Question

In Argand's plane, the point corresponding to $\frac{(1-i\sqrt{3})(1+i)}{(\sqrt{3}+i)}$ lies in

• A. quadrant I
• B. quadrant II
• C. quadrant III
• D. quadrant IV

Answer 1

Answer: (D)

Given,
$\frac{(1-i\sqrt{3})(1+i)}{(\sqrt{3}+i)}=\frac{(1-i\sqrt{3}+i+\sqrt{3})(\sqrt{3}-i)}{(3+1)}$
$\left(\because i^2=-1\right)$
$=\frac{1}{4}\cdot \{(1+\sqrt{3})+i(1-\sqrt{3})\}\cdot (\sqrt{3}-i)$
$=\frac{1}{4}\cdot \{\sqrt{3}(1+\sqrt{3})+i(1-\sqrt{3})\sqrt{3}-(1+\sqrt{3})i+(1-\sqrt{3})\}$
$=\frac{1}{4}\cdot \{\sqrt{3}+3+1-\sqrt{3})+(\sqrt{3}-3-1-\sqrt{3})i\}$
$=\frac{1}{4}\cdot \{4-4i\}=1-i$
The point $(1-i)$ in Arg and plane is $(1,-1)$ which lies in IVth quadrant