Let $A (z_{1})$ and $B (z_{2})$ be two points lying on the curve $z - 3 - 4 i = \frac{25}{¯ ¯ ¯ z - 3 + 4 i}$ where $| z_{1} |$ is maximum. Now, $A (z_{1})$ is rotated about the origin in anti-clockwise direction through $90^{\circ}$ reaching at $P (z_{0}) .$ If $A, B$ and $P$ are collinear, then the value of $| z_{0} - z_{1} | | z_{0} - z_{2} |$ is

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Solution

$z - 3 - 4 i = \frac{25}{¯ ¯ ¯ z - 3 + 4 i}$
$\Rightarrow (z - (3 + 4 i)) ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ (z - (3 + 4 i)) = 25 \Rightarrow | z - (3 + 4 i) | = 5 (∵ z ¯ ¯ ¯ z = | z |^{2})$
This represents a circle with centre at $(3, 4)$ and radius $5.$
Since, $| z_{1} |$ is maximum, $A$ occurs along the diameter.
$∴ z_{1} = 6 + 8 i$

Now, $z_{0} = (6 + 8 i) i = - 8 + 6 i$
$P A \cdot P B = O P^{2} \Rightarrow | z_{0} - z_{1} | | z_{0} - z_{2} | = 100$

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Let A(z1) and B(z2) be two points lying on the curve z−3−4i=25¯¯¯z−3+4i where |z1| is maximum. Now, A(z1) is rotated about the origin in anti-clockwise direction through 90∘ reaching at P(z0). If A,B and P are collinear, then the value of |z0−z1||z0−z2| is