If $(x-i y)^{1 / 3}=2-i \sqrt{3}$ and the point $z=(x, y)$ lies on the line $\frac{x}{2}+\frac{y}{\sqrt{3}}=k$, then $k=$

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Aicharya · Accepted Answer

Given, $(x-i y)^{1 / 3}=2-i \sqrt{3}$ $\Rightarrow \quad x-i y=(2-i \sqrt{3})^3$ $=8+3 i \sqrt{3}-i 2 i \sqrt{3}-18$ $\begin{aligned} & \Rightarrow \quad x-i y=-10-9 i \sqrt{3} \ & \Rightarrow \quad x=-10 	ext { and } y=9 \sqrt{3}\end{aligned}$ Thus, $z=x+i y=-10+9 i \sqrt{3}$ $\because z$ lies on the line $\frac{x}{2}+\frac{y}{\sqrt{3}}=k$ $\Rightarrow \quad \frac{-10}{2}+\frac{9 \sqrt{3}}{\sqrt{3}}=k \Rightarrow k=4$

If $(x-i y)^{1 / 3}=2-i \sqrt{3}$ and the point $z=(x, y)$ lies on the line…

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