Logo
Jeeify
MCQMediumJEE 2024Argand Plane & Geometry

JEE Mathematics 2024 Question with Solution

Let zz be a complex number such that the real part of z−2iz+2i\frac{z-2i}{z+2i} is zero. Then, the maximum value of ∣z−(6+8i)∣|z-(6+8i)| is equal to:

  • A

    1212

  • B

    ∞\infty

  • C

    1010

  • D

    88

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Re⁡(z−2iz+2i)=0\operatorname{Re}\left(\frac{z-2i}{z+2i}\right)=0

Find: The maximum value of ∣z−(6+8i)∣|z-(6+8i)|.

Let

z=x+iyz=x+iy

Then

z−2iz+2i=x+i(y−2)x+i(y+2)\frac{z-2i}{z+2i}=\frac{x+i(y-2)}{x+i(y+2)}

Rationalizing the denominator,

x+i(y−2)x+i(y+2)⋅x−i(y+2)x−i(y+2)=x2+y2−4−4ixx2+(y+2)2\frac{x+i(y-2)}{x+i(y+2)}\cdot \frac{x-i(y+2)}{x-i(y+2)} = \frac{x^2+y^2-4-4ix}{x^2+(y+2)^2}

So its real part is

Re⁡(z−2iz+2i)=x2+y2−4x2+(y+2)2\operatorname{Re}\left(\frac{z-2i}{z+2i}\right)=\frac{x^2+y^2-4}{x^2+(y+2)^2}

Since the real part is zero,

x2+y2−4x2+(y+2)2=0  ⟹  x2+y2−4=0  ⟹  x2+y2=4\frac{x^2+y^2-4}{x^2+(y+2)^2}=0 \implies x^2+y^2-4=0 \implies x^2+y^2=4

Thus, the locus of zz is the circle centered at the origin with radius 22.

Now ∣z−(6+8i)∣|z-(6+8i)| is the distance from the variable point zz on the circle x2+y2=4x^2+y^2=4 to the fixed point (6,8)(6,8).

The distance from the center O(0,0)O(0,0) to (6,8)(6,8) is

OP=62+82=10OP=\sqrt{6^2+8^2}=10

For a point on a circle, the maximum distance from an external point equals distance from the center plus the radius. Hence,

max⁡∣z−(6+8i)∣=10+2=12\max |z-(6+8i)|=10+2=12

Therefore, the correct option is A.

Common mistakes

  • Treating Re⁡(z−2iz+2i)=0\operatorname{Re}\left(\frac{z-2i}{z+2i}\right)=0 as meaning the whole fraction is zero. A zero real part only means the expression is purely imaginary. First extract the real part after rationalizing, then set only that numerator equal to zero.

  • Using the minimum-distance formula instead of the maximum-distance formula for a circle. Here the point (6,8)(6,8) lies outside the circle, so the maximum distance is OP+rOP+r, not OP−rOP-r.

  • Making an algebraic mistake while multiplying by the conjugate of the denominator. An incorrect real part leads to the wrong locus. After rationalization, the real part should reduce to a multiple of x2+y2−4x^2+y^2-4 divided by a positive denominator.

Practice more Argand Plane & Geometry questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Argand Plane & Geometry

Related questions