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If $\mathrm{w}=\frac{\mathrm{z}}{\mathrm{z}-\frac{1}{3} \mathrm{i}}$ and $|\mathrm{w}|=1, \mathrm{i}=\sqrt{-1}$, then $\mathrm{z}$ lies on
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Verified Answer
The correct answer is:
line.
$\begin{aligned}
& w=\frac{z}{z-\frac{1}{3} i} \\
& \Rightarrow w=\frac{3 z}{3 z-i}
\end{aligned}$
Applying mod on both sides, we get
$\begin{aligned}
& |w|=\frac{3|z|}{|3 z-i|} \\
& \Rightarrow 3|z|=|3 z-i| \quad \ldots[|w|=1]
\end{aligned}$
Consider $\mathrm{z}=\mathrm{a}+\mathrm{ib}$
$\begin{aligned}
& \Rightarrow 3|a+i b|=|3 a+3 i b-i| \\
& \Rightarrow 3|a+i b|=|3 a+(3 b-1) i| \\
& \Rightarrow 3\left(\sqrt{a^2+b^2}\right)=\left(\sqrt{9 a^2+(3 b-1)^2}\right) \\
& \Rightarrow 9 a^2+9 b^2=9 a^2+9 b^2-6 b+1 \\
& \Rightarrow 6 b-1=0
\end{aligned}$
$\therefore \quad$ The above equation represents a straight line.
& w=\frac{z}{z-\frac{1}{3} i} \\
& \Rightarrow w=\frac{3 z}{3 z-i}
\end{aligned}$
Applying mod on both sides, we get
$\begin{aligned}
& |w|=\frac{3|z|}{|3 z-i|} \\
& \Rightarrow 3|z|=|3 z-i| \quad \ldots[|w|=1]
\end{aligned}$
Consider $\mathrm{z}=\mathrm{a}+\mathrm{ib}$
$\begin{aligned}
& \Rightarrow 3|a+i b|=|3 a+3 i b-i| \\
& \Rightarrow 3|a+i b|=|3 a+(3 b-1) i| \\
& \Rightarrow 3\left(\sqrt{a^2+b^2}\right)=\left(\sqrt{9 a^2+(3 b-1)^2}\right) \\
& \Rightarrow 9 a^2+9 b^2=9 a^2+9 b^2-6 b+1 \\
& \Rightarrow 6 b-1=0
\end{aligned}$
$\therefore \quad$ The above equation represents a straight line.
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