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What is the square root of $\frac{1}{2}-\mathrm{i} \frac{\sqrt{3}}{2}$ ?
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Verified Answer
The correct answer is:
$\pm\left(\frac{\sqrt{3}}{2}+\frac{\mathrm{i}}{2}\right)$
Let $\sqrt{\frac{1}{2}-\frac{\mathrm{i} \sqrt{3}}{2}}=\mathrm{x}-\mathrm{iy}$
Squaring both the sides,
$\Rightarrow \frac{1}{2}-\frac{i \sqrt{3}}{2}=(x-i y)^{2} \Rightarrow \frac{1}{2}-\frac{i \sqrt{3}}{2}=x^{2}-y^{2}-2 i x y$
Comparing real and imaginary parts, we get
$\mathrm{x}^{2}-\mathrm{y}^{2}=\frac{1}{2}$
and $2 x y=\frac{\sqrt{3}}{2}$
and $\left(x^{2}+y^{2}\right)^{2}=\left(x^{2}-y^{2}\right)^{2}+4 x^{2} y^{2}=\frac{1}{4}+\frac{3}{4}=1$
$\Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}=1$
On solving eqs. (1) and (3) we get
$\mathrm{x}^{2}=\frac{3}{4}$ and $\mathrm{y}^{2}=\frac{1}{4}$
$\Rightarrow x=\pm \frac{\sqrt{3}}{2}, y=\pm \frac{1}{2}$
$\therefore \sqrt{\frac{1}{2}-\frac{\mathrm{i} \sqrt{3}}{2}}=\pm\left(\frac{\sqrt{3}}{2}+\frac{\mathrm{i}}{2}\right)$
Squaring both the sides,
$\Rightarrow \frac{1}{2}-\frac{i \sqrt{3}}{2}=(x-i y)^{2} \Rightarrow \frac{1}{2}-\frac{i \sqrt{3}}{2}=x^{2}-y^{2}-2 i x y$
Comparing real and imaginary parts, we get
$\mathrm{x}^{2}-\mathrm{y}^{2}=\frac{1}{2}$
and $2 x y=\frac{\sqrt{3}}{2}$
and $\left(x^{2}+y^{2}\right)^{2}=\left(x^{2}-y^{2}\right)^{2}+4 x^{2} y^{2}=\frac{1}{4}+\frac{3}{4}=1$
$\Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}=1$
On solving eqs. (1) and (3) we get
$\mathrm{x}^{2}=\frac{3}{4}$ and $\mathrm{y}^{2}=\frac{1}{4}$
$\Rightarrow x=\pm \frac{\sqrt{3}}{2}, y=\pm \frac{1}{2}$
$\therefore \sqrt{\frac{1}{2}-\frac{\mathrm{i} \sqrt{3}}{2}}=\pm\left(\frac{\sqrt{3}}{2}+\frac{\mathrm{i}}{2}\right)$
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