Let S = \ z \in C:|z - 2| \le 1,\,z(1 + i) + \overline z (1 - i) \le 2\ . Let |z - 4i| attains minimum and maximum values, respectively, at z1 \in S and z2 \in S. If 5(|z_1|^2 + |z_2|^2) = \alpha + \beta \sqrt 5 , where \alpha and \beta are integers, then the value of \alpha + \beta is equal to \underline\hspace2cm .

Complex Numbers PYQ — JEE Main 2022 · 29 Jun · Shift 1 · Q36 with Solution

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Correct answer: 26

Interpret the region $S$

Let $z = x+iy$ .

Then $|z-2|\le 1 \implies (x-2)^2+y^2\le 1,$ which is the closed disk centered at $(2,0)$ with radius $1$ .

Now simplify the second condition: $z(1+i)+\overline z(1-i)\le 2.$ Since $z=x+iy$ and $\overline z=x-iy$ , \begin{align*} z(1+i) &= (x+iy)(1+i) = (x-y)+i(x+y),\ \overline z(1-i) &= (x-iy)(1-i) = (x-y)-i(x+y). \end{align*} Adding, $z(1+i)+\overline z(1-i)=2(x-y).$ So the inequality becomes $2(x-y)\le 2 \implies x-y\le 1 \implies y\ge x-1.$

Hence, $S=\{(x,y):(x-2)^2+y^2\le 1,\ y\ge x-1\}.$ This is the part of the disk above the line $y=x-1$ .

Geometric meaning of $|z-4i|$

We need the minimum and maximum of $|z-4i| = \sqrt{x^2+(y-4)^2},$ which is the distance from the point $P=(0,4)$ to points of the region $S$ .

So we need the nearest point $z_1$ in $S$ to $P$ , and the farthest point $z_2$ in $S$ from $P$ .

Understand the boundary of $S$

The circle is $(x-2)^2+y^2=1.$ The line is $y=x-1.$ Find their intersection points: \begin{align*} (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. \end{align*} So intersection points are $(1,0),\quad (2,1).$

Thus the boundary of $S$ consists of:

the chord segment from $(1,0)$ to $(2,1)$ on the line $y=x-1$ ,
the major arc of the circle joining $(1,0)$ and $(2,1)$ lying above the line.

Minimum distance from $P=(0,4)$ to $S$

Since $P$ lies outside the disk, the nearest point in the full disk to $P$ lies on the circle along the line joining the center $C=(2,0)$ to $P$ .

Vector from $C$ to $P$ is $(-2,4),$ whose length is $\sqrt{(-2)^2+4^2}=\sqrt{20}=2\sqrt5.$ Unit vector in this direction is $\left(-\frac1{\sqrt5},\frac2{\sqrt5}\right).$ So the nearest point on the circle is

=\left(2-\frac1{\sqrt5},\frac2{\sqrt5}\right).$$ Check it satisfies $y\ge x-1$: $$\frac2{\sqrt5} \ge 1-\frac1{\sqrt5} \iff \frac3{\sqrt5}\ge 1,$$ which is true. Hence this point belongs to $S$. Therefore the minimum occurs at $$z_1=\left(2-\frac1{\sqrt5}\right)+i\left(\frac2{\sqrt5}\right).$$ Now, \begin{align*} |z_1|^2 &= \left(2-\frac1{\sqrt5}\right)^2+\left(\frac2{\sqrt5}\right)^2 \\ &=4-\frac4{\sqrt5}+\frac15+\frac45 \\ &=5-\frac4{\sqrt5}. \end{align*} --- 5. **Maximum distance from $P=(0,4)$ to $S$** For the full disk, the farthest point from $P$ on the circle is opposite to the nearest point, i.e. along direction from $C$ away from $P$: $$\left(2,0\right)-\left(-\frac1{\sqrt5},\frac2{\sqrt5}\right)=\left(2+\frac1{\sqrt5},-\frac2{\sqrt5}\right).$$ But this point does **not** satisfy $y\ge x-1$, so it is not in $S$. Thus the farthest point in $S$ must occur on the boundary of the allowed region. We compare the two boundary parts. ### (a) On the chord $y=x-1$, between $(1,0)$ and $(2,1)$ Distance squared from $P$ is \begin{align*} D^2 &= x^2+(y-4)^2 = x^2+(x-5)^2 \\ &= 2x^2-10x+25. \end{align*} For $x\in[1,2]$, this is a convex quadratic, so its maximum on the interval occurs at an endpoint. At $(1,0)$: $$D^2=1+16=17.$$ At $(2,1)$: $$D^2=4+9=13.$$ So on the chord, maximum is at $(1,0)$. ### (b) On the circular arc A continuous function on the compact arc attains extrema. Since the unconstrained farthest point on the full circle is excluded, the maximum on the permitted arc occurs at an endpoint of the permitted arc, i.e. again either $(1,0)$ or $(2,1)$. We already computed: $$|(1,0)-(0,4)|^2=17, \qquad |(2,1)-(0,4)|^2=13.$$ So the maximum over all of $S$ is at $$z_2=1.$$ Hence, $$|z_2|^2=1.$$ --- 6. **Compute the required expression** We have $$|z_1|^2=5-\frac4{\sqrt5}, \qquad |z_2|^2=1.$$ Thus \begin{align*} 5\left(|z_1|^2+|z_2|^2\right) &=5\left(5-\frac4{\sqrt5}+1\right) \\ &=5\left(6-\frac4{\sqrt5}\right) \\ &=30-\frac{20}{\sqrt5} \\ &=30-4\sqrt5. \end{align*} So $$\alpha=30,\qquad \beta=-4.$$ Therefore, $$\alpha+\beta=30-4=26.$$ --- 7. **Comparison with stored answer** Derived answer: $26$. Stored correct answer: $26$. They match.

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Correct answer: 26

Interpret the region $S$

Let $z = x+iy$ .

Then $|z-2|\le 1 \implies (x-2)^2+y^2\le 1,$ which is the closed disk centered at $(2,0)$ with radius $1$ .

Hence, $S=\{(x,y):(x-2)^2+y^2\le 1,\ y\ge x-1\}.$ This is the part of the disk above the line $y=x-1$ .

Geometric meaning of $|z-4i|$

We need the minimum and maximum of $|z-4i| = \sqrt{x^2+(y-4)^2},$ which is the distance from the point $P=(0,4)$ to points of the region $S$ .

So we need the nearest point $z_1$ in $S$ to $P$ , and the farthest point $z_2$ in $S$ from $P$ .

Understand the boundary of $S$

Thus the boundary of $S$ consists of:

the chord segment from $(1,0)$ to $(2,1)$ on the line $y=x-1$ ,
the major arc of the circle joining $(1,0)$ and $(2,1)$ lying above the line.

Minimum distance from $P=(0,4)$ to $S$

Since $P$ lies outside the disk, the nearest point in the full disk to $P$ lies on the circle along the line joining the center $C=(2,0)$ to $P$ .

=\left(2-\frac1{\sqrt5},\frac2{\sqrt5}\right).$$ Check it satisfies $y\ge x-1$: $$\frac2{\sqrt5} \ge 1-\frac1{\sqrt5} \iff \frac3{\sqrt5}\ge 1,$$ which is true. Hence this point belongs to $S$. Therefore the minimum occurs at $$z_1=\left(2-\frac1{\sqrt5}\right)+i\left(\frac2{\sqrt5}\right).$$ Now, \begin{align*} |z_1|^2 &= \left(2-\frac1{\sqrt5}\right)^2+\left(\frac2{\sqrt5}\right)^2 \\ &=4-\frac4{\sqrt5}+\frac15+\frac45 \\ &=5-\frac4{\sqrt5}. \end{align*} --- 5. **Maximum distance from $P=(0,4)$ to $S$** For the full disk, the farthest point from $P$ on the circle is opposite to the nearest point, i.e. along direction from $C$ away from $P$: $$\left(2,0\right)-\left(-\frac1{\sqrt5},\frac2{\sqrt5}\right)=\left(2+\frac1{\sqrt5},-\frac2{\sqrt5}\right).$$ But this point does **not** satisfy $y\ge x-1$, so it is not in $S$. Thus the farthest point in $S$ must occur on the boundary of the allowed region. We compare the two boundary parts. ### (a) On the chord $y=x-1$, between $(1,0)$ and $(2,1)$ Distance squared from $P$ is \begin{align*} D^2 &= x^2+(y-4)^2 = x^2+(x-5)^2 \\ &= 2x^2-10x+25. \end{align*} For $x\in[1,2]$, this is a convex quadratic, so its maximum on the interval occurs at an endpoint. At $(1,0)$: $$D^2=1+16=17.$$ At $(2,1)$: $$D^2=4+9=13.$$ So on the chord, maximum is at $(1,0)$. ### (b) On the circular arc A continuous function on the compact arc attains extrema. Since the unconstrained farthest point on the full circle is excluded, the maximum on the permitted arc occurs at an endpoint of the permitted arc, i.e. again either $(1,0)$ or $(2,1)$. We already computed: $$|(1,0)-(0,4)|^2=17, \qquad |(2,1)-(0,4)|^2=13.$$ So the maximum over all of $S$ is at $$z_2=1.$$ Hence, $$|z_2|^2=1.$$ --- 6. **Compute the required expression** We have $$|z_1|^2=5-\frac4{\sqrt5}, \qquad |z_2|^2=1.$$ Thus \begin{align*} 5\left(|z_1|^2+|z_2|^2\right) &=5\left(5-\frac4{\sqrt5}+1\right) \\ &=5\left(6-\frac4{\sqrt5}\right) \\ &=30-\frac{20}{\sqrt5} \\ &=30-4\sqrt5. \end{align*} So $$\alpha=30,\qquad \beta=-4.$$ Therefore, $$\alpha+\beta=30-4=26.$$ --- 7. **Comparison with stored answer** Derived answer: $26$. Stored correct answer: $26$. They match.

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