Question

Let $z_1=1-i$ and $z_2=\sqrt{3}+i$, then $z_1\cdot z_2$ in polar form is

• A. $2(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6})$
• B. $2\sqrt{2}(\cos \frac{5\pi }{12}+i\sin \frac{5\pi }{12})$
• C. $2(\cos (-\frac{\pi }{12})+i\sin (-\frac{\pi }{12}))$
• D. $2\sqrt{2}(\cos (-\frac{\pi }{12})+i\sin (-\frac{\pi }{12}))$

Answer 1

The correct option is D $2\sqrt{2}(\cos (-\frac{\pi }{12})+i\sin (-\frac{\pi }{12}))$

$z_1=1-i$ and $z_2=\sqrt{3}+i$

$\therefore z_1\cdot z_2=(1-i)(\sqrt{3}+i)\Rightarrow z_1\cdot z_2=\sqrt{3}+i-\sqrt{3}i+1\Rightarrow z_1\cdot z_2=(\sqrt{3}+1)-i(\sqrt{3}-1)$
$|z_1\cdot z_2|=\sqrt{(\sqrt{3}+1{)}^2+(\sqrt{3}-1{)}^2}=2\sqrt{2}$

Now, $\tan \alpha =\frac{|\sqrt{3}-1|}{|\sqrt{3}+1|}\Rightarrow \alpha =\frac{\pi }{12}$
$\because z_1\cdot z_2$ lies in fourth quadrant,
$z_1\cdot z_2=2\sqrt{2}\left(\cos \right(-\alpha )+i\sin (-\alpha \left)\right)$
$z_1\cdot z_2=2\sqrt{2}(\cos (-\frac{\pi }{12})+i\sin (-\frac{\pi }{12}))$