If z is a complex number lying in the first quadrant such that `"Re"(z)+"Im"(z)=3`, then the maximum value of `"{Re"(z)}^(2)"Im"(z)`, is

Options

A. 1
B. 2
C. 3
D. 4

Answer

D

Explanation

Let `z=a+ib`, where `a gt 0, b gt 0`, Then,
`"Re"(z)+"Im"(z)=3 rArr a+b=3`
`P=a^(2)(3-a)=3a^(2)-a^(3))`
`therefore (dp)/(da) = 6a-3a^(2)` and `(d^(2)P)/(da^(2))=6-6a`
For maximum or minimum value of P, we must have `(dP)/(da)=0 rArr a=0` and a=2.
Clearly, `(d^(2)P)/(da^(2))=6-12 lt 0` for a=2
For a=2, P=4(3-2)=4.