Question

If $z$ is a complex number lying in the first quadrant such that $Re\left(z\right)+Im\left(z\right)=3$, then the maximum values of ${\left[Re\right(z\left)\right]}^{2}Im\left(z\right)$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is D $4$

z lies in the first quadrant
$\therefore$ if $z=x+iy$ then $x>0,y>0$
$Re\left(z\right)+Im\left(z\right)=3$
$x+y=3⟶1$
We have to maximize ${\left[Re\right(z\left)\right]}^{2}Im\left(z\right)=x^{2}y$
Let $S=x^{2}y$
$S=x^2(3-x)$ (From equation 1)
$S=3x^2-x^3$
$\frac{dS}{dx}=6x-3x^2=0$
$3x(2-x)=0$
$x=0,x=2$
$\frac{d^{2}S}{d{x}^2}=\frac{d}{dx}(6x-3x^2)=6-6x$
${\left[\frac{d^{2}S}{{dx}^2}\right]}_{x=0}=6-0=6>0$ $\therefore$ At $x=0$, S is minimum
${\left[\frac{d^{2}S}{{dx}^2}\right]}_{x=2}=6-12=-6<0$ $\therefore$ At $x=2$, S is maximum
$\therefore {\left[Re\right(z\left)\right]}^{2}Im\left(z\right)=x^{2}y$ is maximum when $x=2$
If $x=2$ & $y=1$ (from equation 1)
$\therefore x^{2}y={\left(2\right)}^2.1=4$