If mu = frac{2z + 5i}{z - 3} and |mu| = 2,the | vedclass

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(A) Given $\mu = \frac{2z + 5i}{z - 3}$ and $|\mu| = 2$.Taking the modulus on both sides: $\left| \frac{2z + 5i}{z - 3} \right| = 2$.This implies $|2z

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Straight line

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(A) Given $\mu = \frac{2z + 5i}{z - 3}$ and $|\mu| = 2$.Taking the modulus on both sides: $\left| \frac{2z + 5i}{z - 3} \right| = 2$.This implies $|2z + 5i| = 2|z - 3|$.Squaring both sides: $|2z + 5i|^2 = 4|z - 3|^2$.Let $z = x + iy$. Then $|2(x + iy) + 5i|^2 = 4|x + iy - 3|^2$.$|2x + i(2y + 5)|^2 = 4|(x - 3) + iy|^2$.$(2x)^2 + (2y + 5)^2 = 4((x - 3)^2 + y^2)$.$4x^2 + 4y^2 + 20y + 25 = 4(x^2 - 6x + 9 + y^2)$.$4x^2 + 4y^2 + 20y + 25 = 4x^2 - 24x + 36 + 4y^2$.$20y + 25 = -24x + 36$.$24x + 20y - 11 = 0$.This is the equation of a straight line.

If $\mu = \frac{2z + 5i}{z - 3}$ and $|\mu| = 2$,then the locus of $z$ is:

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