then,
$$\begin{gathered} \mathrm{tan\gamma }=\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}=\mathrm{tan\theta } \\ \Rightarrow \mathrm{\gamma }=\mathrm{\theta } \end{gathered}$$

∴ The argument of $\left(1+i\sqrt{3}\right)\left(1+i\right)\left(\mathrm{cos\theta }+i\mathrm{sin\theta }\right)=\mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma }=\frac{\mathrm{\pi }}{3}+\frac{\mathrm{\pi }}{4}+\mathrm{\theta }=\frac{7\mathrm{\pi }}{12}+\mathrm{\theta }$

Hence, the argument of $\left(1+i\sqrt{3}\right)\left(1+i\right)\left(\mathrm{cos\theta }+i\mathrm{sin\theta }\right)\mathrm{is}\frac{7\mathrm{\pi }}{12}+\mathrm{\theta }$.