Let C be the set of all complex numbers. Let S1 = z \in C : |z - 2| \le 1 and S2 = z \in C : z(1 + i) + \overline z (1 - i) \ge 4. Then, the maximum value of | z - 5 \over 2 |^2 for z \in S1 \cap S2 is equal to :

Complex Numbers PYQ — JEE Main 2021 · 27 Jul · Shift 2 · Q26 with Solution

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Correct answer: D

Interpret the sets geometrically

Let $z=x+iy, \quad \overline z=x-iy.$

We are given $S_1=\{z\in \mathbb C:|z-2|\le 1\}.$ Since $2$ is the complex number $(2,0)$ , this is the closed disk $(x-2)^2+y^2\le 1.$

Now simplify the condition for $S_2$ : $z(1+i)+\overline z(1-i)\ge 4.$ Substitute $z=x+iy$ and $\overline z=x-iy$ :

$z(1+i)=(x+iy)(1+i)=(x-y)+i(x+y),$ $\overline z(1-i)=(x-iy)(1-i)=(x-y)-i(x+y).$

Adding, $z(1+i)+\overline z(1-i)=2(x-y).$ Hence $2(x-y)\ge 4 \implies x-y\ge 2 \implies y\le x-2.$

So, $S_2=\{(x,y): y\le x-2\},$ which is a half-plane.

Thus $S_1\cap S_2$ is the part of the disk centered at $(2,0)$ , radius $1$ , lying below the line $y=x-2$ .

Objective function

We need the maximum of $\left|z-\frac52\right|^2.$ Since $\frac52$ is the point $\left(\frac52,0\right)$ , this is $\left(x-\frac52\right)^2+y^2.$ So we want the maximum squared distance from the point $P=\left(\frac52,0\right)$ to the region $S_1\cap S_2$ .

Because the feasible region is closed and bounded, the maximum exists, and it will occur on the boundary.

Boundary of the feasible region

The disk boundary is $(x-2)^2+y^2=1.$ The line boundary is $y=x-2.$

Their points of intersection are found by substituting $y=x-2$ into the circle: $(x-2)^2+(x-2)^2=1$ $2(x-2)^2=1$ $(x-2)^2=\frac12$ $x-2=\pm \frac{1}{\sqrt2}.$

Since the feasible side is $y\le x-2$ , both intersection points are on the boundary; the relevant arc is the one satisfying the inequality.

Thus the intersection points are

\qquad B=\left(2-\frac1{\sqrt2},-\frac1{\sqrt2}\right).$$ --- 4. **Check where the farthest point from $P$ lies** The feasible region is a circular segment. Since we seek the farthest point from $P$, it is enough to check: - the circular arc of the disk lying in $y\le x-2$, - the chord on the line $y=x-2$ between $A$ and $B$. ### On the circular arc Parameterize the circle: $$x=2+\cos\theta, \quad y=\sin\theta.$$ Then $$\left|z-\frac52\right|^2=\left(2+\cos\theta-\frac52\right)^2+\sin^2\theta =\left(\cos\theta-\frac12\right)^2+\sin^2\theta.$$ Simplify: $$=\cos^2\theta-\cos\theta+\frac14+\sin^2\theta =1-\cos\theta+\frac14 =\frac54-\cos\theta.$$ So we must maximize $\frac54-\cos\theta$, i.e. minimize $\cos\theta$ on the allowed arc. The condition $y\le x-2$ becomes $$\sin\theta\le \cos\theta \iff \sin\theta-\cos\theta\le 0 \iff \sqrt2\sin\left(\theta-\frac\pi4\right)\le 0.$$ This describes the arc from $\theta=-\frac{3\pi}{4}$ to $\theta=\frac\pi4$ (mod $2\pi$), i.e. the lower-right major arc between the intersection points. On this arc, the minimum value of $\cos\theta$ occurs at $$\theta=-\frac{3\pi}{4},$$ where $$\cos\theta=-\frac{1}{\sqrt2}.$$ Hence the maximum value on the arc is $$\frac54-\left(-\frac1{\sqrt2}\right)=\frac54+\frac1{\sqrt2} =\frac{5+2\sqrt2}{4}.$$ This occurs at point $$B=\left(2-\frac1{\sqrt2},-\frac1{\sqrt2}\right).$$ --- ### On the chord $AB$ The chord lies on $$y=x-2.$$ Let $t=x-2$, so points on the chord are $$(x,y)=(2+t,t), \qquad -\frac1{\sqrt2}\le t\le \frac1{\sqrt2}.$$ Then $$\left|z-\frac52\right|^2=\left(2+t-\frac52\right)^2+t^2 =\left(t-\frac12\right)^2+t^2 =2t^2-t+\frac14.$$ This is a convex quadratic, so its maximum on the interval occurs at an endpoint. At $$t=\frac1{\sqrt2}: \quad 2t^2-t+\frac14=1-\frac1{\sqrt2}+\frac14=\frac54-\frac1{\sqrt2},$$ At $$t=-\frac1{\sqrt2}: \quad 2t^2-t+\frac14=1+\frac1{\sqrt2}+\frac14=\frac54+\frac1{\sqrt2}= rac{5+2\sqrt2}{4}.$$ So the maximum on the chord is also $$\frac{5+2\sqrt2}{4}.$$ --- 5. **Final answer** Therefore, $$\max_{z\in S_1\cap S_2}\left|z-\frac52\right|^2=\frac{5+2\sqrt2}{4}.$$ So the correct option is $$\boxed{\text{D}}.$$ --- 6. **Comparison with stored answer** Stored correct answer: **D** Our derived answer: **D** They agree.

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View written solutionFree

Correct answer: D

Interpret the sets geometrically

Let $z=x+iy, \quad \overline z=x-iy.$

We are given $S_1=\{z\in \mathbb C:|z-2|\le 1\}.$ Since $2$ is the complex number $(2,0)$ , this is the closed disk $(x-2)^2+y^2\le 1.$

Now simplify the condition for $S_2$ : $z(1+i)+\overline z(1-i)\ge 4.$ Substitute $z=x+iy$ and $\overline z=x-iy$ :

$z(1+i)=(x+iy)(1+i)=(x-y)+i(x+y),$ $\overline z(1-i)=(x-iy)(1-i)=(x-y)-i(x+y).$

Adding, $z(1+i)+\overline z(1-i)=2(x-y).$ Hence $2(x-y)\ge 4 \implies x-y\ge 2 \implies y\le x-2.$

So, $S_2=\{(x,y): y\le x-2\},$ which is a half-plane.

Thus $S_1\cap S_2$ is the part of the disk centered at $(2,0)$ , radius $1$ , lying below the line $y=x-2$ .

Objective function

Because the feasible region is closed and bounded, the maximum exists, and it will occur on the boundary.

Boundary of the feasible region

The disk boundary is $(x-2)^2+y^2=1.$ The line boundary is $y=x-2.$

Their points of intersection are found by substituting $y=x-2$ into the circle: $(x-2)^2+(x-2)^2=1$ $2(x-2)^2=1$ $(x-2)^2=\frac12$ $x-2=\pm \frac{1}{\sqrt2}.$

Since the feasible side is $y\le x-2$ , both intersection points are on the boundary; the relevant arc is the one satisfying the inequality.

Thus the intersection points are

\qquad B=\left(2-\frac1{\sqrt2},-\frac1{\sqrt2}\right).$$ --- 4. **Check where the farthest point from $P$ lies** The feasible region is a circular segment. Since we seek the farthest point from $P$, it is enough to check: - the circular arc of the disk lying in $y\le x-2$, - the chord on the line $y=x-2$ between $A$ and $B$. ### On the circular arc Parameterize the circle: $$x=2+\cos\theta, \quad y=\sin\theta.$$ Then $$\left|z-\frac52\right|^2=\left(2+\cos\theta-\frac52\right)^2+\sin^2\theta =\left(\cos\theta-\frac12\right)^2+\sin^2\theta.$$ Simplify: $$=\cos^2\theta-\cos\theta+\frac14+\sin^2\theta =1-\cos\theta+\frac14 =\frac54-\cos\theta.$$ So we must maximize $\frac54-\cos\theta$, i.e. minimize $\cos\theta$ on the allowed arc. The condition $y\le x-2$ becomes $$\sin\theta\le \cos\theta \iff \sin\theta-\cos\theta\le 0 \iff \sqrt2\sin\left(\theta-\frac\pi4\right)\le 0.$$ This describes the arc from $\theta=-\frac{3\pi}{4}$ to $\theta=\frac\pi4$ (mod $2\pi$), i.e. the lower-right major arc between the intersection points. On this arc, the minimum value of $\cos\theta$ occurs at $$\theta=-\frac{3\pi}{4},$$ where $$\cos\theta=-\frac{1}{\sqrt2}.$$ Hence the maximum value on the arc is $$\frac54-\left(-\frac1{\sqrt2}\right)=\frac54+\frac1{\sqrt2} =\frac{5+2\sqrt2}{4}.$$ This occurs at point $$B=\left(2-\frac1{\sqrt2},-\frac1{\sqrt2}\right).$$ --- ### On the chord $AB$ The chord lies on $$y=x-2.$$ Let $t=x-2$, so points on the chord are $$(x,y)=(2+t,t), \qquad -\frac1{\sqrt2}\le t\le \frac1{\sqrt2}.$$ Then $$\left|z-\frac52\right|^2=\left(2+t-\frac52\right)^2+t^2 =\left(t-\frac12\right)^2+t^2 =2t^2-t+\frac14.$$ This is a convex quadratic, so its maximum on the interval occurs at an endpoint. At $$t=\frac1{\sqrt2}: \quad 2t^2-t+\frac14=1-\frac1{\sqrt2}+\frac14=\frac54-\frac1{\sqrt2},$$ At $$t=-\frac1{\sqrt2}: \quad 2t^2-t+\frac14=1+\frac1{\sqrt2}+\frac14=\frac54+\frac1{\sqrt2}= rac{5+2\sqrt2}{4}.$$ So the maximum on the chord is also $$\frac{5+2\sqrt2}{4}.$$ --- 5. **Final answer** Therefore, $$\max_{z\in S_1\cap S_2}\left|z-\frac52\right|^2=\frac{5+2\sqrt2}{4}.$$ So the correct option is $$\boxed{\text{D}}.$$ --- 6. **Comparison with stored answer** Stored correct answer: **D** Our derived answer: **D** They agree.

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