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Mathematics Question on Complex Numbers and Quadratic Equations

In Argand's plane, the point corresponding to (1−i3)(1+i)(3+i)\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

quadrant IV

Explanation

Solution

Given,
(1−i3)(1+i)(3+i)=(1−i3+i+3)(3−i)(3+1)\frac{(1-i \sqrt{3})(1+i)}{(\sqrt{3}+i)}=\frac{(1-i \sqrt{3}+i+\sqrt{3})(\sqrt{3}-i)}{(3+1)}
(∵i2=−1)\left(\because i^{2}=-1\right)
=14⋅(1+3)+i(1−3)⋅(3−i)=\frac{1}{4} \cdot\\{(1+\sqrt{3})+i(1-\sqrt{3})\\} \cdot(\sqrt{3}-i)
=14⋅3(1+3)+i(1−3)3−(1+3)i+(1−3)=\frac{1}{4} \cdot\\{\sqrt{3}(1+\sqrt{3})+i(1-\sqrt{3}) \sqrt{3} -(1+\sqrt{3}) i+(1-\sqrt{3})\\}
\left.=\frac{1}{4} \cdot\\{\sqrt{3}+ 3+1-\sqrt{3})+(\sqrt{3}-3-1-\sqrt{3}) i\right\\}
=14⋅4−4i=1−i=\frac{1}{4} \cdot\\{4-4 i\\}=1-i
The point (1−i)(1-i) in Arg and plane is (1,−1)(1,-1) which lies in IVth quadrant