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If $z, \bar{z},-z,-\bar{z}$ forms a rectangle of area $2 \sqrt{3}$ square units, then one such $z$ is
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1890 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{2}+\sqrt{3} i$
Let $z=x+i y$
Then, vertices of rectangle are $(x, y),(x,-y)$, $(-x,-y),(-x, y)$.
Now, area of rectangle $=(2 x)(2 y)=4 x y$
It is given that,
$\begin{aligned}
& 4 x y=2 \sqrt{3} \Rightarrow 2 x y=\sqrt{3} \\
& \therefore \quad x=\frac{1}{2}, y=\sqrt{3} \quad \therefore \quad z=\frac{1}{2}+\sqrt{3} i
\end{aligned}$
Then, vertices of rectangle are $(x, y),(x,-y)$, $(-x,-y),(-x, y)$.
Now, area of rectangle $=(2 x)(2 y)=4 x y$
It is given that,
$\begin{aligned}
& 4 x y=2 \sqrt{3} \Rightarrow 2 x y=\sqrt{3} \\
& \therefore \quad x=\frac{1}{2}, y=\sqrt{3} \quad \therefore \quad z=\frac{1}{2}+\sqrt{3} i
\end{aligned}$
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