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Complex Numbers question

2023 · 29 Jan · Shift 2 · Q41
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  5. /2023 · 29 Jan · Shift 2 · Q41

Complex Numbers question

2023 · 29 Jan · Shift 2 · Q41

JEE MainMathematicsComplex NumbersNumerical+4 / −1
Let α=8−14i,A={z∈c:αz−α‾z‾z2−(z‾)2−112i=1}\alpha = 8 - 14i,A = \left\{ {z \in c:{{\alpha z - \overline \alpha \overline z } \over {{z^2} - {{\left( {\overline z } \right)}^2} - 112i}}=1} \right\}α=8−14i,A={z∈c:z2−(z)2−112iαz−αz​=1} and B={z∈c:∣z+3i∣=4}B = \left\{ {z \in c:\left| {z + 3i} \right| = 4} \right\}B={z∈c:∣z+3i∣=4}. Then ∑z∈A∩B(Reolimitsz−Imolimitsz)\sum\limits_{z \in A \cap B} {({\mathop{\rm Re} olimits} z - {\mathop{\rm Im} olimits} z)}z∈A∩B∑​(Reolimitsz−Imolimitsz) is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 14

  1. Given data

We have α=8−14i,α‾=8+14i.\alpha = 8-14i, \qquad \overline{\alpha}=8+14i.α=8−14i,α=8+14i.

Set A={z∈C:αz−α‾ z‾z2−(z‾)2−112i=1}A=\left\{z\in \mathbb C:\frac{\alpha z-\overline\alpha\,\overline z}{z^2-(\overline z)^2-112i}=1\right\}A={z∈C:z2−(z)2−112iαz−αz​=1} and B={z∈C:∣z+3i∣=4}.B=\left\{z\in\mathbb C:|z+3i|=4\right\}.B={z∈C:∣z+3i∣=4}.

We need ∑z∈A∩B(Re⁡z−Im⁡z).\sum_{z\in A\cap B}(\operatorname{Re}z-\operatorname{Im}z).∑z∈A∩B​(Rez−Imz).

  1. Write z=x+iyz=x+iyz=x+iy

Let z=x+iy,z‾=x−iy.z=x+iy, \qquad \overline z=x-iy.z=x+iy,z=x−iy. Then Re⁡z=x,Im⁡z=y.\operatorname{Re}z=x, \qquad \operatorname{Im}z=y.Rez=x,Imz=y.

  1. Simplify the condition for set AAA

We first compute the numerator: αz−α‾ z‾=(8−14i)(x+iy)−(8+14i)(x−iy).\alpha z-\overline\alpha\,\overline z=(8-14i)(x+iy)-(8+14i)(x-iy).αz−αz=(8−14i)(x+iy)−(8+14i)(x−iy).

Now, (8−14i)(x+iy)=8x+8iy−14ix−14i2y=(8x+14y)+i(8y−14x),(8-14i)(x+iy)=8x+8iy-14ix-14i^2y=(8x+14y)+i(8y-14x),(8−14i)(x+iy)=8x+8iy−14ix−14i2y=(8x+14y)+i(8y−14x),

and (8+14i)(x−iy)=8x−8iy+14ix−14i2y=(8x+14y)+i(14x−8y).(8+14i)(x-iy)=8x-8iy+14ix-14i^2y=(8x+14y)+i(14x-8y).(8+14i)(x−iy)=8x−8iy+14ix−14i2y=(8x+14y)+i(14x−8y).

So,

= i\big[(8y-14x)-(14x-8y)\big] = i(16y-28x).$$ Thus, $$\alpha z-\overline\alpha\,\overline z=4i(4y-7x).$$ Now compute $$z^2-(\overline z)^2=(x+iy)^2-(x-iy)^2=4ixy.$$ Hence denominator is $$z^2-(\overline z)^2-112i=4ixy-112i=4i(xy-28).$$ The condition $$\frac{\alpha z-\overline\alpha\,\overline z}{z^2-(\overline z)^2-112i}=1$$ becomes $$\frac{4i(4y-7x)}{4i(xy-28)}=1.$$ Provided denominator is nonzero, this gives $$4y-7x=xy-28.$$ So $$xy-4y+7x-28=0.$$ Factorizing, $$y(x-4)+7(x-4)=0,$$ $$ (x-4)(y+7)=0.$$ Therefore set $A$ corresponds to $$x=4 \quad \text{or} \quad y=-7,$$ with the original denominator nonzero. Since denominator is $4i(xy-28)$, we need $$xy\ne 28.$$ --- 4. **Simplify the condition for set $B$** Given $$|z+3i|=4.$$ Since $$z+3i=x+i(y+3),$$ we get $$x^2+(y+3)^2=16.$$ So $B$ is the circle centered at $(0,-3)$ with radius $4$. --- 5. **Find $A\cap B$** Since $A$ means either $x=4$ or $y=-7$, intersect these with the circle. ### Case 1: $x=4$ Substitute into $$x^2+(y+3)^2=16:$$ $$16+(y+3)^2=16,$$ $$(y+3)^2=0,$$ $$y=-3.$$ So one point is $$z=4-3i.$$ Check denominator condition: $$xy=4(-3)=-12\ne 28,$$ so this point is valid. ### Case 2: $y=-7$ Substitute into the circle: $$x^2+(-7+3)^2=16,$$ $$x^2+16=16,$$ $$x=0.$$ So another point is $$z=-7i.$$ Check denominator condition: $$xy=0\ne 28,$$ so this point is valid. Hence $$A\cap B=\{4-3i,\,-7i\}.$$ --- 6. **Compute the required sum** We need $$\sum_{z\in A\cap B}(\operatorname{Re}z-\operatorname{Im}z).$$ For $z=4-3i$, $$\operatorname{Re}z-\operatorname{Im}z=4-(-3)=7.$$ For $z=-7i$, $$\operatorname{Re}z-\operatorname{Im}z=0-(-7)=7.$$ Therefore, $$7+7=14.$$ --- 7. **Final answer** $$\boxed{14}$$ The derived answer matches the stored correct answer.
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