Question

The principal argument of $\frac{\left[\right(1+i{)}^5(1+\sqrt{3}i{)}^2]}{-2i[\sqrt{3}-1-(\sqrt{3}+1\left)i\right]}$ is

• A. $\frac{5\pi }{6}$
• B. $\frac{7\pi }{12}$
• C. $\frac{\pi }{2}$
• D. $-\frac{5\pi }{12}$

Answer 1

The correct option is A $\frac{5\pi }{6}$

Let $z=\frac{\left[\right(1+i{)}^5(1+\sqrt{3}i{)}^2]}{-2i[\sqrt{3}-1-(\sqrt{3}+1\left)i\right]}$
and $z_1=1+i,z_2=1+\sqrt{3}i,z_3=-2(\sqrt{3}+1)-2(\sqrt{3}-1)i$
Then $z=\frac{z_1^5{z}_2^2}{z_3}$
$\therefore \arg \left(z\right)=\arg \left(\frac{z_1^5{z}_2^2}{z_3}\right)\Rightarrow \arg \left(z\right)=\arg \left(z_1^5\right)+\arg \left(z_2^2\right)-\arg \left(z_3\right)\Rightarrow \arg \left(z\right)=5\arg \left(z_1\right)+2\arg \left(z_2\right)-\arg \left(z_3\right)\Rightarrow \arg \left(z\right)=5\left(\frac{\pi }{4}\right)+2\left(\frac{\pi }{3}\right)-(\frac{\pi }{12}-\pi )+2k\pi ,k\in Z\Rightarrow \arg \left(z\right)=\frac{17\pi }{6}+2k\pi ,k\in Z$
At $k=-1$ , $\arg \left(z\right)=\frac{17\pi }{6}-2\pi$
$\Rightarrow \arg \left(z\right)=\frac{5\pi }{6}\in (-\pi ,\pi ]$