If z_1, z_2 are two complex numbers satisfyin | vedclass

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(A) Given that $\left|\frac{z_1-3 z_2}{3-z_1 \bar{z}_2}ight|=1$ and $\left|z_1ight| 
eq 3$. Squaring both sides,we get $\left|z_1-3 z_2ight|^2

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(A) Given that $\left|\frac{z_1-3 z_2}{3-z_1 \bar{z}_2}ight|=1$ and $\left|z_1ight| 
eq 3$. Squaring both sides,we get $\left|z_1-3 z_2ight|^2 = \left|3-z_1 \bar{z}_2ight|^2$. Using the property $|z|^2 = z \bar{z}$,we have $(z_1-3 z_2)(\bar{z}_1-3 \bar{z}_2) = (3-z_1 \bar{z}_2)(3-\bar{z}_1 z_2)$. Expanding both sides: $|z_1|^2 - 3z_1 \bar{z}_2 - 3z_2 \bar{z}_1 + 9|z_2|^2 = 9 - 3\bar{z}_1 z_2 - 3z_1 \bar{z}_2 + |z_1|^2 |z_2|^2$. Canceling common terms $-3z_1 \bar{z}_2$ and $-3z_2 \bar{z}_1$ from both sides: $|z_1|^2 + 9|z_2|^2 = 9 + |z_1|^2 |z_2|^2$. Rearranging the terms: $|z_1|^2 - 9 - |z_1|^2 |z_2|^2 + 9|z_2|^2 = 0$. Factorizing: $(|z_1|^2 - 9) - |z_2|^2 (|z_1|^2 - 9) = 0$. $(|z_1|^2 - 9)(1 - |z_2|^2) = 0$. Since $|z_1| 
eq 3$,we must have $|z_2|^2 = 1$,which implies $|z_2| = 1$.

If $z_1, z_2$ are two complex numbers satisfying $\left|\frac{z_1-3 z_2}{3-z_1 \bar{z}_2}\right|=1$ and $\left|z_1\right| \neq 3$,then $\left|z_2\right|$ is equal to

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