Let z Be a Complex Number Such That the Real Part of (z − 2i)/(z + 2i) Is Zero. Then the Maximum Value of |z − (6 + 8i)| Is Equal To

JEE Main 2024 — 9 April, Evening Shift. Previous Year Question.

Problem

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

Key insight. A condition like “the real part of a ratio of complex numbers is zero” is really a disguised circle equation. Once $z=x+iy$ is substituted and the fraction is rationalised, the real part boils down to a single quadratic in $x$ and $y$ — set it to zero and the circle reveals itself directly.

Watch this explained step by step →  ·  More on the Shiwam’s Classes channel

Approach

Write $z=x+iy$, rationalise $\dfrac{z-2i}{z+2i}$ by multiplying by the conjugate of the denominator, and extract the real part. Setting it to zero gives a circle equation for $z$. Then find the maximum distance from a point on that circle to the fixed point $6+8i$, which is simply the distance from the circle’s centre to that point, plus the radius.

Solution

Step 1 — Rationalise (z−2i)/(z+2i)

With $z=x+iy$:

$$\frac{z-2i}{z+2i} = \frac{x+i(y-2)}{x+i(y+2)} = \frac{[x+i(y-2)][x-i(y+2)]}{x^2+(y+2)^2}$$

Step 2 — Expand the numerator

$$[x+i(y-2)][x-i(y+2)] = x^2 – ix(y+2) + ix(y-2) + (y-2)(y+2)$$

$$= x^2+(y^2-4) + ix[(y-2)-(y+2)] = (x^2+y^2-4) – 4ix$$

Step 3 — Set the real part to zero

$$\operatorname{Re}\left(\frac{z-2i}{z+2i}\right) = \frac{x^2+y^2-4}{x^2+(y+2)^2} = 0 \implies x^2+y^2-4=0$$

$$x^2+y^2=4$$

This is a circle centred at the origin with radius $2$.

Step 4 — Find the distance from the circle’s centre to (6,8)

$$\sqrt{6^2+8^2} = \sqrt{36+64} = \sqrt{100} = 10$$

Step 5 — Add the radius to get the maximum distance

The farthest a point on the circle can be from $(6,8)$ is the distance from the centre plus the radius:

$$10+2 = 12$$

Answer

Common mistakes

• Forgetting to exclude the case where the denominator itself is zero. Setting only the numerator’s real part to zero technically requires the denominator $x^2+(y+2)^2\neq0$ — automatically true here since that would require $z=-2i$, which isn’t on the resulting circle anyway.
• Adding the radius when the point is inside the circle, or subtracting when it should be added. Since $(6,8)$ is at distance $10$ from the origin — well outside the radius-$2$ circle — the maximum distance is centre-distance plus radius; if the external point were instead inside the circle, the reasoning would need revisiting entirely.

Practise next

• Let $z$ satisfy $\operatorname{Re}\left(\dfrac{z-3i}{z+3i}\right)=0$. Find the maximum value of $|z-(8+6i)|$, using the same rationalise-and-geometry approach.