If z = x + iy and omega = frac{1 - iz}{z - i} | vedclass

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(B) Given $\omega = \frac{1 - iz}{z - i}$ and $|\omega| = 1$. $|\frac{1 - iz}{z - i}| = 1$ $|1 - iz| = |z - i|$ Substitute $z = x + iy$: $|1 - i(x + i

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$z$ lies on the real axis

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(B) Given $\omega = \frac{1 - iz}{z - i}$ and $|\omega| = 1$. $|\frac{1 - iz}{z - i}| = 1$ $|1 - iz| = |z - i|$ Substitute $z = x + iy$: $|1 - i(x + iy)| = |x + iy - i|$ $|1 - ix + y| = |x + i(y - 1)|$ $|(1 + y) - ix| = |x + i(y - 1)|$ Squaring both sides: $(1 + y)^2 + (-x)^2 = x^2 + (y - 1)^2$ $1 + y^2 + 2y + x^2 = x^2 + y^2 + 1 - 2y$ $2y = -2y$ $4y = 0 \implies y = 0$. Since $z = x + iy$ and $y = 0$,$z = x$,which means $z$ lies on the real axis.

If $z = x + iy$ and $\omega = \frac{1 - iz}{z - i}$,then $|\omega| = 1$ shows that in the complex plane:

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