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Complex Numbers question

2021 · 31 Aug · Shift 1 · Q38
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  5. /2021 · 31 Aug · Shift 1 · Q38

Complex Numbers question

2021 · 31 Aug · Shift 1 · Q38

JEE MainMathematicsComplex NumbersNumerical+4 / −1
A point z moves in the complex plane such that arg⁡(z−2z+2)=π4\arg \left( {{{z - 2} \over {z + 2}}} \right) = {\pi \over 4}arg(z+2z−2​)=4π​, then the minimum value of ∣z−92−2i∣2{\left| {z - 9\sqrt 2 - 2i} \right|^2}​z−92​−2i​2 is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 98

  1. Let z=x+iyz=x+iyz=x+iy where x,y∈Rx,y\in\mathbb Rx,y∈R.

  2. We are given arg⁡(z−2z+2)=π4.\arg\left(\frac{z-2}{z+2}\right)=\frac{\pi}{4}.arg(z+2z−2​)=4π​. Write this in geometric form.

    The points corresponding to z−2z-2z−2 and z+2z+2z+2 are vectors from 222 and −2-2−2 respectively to the moving point zzz. Hence, arg⁡(z−2z+2)=arg⁡(z−2)−arg⁡(z+2)=π4.\arg\left(\frac{z-2}{z+2}\right)=\arg(z-2)-\arg(z+2)=\frac{\pi}{4}.arg(z+2z−2​)=arg(z−2)−arg(z+2)=4π​.

    So the angle subtended by the segment joining −2-2−2 and 222 at the point zzz is π4\frac{\pi}{4}4π​.

  3. Therefore, the locus of zzz is a circle passing through −2-2−2 and 222 such that chord [−2,2][-2,2][−2,2] subtends angle π4\frac{\pi}{4}4π​ at any point on the circle.

    Let the radius be RRR. For a chord of length 444 subtending angle θ=π4\theta=\frac{\pi}{4}θ=4π​ at the circumference, chord length=2Rsin⁡θ\text{chord length}=2R\sin\thetachord length=2Rsinθ so 4=2Rsin⁡π4=2R⋅12.4=2R\sin\frac{\pi}{4}=2R\cdot \frac{1}{\sqrt2}.4=2Rsin4π​=2R⋅2​1​. Thus, R=22.R=2\sqrt2.R=22​.

  4. The center lies on the perpendicular bisector of the chord from −2-2−2 to 222, i.e. on the imaginary axis.

    Since the chord endpoints are (−2,0)(-2,0)(−2,0) and (2,0)(2,0)(2,0), if center is (0,k)(0,k)(0,k) then R2=4+k2.R^2=4+k^2.R2=4+k2. Using R=22R=2\sqrt2R=22​, 8=4+k2  ⟹  k2=4  ⟹  k=±2.8=4+k^2 \implies k^2=4 \implies k=\pm 2.8=4+k2⟹k2=4⟹k=±2.

    Now we must choose the correct circle based on arg⁡(z−2z+2)=+π4,\arg\left(\frac{z-2}{z+2}\right)=+\frac{\pi}{4},arg(z+2z−2​)=+4π​, not −π4-\frac{\pi}{4}−4π​.

    Testing the upper point z=2iz=2iz=2i: 2i−22i+2=−2+2i2+2i=i,\frac{2i-2}{2i+2}=\frac{-2+2i}{2+2i}=i,2i+22i−2​=2+2i−2+2i​=i, whose argument is π2\frac{\pi}{2}2π​, not suitable.

    Testing the lower point z=−2iz=-2iz=−2i: −2i−2−2i+2=−2−2i2−2i=−i,\frac{-2i-2}{-2i+2}=\frac{-2-2i}{2-2i}=-i,−2i+2−2i−2​=2−2i−2−2i​=−i, whose argument is −π2-\frac{\pi}{2}−2π​.

    More systematically, the required arc corresponds to the circle with center (0,−2)(0,-2)(0,−2), i.e. x2+(y+2)2=8.x^2+(y+2)^2=8.x2+(y+2)2=8.

  5. We need the minimum of ∣z−(92+2i)∣2.|z-(9\sqrt2+2i)|^2.∣z−(92​+2i)∣2. The fixed point is P=(92,2).P=(9\sqrt2,2).P=(92​,2).

    The circle has center C=(0,−2),R=22.C=(0,-2), \quad R=2\sqrt2.C=(0,−2),R=22​.

    The minimum distance from point PPP to the circle is PC−RPC-RPC−R provided PPP is outside the circle.

    Compute:

    =\sqrt{162+16} =\sqrt{178}.$$ So minimum squared distance would be $$\left(\sqrt{178}-2\sqrt2\right)^2 =178+8-8\sqrt{89} =186-8\sqrt{89},$$ which is not an integer. This suggests we should instead derive the locus algebraically and check carefully.

  6. Algebraic derivation of the locus:

    Let z−2z+2=u.\frac{z-2}{z+2}=u.z+2z−2​=u. Since arg⁡u=π4\arg u=\frac{\pi}{4}argu=4π​, we can write u=t(1+i),t>0.u=t(1+i), \quad t>0.u=t(1+i),t>0.

    Hence z−2=t(1+i)(z+2).z-2=t(1+i)(z+2).z−2=t(1+i)(z+2). Put z=x+iyz=x+iyz=x+iy and use the condition that z−2z+2\frac{z-2}{z+2}z+2z−2​ has equal real and imaginary parts with positive ratio.

    Let z−2z+2=(x−2)+iy(x+2)+iy.\frac{z-2}{z+2}=\frac{(x-2)+iy}{(x+2)+iy}.z+2z−2​=(x+2)+iy(x−2)+iy​. Multiply numerator and denominator by (x+2)−iy(x+2)-iy(x+2)−iy:

    \frac{(x-2+iy)(x+2-iy)}{(x+2)^2+y^2}.$$ Numerator: $$(x-2)(x+2)+y^2+i\big(y(x+2)-y(x-2)\big) =x^2+y^2-4+4iy.$$ So $$\frac{z-2}{z+2}=\frac{x^2+y^2-4}{(x+2)^2+y^2}+i\frac{4y}{(x+2)^2+y^2}.$$ Since argument is $\frac{\pi}{4}$, real part $=$ imaginary part and both are positive: $$x^2+y^2-4=4y.$$ Therefore, $$x^2+y^2-4y-4=0$$ $$x^2+(y-2)^2=8.$$ So the correct locus is the circle centered at $$C=(0,2)$$ with radius $$R=2\sqrt2.$$

  7. Now the fixed point is P=(92,2).P=(9\sqrt2,2).P=(92​,2). Since the center is also at y=2y=2y=2, PC=92.PC=9\sqrt2.PC=92​.

    Hence the minimum distance from PPP to the circle is PC−R=92−22=72.PC-R=9\sqrt2-2\sqrt2=7\sqrt2.PC−R=92​−22​=72​.

    Therefore, min⁡∣z−(92+2i)∣2=(72)2=98.\min |z-(9\sqrt2+2i)|^2=(7\sqrt2)^2=98.min∣z−(92​+2i)∣2=(72​)2=98.

  8. Final answer: 98\boxed{98}98​

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