Let z be the complex number satisfying |z-5 | 3 and having maximum positive principal argument. Then 34 ( 5z-125 iz+16 )^2 is equal to:

Question

Accepted Answer

Correct answer: C. (c) The region |z-5 | 3 is a disk centered at (5 ,0 ) with radius 3. Maximum positive principal argument from origin occurs at the tangent point from origin to the circle. sinθ= 35 , cosθ= 45. The tangent point is the foot of perpendicular from (5 ,0 ) to the line 3x-4y=0, giving z= 165+ 125i. 5z-12=4+12i ⇒ |5z-12 |^2=160 5 iz+16=4+16i ⇒ |5 iz+16 |^2=272 34 ( 5z-125 iz+16 )^2=34 160272=34 1017=20.

$Let z be the complex number satisfying | z - 5 | \leq 3 and having maximum positive principal argument. Then 34 (\frac{5 z - 12}{5 iz + 16})^{2} is equal to:$

Solution

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