Let the set of all values of k R such that the equation z(z+2+i)+k(2+3i)=0,z C, has at lea…

Question

Let the set of all values of $k\in R$ such that the equation $z(\vec{z}+2+i)+k(2+3i)=0,z\in C$, has at least one solution, be the interval $[\alpha ,\beta ]$. Then $9(\alpha +\beta )$ is equal to:

Accepted Answer

Let

$z=x+iy\, x,y\in R.$

Then

$\vec{z}=x-iy.$

The given equation is

$z(\vec{z}+2+i)+k(2+3i)=0.$

We have to find all real values of $k$ for which this equation has at least one complex solution $z$.

First, simplify $z(\vec{z}+2+i)$:

$z(\vec{z}+2+i)=z\vec{z}+z(2+i).$

Now,

$z\vec{z}=|z{|}^{2}={x}^{2}+{y}^{2}$

and

$z(2+i)=(x+iy)(2+i).$

Let us multiply:

$(x+iy)(2+i)=2x+ix+2iy+{i}^{2}y$

Since ${i}^{2}=-1$,

$(x+iy)(2+i)=(2x-y)+i(x+2y).$

So,

$z(\vec{z}+2+i)={x}^{2}+{y}^{2}+(2x-y)+i(x+2y).$

Hence the given equation becomes

$({x}^{2}+{y}^{2}+2x-y)+i(x+2y)+k(2+3i)=0.$

Now separate real and imaginary parts:

$({x}^{2}+{y}^{2}+2x-y+2k)+i(x+2y+3k)=0.$

Therefore,

${x}^{2}+{y}^{2}+2x-y+2k=0\, \, x+2y+3k=0.$

From the second equation,

$3k=-(x+2y)$

so

$k=-\frac{x+2y}{3}.$

Substitute this into the first equation:

${x}^{2}+{y}^{2}+2x-y+2(-\frac{x+2y}{3})=0.$

Multiply by $3$:

$3{x}^{2}+3{y}^{2}+6x-3y-2x-4y=0$

$3{x}^{2}+3{y}^{2}+4x-7y=0.$

Now complete squares:

$3({x}^{2}+\frac{4}{3}x)+3({y}^{2}-\frac{7}{3}y)=0.$

$3[{(x+\frac{2}{3})}^{2}-\frac{4}{9}]+3[{(y-\frac{7}{6})}^{2}-\frac{49}{36}]=0.$

$3{(x+\frac{2}{3})}^{2}-\frac{4}{3}+3{(y-\frac{7}{6})}^{2}-\frac{49}{12}=0.$

$3{(x+\frac{2}{3})}^{2}+3{(y-\frac{7}{6})}^{2}=\frac{4}{3}+\frac{49}{12}.$

$3{(x+\frac{2}{3})}^{2}+3{(y-\frac{7}{6})}^{2}=\frac{16+49}{12}=\frac{65}{12}.$

So,

${(x+\frac{2}{3})}^{2}+{(y-\frac{7}{6})}^{2}=\frac{65}{36}.$

This is a circle with centre

$(-\frac{2}{3},\frac{7}{6})$

and radius

$\frac{\sqrt{65}}{6}.$

Now,

$k=-\frac{x+2y}{3}.$

So we need the range of the linear expression $x+2y$ on this circle.

Let

$S=x+2y.$

For a circle, maximum and minimum of a linear expression occur at the extreme points in the direction of its normal vector.

At the centre,

${S}_{c}=-\frac{2}{3}+2\cdot \frac{7}{6}=-\frac{2}{3}+\frac{7}{3}=\frac{5}{3}.$

The vector corresponding to $x+2y$ is $(1,2)$, whose magnitude is

$\sqrt{{1}^{2}+{2}^{2}}=\sqrt{5}.$

Hence the maximum variation from the centre is

$	imes \sqrt{5}=\frac{\sqrt{65}}{6}\cdot \sqrt{5}=\frac{\sqrt{325}}{6}=\frac{5\sqrt{13}}{6}.$

Therefore,

${S}_{min}=\frac{5}{3}-\frac{5\sqrt{13}}{6},\, {S}_{max}=\frac{5}{3}+\frac{5\sqrt{13}}{6}.$

Since

$k=-\frac{S}{3},$

the interval for $k$ is

$[-\frac{{S}_{max}}{3},-\frac{{S}_{min}}{3}].$

So if this interval is $[\alpha ,\beta ]$, then

$\alpha +\beta =-\frac{{S}_{max}+{S}_{min}}{3}=-\frac{2\cdot \frac{5}{3}}{3}=-\frac{10}{9}.$

Hence,

$9(\alpha +\beta )=9(-\frac{10}{9})=-10.$

Therefore, the correct answer is

$-10.$

So, Option A is correct.

Solutions · 1