MHT CET · Maths · Complex Number
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
- A circle.
- B line.
- C parabola.
- D ellipse.
Answer & Solution
Correct Answer
(B) line.
Step-by-step Solution
Detailed explanation
\(w=\frac{z}{z-\frac{1}{3} i}\)
\(\Rightarrow w=\frac{3 z}{3 z-i}\)
Applying mod on both sides, we get
\(|w|=\frac{3|z|}{|3 z-i|}\)
\(\Rightarrow 3|z|=|3 z-i| \quad \ldots[|w|=1]\)
Consider \(\mathrm{z}=\mathrm{a}+\mathrm{ib}\)
\(\Rightarrow 3|a+i b|=|3 a+3 i b-i| \)
\( \Rightarrow 3|a+i b|=|3 a+(3 b-1) i| \)
\( \Rightarrow 3\left(\sqrt{a^2+b^2}\right)=\left(\sqrt{9 a^2+(3 b-1)^2}\right) \)
\( \Rightarrow 9 a^2+9 b^2=9 a^2+9 b^2-6 b+1 \)
\( \Rightarrow 6 b-1=0\)
\(\therefore\) The above equation represents a straight line.
\(\Rightarrow w=\frac{3 z}{3 z-i}\)
Applying mod on both sides, we get
\(|w|=\frac{3|z|}{|3 z-i|}\)
\(\Rightarrow 3|z|=|3 z-i| \quad \ldots[|w|=1]\)
Consider \(\mathrm{z}=\mathrm{a}+\mathrm{ib}\)
\(\Rightarrow 3|a+i b|=|3 a+3 i b-i| \)
\( \Rightarrow 3|a+i b|=|3 a+(3 b-1) i| \)
\( \Rightarrow 3\left(\sqrt{a^2+b^2}\right)=\left(\sqrt{9 a^2+(3 b-1)^2}\right) \)
\( \Rightarrow 9 a^2+9 b^2=9 a^2+9 b^2-6 b+1 \)
\( \Rightarrow 6 b-1=0\)
\(\therefore\) The above equation represents a straight line.
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