Let z be the complex number satisfying |z − 5| ≤ 3 and having maximum positive principal argument.
Then 34|5z-125iz+16|2 is equal to: [2026]
(3)
|z-5|≤3
For arg(z) to be maximum, z lies at P.
z≡(4cosθ, 4sinθ)
≡(4·45, 4·35)=(165, 125)=165+12i5
Now, 34|5z-125iz+16|2=34|(16+12i)-12(16i-12)+16|2
=34|4+12i16i+4|2
=34(16+144256+16)
=34(160272)=20