Q.

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]

1 16  
2 12  
3 20  
4 26  

Ans.

(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