Let z be those complex numbers which satisfy | z + 5 | \le 4 and z(1 + i) + \overline z (1 - i) \ge - 10, i = \sqrt - 1 . If the maximum value of | z + 1 |2 is \alpha + \beta\sqrt 2 , then the value of ( \alpha + \beta ) is \underline\hspace2cm .

Complex Numbers PYQ — JEE Main 2021 · 26 Feb · Shift 2 · Q43 with Solution

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Correct answer: 48

Let

\quad \overline z=x-iy$$ where $x,y\in\mathbb R$. We must maximize $$|z+1|^2$$ subject to 1. $$|z+5|\le 4$$ 2. $$z(1+i)+\overline z(1-i)\ge -10.$$ --- ## 1. Convert the conditions into Cartesian form ### Condition 1 Since $$z+5=(x+5)+iy,$$ we get $$|z+5|\le 4 \iff (x+5)^2+y^2\le 16.$$ This is a closed disk with center $(-5,0)$ and radius $4$. ### Condition 2 Compute: $$z(1+i)=(x+iy)(1+i)=x+ix+iy+i^2y=(x-y)+i(x+y),$$ and $$\overline z(1-i)=(x-iy)(1-i)=x-ix-iy+(-i)(-i)y=(x-y)-i(x+y).$$ Adding, $$z(1+i)+\overline z(1-i)=2(x-y).$$ So the inequality becomes $$2(x-y)\ge -10 \iff x-y\ge -5 \iff y\le x+5.$$ Thus the feasible region is the part of the disk $$(x+5)^2+y^2\le 16$$ lying below the line $$y=x+5.$$ --- ## 2. Objective function We need the maximum of $$|z+1|^2=|(x+1)+iy|^2=(x+1)^2+y^2.$$ Geometrically, this is the square of the distance from the point $(-1,0)$. So we need the farthest feasible point from $(-1,0)$. --- ## 3. First check the full circle maximum For the full disk centered at $C(-5,0)$ with radius $4$, the point farthest from $P(-1,0)$ would lie on the circle in the direction opposite to $P$ from the center. Distance between center and $P$ is $$CP=4.$$ Hence the maximum possible distance from $P$ on the full circle is $$CP+r=4+4=8,$$ so the corresponding square would be $$8^2=64.$$ That farthest point is $$(-9,0).$$ Check whether it satisfies the half-plane condition: $$y\le x+5 \iff 0\le -9+5=-4,$$ which is false. So the unconstrained farthest point is not feasible. --- ## 4. Feasible boundary where maximum occurs Since the disk is compact and the objective is continuous, the maximum exists on the boundary of the feasible region. The boundary consists of: 1. the circular arc $$(x+5)^2+y^2=16$$ with $y\le x+5$, 2. the chord on the line $$y=x+5$$ inside the circle. The maximum should occur at an extreme boundary point. Let us find the intersection points of the line and circle. Substitute $y=x+5$ into the circle: $$(x+5)^2+(x+5)^2=16$$ $$2(x+5)^2=16$$ $$(x+5)^2=8$$ $$x+5=\pm 2\sqrt2.$$ Since $y=x+5$, the intersection points are: - If $x+5=2\sqrt2$: $$x=-5+2\sqrt2,\quad y=2\sqrt2.$$ - If $x+5=-2\sqrt2$: $$x=-5-2\sqrt2,\quad y=-2\sqrt2.$$ So the two points are $$A\left(-5+2\sqrt2,\ 2\sqrt2\right), \qquad B\left(-5-2\sqrt2,\ -2\sqrt2\right).$$ --- ## 5. Check the value of $|z+1|^2$ at these extreme points ### At $A$ $$|z+1|^2=(x+1)^2+y^2$$ $$=(-5+2\sqrt2+1)^2+(2\sqrt2)^2$$ $$=(-4+2\sqrt2)^2+8$$ $$=16-16\sqrt2+8+8$$ $$=32-16\sqrt2.$$ ### At $B$ $$|z+1|^2=(-5-2\sqrt2+1)^2+(-2\sqrt2)^2$$ $$=(-4-2\sqrt2)^2+8$$ $$=16+16\sqrt2+8+8$$ $$=32+16\sqrt2.$$ So among these two, the larger value is $$32+16\sqrt2.$$ --- ## 6. Verify that this is the actual maximum Along the line segment $y=x+5$ inside the disk, the objective is a convex quadratic, so its maximum on the segment occurs at an endpoint; we already checked both endpoints. Along the circular arc, the farthest feasible point from $(-1,0)$ occurs where the line constraint first becomes active, i.e. at the endpoint $B$, because the unconstrained farthest point $(-9,0)$ is cut off by the half-plane. Hence the global maximum is indeed $$|z+1|^2_{\max}=32+16\sqrt2.$$ Thus $$\alpha=32,\qquad \beta=16.$$ So $$\alpha+\beta=48.$$ --- ## 7. Comparison with stored answer Derived answer: $48$. This matches the stored correct answer.

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View written solutionFree

Correct answer: 48

Let

\quad \overline z=x-iy$$ where $x,y\in\mathbb R$. We must maximize $$|z+1|^2$$ subject to 1. $$|z+5|\le 4$$ 2. $$z(1+i)+\overline z(1-i)\ge -10.$$ --- ## 1. Convert the conditions into Cartesian form ### Condition 1 Since $$z+5=(x+5)+iy,$$ we get $$|z+5|\le 4 \iff (x+5)^2+y^2\le 16.$$ This is a closed disk with center $(-5,0)$ and radius $4$. ### Condition 2 Compute: $$z(1+i)=(x+iy)(1+i)=x+ix+iy+i^2y=(x-y)+i(x+y),$$ and $$\overline z(1-i)=(x-iy)(1-i)=x-ix-iy+(-i)(-i)y=(x-y)-i(x+y).$$ Adding, $$z(1+i)+\overline z(1-i)=2(x-y).$$ So the inequality becomes $$2(x-y)\ge -10 \iff x-y\ge -5 \iff y\le x+5.$$ Thus the feasible region is the part of the disk $$(x+5)^2+y^2\le 16$$ lying below the line $$y=x+5.$$ --- ## 2. Objective function We need the maximum of $$|z+1|^2=|(x+1)+iy|^2=(x+1)^2+y^2.$$ Geometrically, this is the square of the distance from the point $(-1,0)$. So we need the farthest feasible point from $(-1,0)$. --- ## 3. First check the full circle maximum For the full disk centered at $C(-5,0)$ with radius $4$, the point farthest from $P(-1,0)$ would lie on the circle in the direction opposite to $P$ from the center. Distance between center and $P$ is $$CP=4.$$ Hence the maximum possible distance from $P$ on the full circle is $$CP+r=4+4=8,$$ so the corresponding square would be $$8^2=64.$$ That farthest point is $$(-9,0).$$ Check whether it satisfies the half-plane condition: $$y\le x+5 \iff 0\le -9+5=-4,$$ which is false. So the unconstrained farthest point is not feasible. --- ## 4. Feasible boundary where maximum occurs Since the disk is compact and the objective is continuous, the maximum exists on the boundary of the feasible region. The boundary consists of: 1. the circular arc $$(x+5)^2+y^2=16$$ with $y\le x+5$, 2. the chord on the line $$y=x+5$$ inside the circle. The maximum should occur at an extreme boundary point. Let us find the intersection points of the line and circle. Substitute $y=x+5$ into the circle: $$(x+5)^2+(x+5)^2=16$$ $$2(x+5)^2=16$$ $$(x+5)^2=8$$ $$x+5=\pm 2\sqrt2.$$ Since $y=x+5$, the intersection points are: - If $x+5=2\sqrt2$: $$x=-5+2\sqrt2,\quad y=2\sqrt2.$$ - If $x+5=-2\sqrt2$: $$x=-5-2\sqrt2,\quad y=-2\sqrt2.$$ So the two points are $$A\left(-5+2\sqrt2,\ 2\sqrt2\right), \qquad B\left(-5-2\sqrt2,\ -2\sqrt2\right).$$ --- ## 5. Check the value of $|z+1|^2$ at these extreme points ### At $A$ $$|z+1|^2=(x+1)^2+y^2$$ $$=(-5+2\sqrt2+1)^2+(2\sqrt2)^2$$ $$=(-4+2\sqrt2)^2+8$$ $$=16-16\sqrt2+8+8$$ $$=32-16\sqrt2.$$ ### At $B$ $$|z+1|^2=(-5-2\sqrt2+1)^2+(-2\sqrt2)^2$$ $$=(-4-2\sqrt2)^2+8$$ $$=16+16\sqrt2+8+8$$ $$=32+16\sqrt2.$$ So among these two, the larger value is $$32+16\sqrt2.$$ --- ## 6. Verify that this is the actual maximum Along the line segment $y=x+5$ inside the disk, the objective is a convex quadratic, so its maximum on the segment occurs at an endpoint; we already checked both endpoints. Along the circular arc, the farthest feasible point from $(-1,0)$ occurs where the line constraint first becomes active, i.e. at the endpoint $B$, because the unconstrained farthest point $(-9,0)$ is cut off by the half-plane. Hence the global maximum is indeed $$|z+1|^2_{\max}=32+16\sqrt2.$$ Thus $$\alpha=32,\qquad \beta=16.$$ So $$\alpha+\beta=48.$$ --- ## 7. Comparison with stored answer Derived answer: $48$. This matches the stored correct answer.

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