KCET · Maths · Complex Number
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 & Solution
Correct Answer
(D) quadrant IV
Step-by-step Solution
Detailed explanation
Given,
\[
\begin{aligned}
&\frac{(1-i \sqrt{3})(1+i)}{(\sqrt{3}+i)}=\frac{(1-i \sqrt{3}+i+\sqrt{3})(\sqrt{3}-i)}{(3+1)} \\
&=\frac{1}{4} \cdot\{(1+\sqrt{3})+i(1-\sqrt{3})\} \cdot(\sqrt{3}-i) \\
&=\frac{1}{4} \cdot\{\sqrt{3}(1+\sqrt{3})+i(1-\sqrt{3}) \sqrt{3} \\
&\left.=\frac{1}{4} \cdot\{\sqrt{3}+3+1-\sqrt{3})+(\sqrt{3}-3-1-\sqrt{3}) i\right\} \\
&=\frac{1}{4} \cdot\{4-4 i\}=1-i
\end{aligned}
\]
The point \((1-i)\) in Arg and plane is \((1,-1)\) which lies in IVth quadrant.
\[
\begin{aligned}
&\frac{(1-i \sqrt{3})(1+i)}{(\sqrt{3}+i)}=\frac{(1-i \sqrt{3}+i+\sqrt{3})(\sqrt{3}-i)}{(3+1)} \\
&=\frac{1}{4} \cdot\{(1+\sqrt{3})+i(1-\sqrt{3})\} \cdot(\sqrt{3}-i) \\
&=\frac{1}{4} \cdot\{\sqrt{3}(1+\sqrt{3})+i(1-\sqrt{3}) \sqrt{3} \\
&\left.=\frac{1}{4} \cdot\{\sqrt{3}+3+1-\sqrt{3})+(\sqrt{3}-3-1-\sqrt{3}) i\right\} \\
&=\frac{1}{4} \cdot\{4-4 i\}=1-i
\end{aligned}
\]
The point \((1-i)\) in Arg and plane is \((1,-1)\) which lies in IVth quadrant.
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