Question

If $z=\sqrt{3}+i$ and $w=3i$
Then $\arg \left(zw\right)$ and $\arg \frac{w}{z}$ is

• A. $\arg \left(zw\right)=\frac{2\pi }{3},\arg \left(\frac{w}{z}\right)=\frac{\pi }{3}$
• B. $\arg \left(zw\right)=\frac{2\pi }{3},\arg \left(\frac{w}{z}\right)=\frac{2\pi }{3}$
• C. $\arg \left(zw\right)=\frac{\pi }{3},\arg \left(\frac{w}{z}\right)=\frac{\pi }{3}$
• D. $\arg \left(zw\right)=\frac{\pi }{3},\arg \left(\frac{w}{z}\right)=\frac{\pi }{5}$

Answer 1

The correct option is A $\arg \left(zw\right)=\frac{2\pi }{3},\arg \left(\frac{w}{z}\right)=\frac{\pi }{3}$

$z=\sqrt{3}+i$ and $w=3i$
$\tan \alpha =\frac{|1|}{|\sqrt{3}|}\Rightarrow \alpha =\frac{\pi }{6}$
As $z$ lies in the first quadrant So, $\arg z=\alpha =\frac{\pi }{6}$
$\because w$ lies on positive $y-$ axis so
$\arg \left(w\right)=\frac{\pi }{2}$
Now
$\because \arg \left(z_1{z}_2\right)=\arg z_1+\arg z_2+2k\pi ,k\in Z$
and
$\because \arg \frac{z_1}{z_2}=\arg z_1-\arg z_2+2k\pi ,k\in Z$
$\therefore \arg wz=\arg z+\arg w+2k\pi ,k\in Z$
$=\frac{\pi }{6}+\frac{\pi }{2}+2k\pi ,k\in Z$
At $k=0,\arg \left(wz\right)=\frac{2\pi }{3}\in (-\pi ,\pi ]$
So, principal argument of $wz$ is $\frac{2\pi }{3}$
and
$\arg \frac{w}{z}=\arg w-\arg z+2k\pi ,k\in Z$
$=\frac{\pi }{2}-\frac{\pi }{6}+2k\pi ,k\in Z$
At $k=0,\arg \frac{w}{z}=\frac{\pi }{3}\in (-\pi ,\pi ]$
So, principal argument of $\frac{w}{z}$ is $\frac{\pi }{3}$