If \omega = \frac{z}{z-\left ( 1/3 \right )i} and \left | \omega \right |= 1,then z lies on

Answers (1)
S Sayak

\omega = \frac{z}{z-\left ( 1/3 \right )i}

\left | \omega \right |= 1

Therefore

\\\left | \frac{z}{z-(1/3)i} \right |=1\\ |z|=|z-(1/3)i|

Let z = a + ib

\\|z|=|z-(1/3)i|\\ \sqrt{a^{2}+b^{2}}=\sqrt{a^{2}+(b-1/3)^{2}}\\ a^{2}+b^{2}=a^{2}+b^{2}-\frac{2b}{3}+\frac{1}{9}\\ b=\frac{1}{6}

a\epsilon \mathbb{R}

z lies on the liney=\frac{1}{6}

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