Question:
In Argand's plane, the point corresponding to $\frac {(1-i\sqrt {3})(1+i)} {(\sqrt {3}+i)}$ lies in
In Argand's plane, the point corresponding to $\frac {(1-i\sqrt {3})(1+i)} {(\sqrt {3}+i)}$ lies in
Updated On: Feb 21, 2024
- quadrant I
- quadrant II
- quadrant III
- quadrant IV
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The Correct Option is D
Solution and Explanation
Given,
$\frac{(1-i \sqrt{3})(1+i)}{(\sqrt{3}+i)}=\frac{(1-i \sqrt{3}+i+\sqrt{3})(\sqrt{3}-i)}{(3+1)}$
$\left(\because i^{2}=-1\right)$
$=\frac{1}{4} \cdot\{(1+\sqrt{3})+i(1-\sqrt{3})\} \cdot(\sqrt{3}-i)$
$=\frac{1}{4} \cdot\{\sqrt{3}(1+\sqrt{3})+i(1-\sqrt{3}) \sqrt{3} -(1+\sqrt{3}) i+(1-\sqrt{3})\}$
$\left.=\frac{1}{4} \cdot\{\sqrt{3}+ 3+1-\sqrt{3})+(\sqrt{3}-3-1-\sqrt{3}) i\right\}$
$=\frac{1}{4} \cdot\{4-4 i\}=1-i$
The point $(1-i)$ in Arg and plane is $(1,-1)$ which lies in IVth quadrant
$\frac{(1-i \sqrt{3})(1+i)}{(\sqrt{3}+i)}=\frac{(1-i \sqrt{3}+i+\sqrt{3})(\sqrt{3}-i)}{(3+1)}$
$\left(\because i^{2}=-1\right)$
$=\frac{1}{4} \cdot\{(1+\sqrt{3})+i(1-\sqrt{3})\} \cdot(\sqrt{3}-i)$
$=\frac{1}{4} \cdot\{\sqrt{3}(1+\sqrt{3})+i(1-\sqrt{3}) \sqrt{3} -(1+\sqrt{3}) i+(1-\sqrt{3})\}$
$\left.=\frac{1}{4} \cdot\{\sqrt{3}+ 3+1-\sqrt{3})+(\sqrt{3}-3-1-\sqrt{3}) i\right\}$
$=\frac{1}{4} \cdot\{4-4 i\}=1-i$
The point $(1-i)$ in Arg and plane is $(1,-1)$ which lies in IVth quadrant
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Concepts Used:
Complex Numbers and Quadratic Equations
Complex Number: Any number that is formed as a+ib is called a complex number. For example: 9+3i,7+8i are complex numbers. Here i = -1. With this we can say that i² = 1. So, for every equation which does not have a real solution we can use i = -1.
Quadratic equation: A polynomial that has two roots or is of the degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b and c are the real numbers.
