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The area of the polygon, whose vertices are the non-real roots of the equation $\bar{z}=i z^{2} \: is :$
Option: 1
$\frac{3 \sqrt{3}}{4}$

Option: 2
$\frac{3 \sqrt{3}}{2}$

Option: 3
$\frac{3}{2}$

Option: 4
$\frac{3}{4}$

Answers (1)

best_answer

$\mathrm{\text { Let } z =x+i y} \\$

$\mathrm{\text { so } \bar{z} =x-i y} \\$

$\mathrm{\bar{z}=i z^{2} \Rightarrow x-i y=i\left(x^{2}-y^{2}+2 x y i\right)} \\$

$\mathrm{\Rightarrow x-i y=-2 x y+\left(x^{2}-y^{2}\right) i}$

Comparing: $\mathrm{ x=-2 x y \quad \& \quad-y=x^{2}-y^{2} }\\$

$\mathrm{\Rightarrow x(1+2 y)=0 \quad \text { \& } x^{2}-y^{2}+y=0 }$

So either $\mathrm{x=0 \quad \text { or } \quad y=-1 / 2 }$

$\mathrm{If \: x=0 ;-y^{2}+y=0 \Rightarrow y=0 \: or \: 1 \Rightarrow(0,0),(0,1)}$

$\mathrm{\text { If } y=-1 / 2, x^{2}=\frac{1}{4}+\frac{1}{2}=\frac{3}{4} \Rightarrow x=\pm \frac{\sqrt{3}}{2}} \\$

$\mathrm{\Rightarrow\left(\frac{\sqrt{3}}{2},-\frac{1}{2}\right),\left(-\frac{\sqrt{3}}{2},-\frac{1}{2}\right)}$

So the non-real solutions are $\mathrm{z=i, \frac{\sqrt{3}}{2}-\frac{i}{2},-\frac{\sqrt{3}}{2}-\frac{i}{2}}$

It is a triangle whose area $\mathrm{=\frac{1}{2} \times \frac{2 \sqrt{3}}{2} \times \frac{3}{2}=\frac{3 \sqrt{3}}{4}}$

Hence the correct answer is option 1

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Nehul

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