Let P=\z \in \mathbbC:|z+2-3 i| \leq 1\ and Q=\z \in \mathbbC: z(1+i)+\barz(1-i) \leq-8\ . Let in P \cap Q , |z-3+2 i| be maximum and minimum at z_1 and z_2 respectively. If |z_1|^2+2|z_2|^2=\alpha+\beta \sqrt2 , where \alpha, \beta are integers, then \alpha+\beta equals \underline\hspace2cm .

Complex Numbers PYQ — JEE Main 2024 · 1 Feb · Shift 1 · Q57 with Solution

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Correct answer: 36

Interpret the sets geometrically

Let $z=x+iy$ .

Set $P$

Given $|z+2-3i|\le 1$ we write $z+2-3i=(x+2)+i(y-3).$ So, $(x+2)^2+(y-3)^2\le 1.$ This is the closed disk centered at $C_1=(-2,3)$ with radius $1$ .

Set $Q$

Given $z(1+i)+\bar z(1-i)\le -8.$ Now, $z=x+iy,\quad \bar z=x-iy.$ Then $z(1+i)=(x+iy)(1+i)=(x-y)+i(x+y),$ $\bar z(1-i)=(x-iy)(1-i)=(x-y)-i(x+y).$ Adding, $z(1+i)+\bar z(1-i)=2(x-y).$ Hence the condition becomes $2(x-y)\le -8 \implies x-y\le -4 \implies y-x\ge 4.$ So $Q$ is the half-plane $y-x\ge 4.$

Find $P\cap Q$

The circle of $P$ has center $(-2,3)$ . Check its position relative to the line $y-x=4$ : $3-(-2)=5>4,$ so the center lies in the half-plane $Q$ .

Distance from center $(-2,3)$ to the line $y-x-4=0$ is $\frac{|(-1)(-2)+(1)(3)-4|}{\sqrt{(-1)^2+1^2}}=\frac{|2+3-4|}{\sqrt2}=\frac{1}{\sqrt2}<1.$ So the line cuts the disk, and $P\cap Q$ is the part of the disk on the side containing the center.

Quantity to maximize/minimize

We need extrema of $|z-3+2i|=|(x-3)+i(y+2)|.$ This is the distance from point $(x,y)$ to $A=(3,-2).$ Thus, in the region $P\cap Q$ , we seek the farthest and nearest points from $A$ .

Useful geometry: line of constraint cuts the circle at two points

The boundary line is $y-x=4.$ Its intersection with the circle $(x+2)^2+(y-3)^2=1$ will be important.

Put $y=x+4$ into the circle: $(x+2)^2+(x+1)^2=1$ $x^2+4x+4+x^2+2x+1=1$ $2x^2+6x+4=0$ $x^2+3x+2=0$ $(x+1)(x+2)=0.$ So the intersection points are:

for $x=-1$ , $y=3$ giving $(-1,3)$ ,
for $x=-2$ , $y=2$ giving $(-2,2)$ .

Minimum distance from $A=(3,-2)$ to $P\cap Q$

Since $A$ lies outside the disk and the nearest point on the full disk would be on the line joining center to $A$ , we must check whether that point lies in $Q$ .

Distance from $A$ to center $C_1=(-2,3)$ is $AC_1=\sqrt{(3+2)^2+(-2-3)^2}=\sqrt{25+25}=5\sqrt2.$ The nearest point on the full disk would be along direction from center to $A$ , i.e. toward $(1,-1)$ , which decreases $y-x$ , so it lies outside $Q$ . Hence the constrained minimum occurs on the boundary segment of the line $y-x=4$ .

So compare distances from $A$ to the two boundary endpoints:

For $(-1,3)$ : $d^2=( -1-3)^2+(3+2)^2=16+25=41.$

For $(-2,2)$ : $d^2=( -2-3)^2+(2+2)^2=25+16=41.$ Both are equal. In fact the segment lies on the perpendicular bisector setup here, so the minimum is attained at both endpoints. Take either as $z_2$ .

Thus $|z_2|^2 = (-1)^2+3^2=10$ if $z_2=-1+3i$ , or $|z_2|^2=(-2)^2+2^2=8$ if $z_2=-2+2i$ .

But we must identify where the minimum of $|z-3+2i|$ over the feasible set occurs more carefully.

Let us parametrize the boundary segment between $(-2,2)$ and $(-1,3)$ :

\quad (x+2)^2+(x+1)^2\le 1.$$ Distance squared from $A$ is $$f(x)=(x-3)^2+(x+6)^2=2x^2+6x+45.$$ On $x\in[-2,-1]$, $$f'(x)=4x+6.$$ This vanishes at $$x=-\frac32,$$ which lies in the interval. Then $$y=\frac52.$$ Check this point is in the disk: $$\left(-\frac32+2\right)^2+\left(\frac52-3\right)^2=\left(\frac12\right)^2+\left(-\frac12\right)^2=\frac12<1.$$ So this point belongs to $P\cap Q$. Thus the minimum occurs at $$z_2=-\frac32+\frac52 i.$$ Now $$|z_2|^2=\left(-\frac32\right)^2+\left(\frac52\right)^2=\frac94+\frac{25}4=\frac{34}{4}=\frac{17}{2}.$$ --- 6. **Maximum distance from $A$ to $P\cap Q$** The farthest point on the full disk from $A$ lies on the line through center away from $A$. Vector from $A$ to center is $$C_1-A=(-5,5)=5(-1,1).$$ Unit vector in this direction is $$\frac{1}{\sqrt2}(-1,1).$$ So the farthest point on the circle is $$z_1: (-2,3)+1\cdot \frac{1}{\sqrt2}(-1,1)=\left(-2-\frac1{\sqrt2},\ 3+\frac1{\sqrt2}\right).$$ Check it satisfies $Q$: $$y-x = 3+\frac1{\sqrt2}-\left(-2-\frac1{\sqrt2}\right)=5+\sqrt2>4,$$ so it is feasible. Hence this is the maximizing point. Now $$|z_1|^2=\left(-2-\frac1{\sqrt2}\right)^2+\left(3+\frac1{\sqrt2}\right)^2.$$ Compute: $$\left(-2-\frac1{\sqrt2}\right)^2=4+2\sqrt2+\frac12=\frac92+2\sqrt2,$$ $$\left(3+\frac1{\sqrt2}\right)^2=9+3\sqrt2+\frac12=\frac{19}{2}+3\sqrt2.$$ Thus $$|z_1|^2=14+5\sqrt2.$$ --- 7. **Evaluate the required expression** $$|z_1|^2+2|z_2|^2=(14+5\sqrt2)+2\cdot \frac{17}{2}=14+5\sqrt2+17=31+5\sqrt2.$$ So $$\alpha=31,\quad \beta=5.$$ Therefore, $$\alpha+\beta=36.$$ --- 8. **Comparison with stored answer** Derived answer is $36$, which matches the stored correct answer.

More from Complex Numbers

View written solutionFree

Correct answer: 36

Interpret the sets geometrically

Let $z=x+iy$ .

Set $P$

Given $|z+2-3i|\le 1$ we write $z+2-3i=(x+2)+i(y-3).$ So, $(x+2)^2+(y-3)^2\le 1.$ This is the closed disk centered at $C_1=(-2,3)$ with radius $1$ .

Set $Q$

Find $P\cap Q$

The circle of $P$ has center $(-2,3)$ . Check its position relative to the line $y-x=4$ : $3-(-2)=5>4,$ so the center lies in the half-plane $Q$ .

Quantity to maximize/minimize

We need extrema of $|z-3+2i|=|(x-3)+i(y+2)|.$ This is the distance from point $(x,y)$ to $A=(3,-2).$ Thus, in the region $P\cap Q$ , we seek the farthest and nearest points from $A$ .

Useful geometry: line of constraint cuts the circle at two points

The boundary line is $y-x=4.$ Its intersection with the circle $(x+2)^2+(y-3)^2=1$ will be important.

Put $y=x+4$ into the circle: $(x+2)^2+(x+1)^2=1$ $x^2+4x+4+x^2+2x+1=1$ $2x^2+6x+4=0$ $x^2+3x+2=0$ $(x+1)(x+2)=0.$ So the intersection points are:

for $x=-1$ , $y=3$ giving $(-1,3)$ ,
for $x=-2$ , $y=2$ giving $(-2,2)$ .

Minimum distance from $A=(3,-2)$ to $P\cap Q$

Since $A$ lies outside the disk and the nearest point on the full disk would be on the line joining center to $A$ , we must check whether that point lies in $Q$ .

So compare distances from $A$ to the two boundary endpoints:

For $(-1,3)$ : $d^2=( -1-3)^2+(3+2)^2=16+25=41.$

Thus $|z_2|^2 = (-1)^2+3^2=10$ if $z_2=-1+3i$ , or $|z_2|^2=(-2)^2+2^2=8$ if $z_2=-2+2i$ .

But we must identify where the minimum of $|z-3+2i|$ over the feasible set occurs more carefully.

Let us parametrize the boundary segment between $(-2,2)$ and $(-1,3)$ :

\quad (x+2)^2+(x+1)^2\le 1.$$ Distance squared from $A$ is $$f(x)=(x-3)^2+(x+6)^2=2x^2+6x+45.$$ On $x\in[-2,-1]$, $$f'(x)=4x+6.$$ This vanishes at $$x=-\frac32,$$ which lies in the interval. Then $$y=\frac52.$$ Check this point is in the disk: $$\left(-\frac32+2\right)^2+\left(\frac52-3\right)^2=\left(\frac12\right)^2+\left(-\frac12\right)^2=\frac12<1.$$ So this point belongs to $P\cap Q$. Thus the minimum occurs at $$z_2=-\frac32+\frac52 i.$$ Now $$|z_2|^2=\left(-\frac32\right)^2+\left(\frac52\right)^2=\frac94+\frac{25}4=\frac{34}{4}=\frac{17}{2}.$$ --- 6. **Maximum distance from $A$ to $P\cap Q$** The farthest point on the full disk from $A$ lies on the line through center away from $A$. Vector from $A$ to center is $$C_1-A=(-5,5)=5(-1,1).$$ Unit vector in this direction is $$\frac{1}{\sqrt2}(-1,1).$$ So the farthest point on the circle is $$z_1: (-2,3)+1\cdot \frac{1}{\sqrt2}(-1,1)=\left(-2-\frac1{\sqrt2},\ 3+\frac1{\sqrt2}\right).$$ Check it satisfies $Q$: $$y-x = 3+\frac1{\sqrt2}-\left(-2-\frac1{\sqrt2}\right)=5+\sqrt2>4,$$ so it is feasible. Hence this is the maximizing point. Now $$|z_1|^2=\left(-2-\frac1{\sqrt2}\right)^2+\left(3+\frac1{\sqrt2}\right)^2.$$ Compute: $$\left(-2-\frac1{\sqrt2}\right)^2=4+2\sqrt2+\frac12=\frac92+2\sqrt2,$$ $$\left(3+\frac1{\sqrt2}\right)^2=9+3\sqrt2+\frac12=\frac{19}{2}+3\sqrt2.$$ Thus $$|z_1|^2=14+5\sqrt2.$$ --- 7. **Evaluate the required expression** $$|z_1|^2+2|z_2|^2=(14+5\sqrt2)+2\cdot \frac{17}{2}=14+5\sqrt2+17=31+5\sqrt2.$$ So $$\alpha=31,\quad \beta=5.$$ Therefore, $$\alpha+\beta=36.$$ --- 8. **Comparison with stored answer** Derived answer is $36$, which matches the stored correct answer.

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