If z_1 and z_2 are two distinct complex numbe | vedclass

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(A) Given $\left|\frac{z_1-2 z_2}{\frac{1}{2}-z_1 \bar{z}_2}\right|=2$. Squaring both sides,we get $\left|\frac{z_1-2 z_2}{\frac{1}{2}-z_1 \bar{z}_2}\

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either $z_1$ lies on a circle of radius $1$ or $z_2$ lies on a circle of radius $\frac{1}{2}$.

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(A) Given $\left|\frac{z_1-2 z_2}{\frac{1}{2}-z_1 \bar{z}_2}\right|=2$. Squaring both sides,we get $\left|\frac{z_1-2 z_2}{\frac{1}{2}-z_1 \bar{z}_2}\right|^2=4$. Using the property $|z|^2 = z \bar{z}$,we have $\frac{(z_1-2 z_2)(\bar{z}_1-2 \bar{z}_2)}{(\frac{1}{2}-z_1 \bar{z}_2)(\frac{1}{2}-\bar{z}_1 z_2)}=4$. Expanding the numerator: $z_1 \bar{z}_1 - 2 z_1 \bar{z}_2 - 2 z_2 \bar{z}_1 + 4 z_2 \bar{z}_2 = |z_1|^2 - 2 z_1 \bar{z}_2 - 2 \bar{z}_1 z_2 + 4 |z_2|^2$. Expanding the denominator: $\frac{1}{4} - \frac{1}{2} \bar{z}_1 z_2 - \frac{1}{2} z_1 \bar{z}_2 + |z_1|^2 |z_2|^2$. So,$|z_1|^2 - 2 z_1 \bar{z}_2 - 2 \bar{z}_1 z_2 + 4 |z_2|^2 = 4 (\frac{1}{4} - \frac{1}{2} \bar{z}_1 z_2 - \frac{1}{2} z_1 \bar{z}_2 + |z_1|^2 |z_2|^2)$. $|z_1|^2 - 2 z_1 \bar{z}_2 - 2 \bar{z}_1 z_2 + 4 |z_2|^2 = 1 - 2 \bar{z}_1 z_2 - 2 z_1 \bar{z}_2 + 4 |z_1|^2 |z_2|^2$. Canceling $-2 z_1 \bar{z}_2 - 2 \bar{z}_1 z_2$ from both sides: $|z_1|^2 + 4 |z_2|^2 = 1 + 4 |z_1|^2 |z_2|^2$. Rearranging: $|z_1|^2 - 1 - 4 |z_2|^2 + 4 |z_1|^2 |z_2|^2 = 0$. $(|z_1|^2 - 1) - 4 |z_2|^2 (1 - |z_1|^2) = 0$. $(|z_1|^2 - 1) + 4 |z_2|^2 (|z_1|^2 - 1) = 0$. $(|z_1|^2 - 1)(1 + 4 |z_2|^2) = 0$. Since $1 + 4 |z_2|^2$ cannot be $0$,we must have $|z_1|^2 - 1 = 0$,which means $|z_1| = 1$. Wait,checking the original equation again,if we write $4|z_2|^2$ as $|2z_2|^2$,the expression factors as $(|z_1|^2 - 1)(1 - |2z_2|^2) = 0$. Thus,$|z_1| = 1$ or $|2z_2| = 1$,which means $|z_1| = 1$ or $|z_2| = \frac{1}{2}$.

If $z_1$ and $z_2$ are two distinct complex numbers such that $\left|\frac{z_1-2 z_2}{\frac{1}{2}-z_1 \bar{z}_2}\right|=2$,then:

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