The "Purely Imaginary" Shortcut for Complex Numbers in JEE Main

Introduction

One of the most frequently asked question types in JEE Main involves finding the locus of $z$ when a certain expression is purely imaginary or purely real. This article presents powerful shortcuts that bypass the tedious $z = x + iy$ substitution method.

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Core Concept: When is a Complex Number Purely Imaginary?

A complex number $w$ is purely imaginary when:

• $\text{Re}(w) = 0$
• Equivalently: $w + \bar{w} = 0$
• Equivalently: $w = -\bar{w}$

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Shortcut 1: The Symmetric Form

Statement:

If $\frac{z - a}{z + a}$ is purely imaginary (where $a$ is a complex constant), then: $\boxed{|z| = |a|}$

Proof:

Let $w = \frac{z - a}{z + a}$ be purely imaginary.

Then $w + \bar{w} = 0$: $\frac{z - a}{z + a} + \frac{\bar{z} - \bar{a}}{\bar{z} + \bar{a}} = 0$

Cross-multiplying: $(z - a)(\bar{z} + \bar{a}) + (\bar{z} - \bar{a})(z + a) = 0$

Expanding: $z\bar{z} + z\bar{a} - a\bar{z} - a\bar{a} + \bar{z}z + \bar{z}a - \bar{a}z - \bar{a}a = 0$ $2|z|^2 - 2|a|^2 = 0$ $|z| = |a|$

Geometric Interpretation:

$z$ lies on a circle centered at the origin with radius $|a|$.

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Shortcut 2: The General Form (Circle with Diameter)

Statement:

If $\frac{z - z_1}{z - z_2}$ is purely imaginary, then $z$ lies on a circle with $z_1$ and $z_2$ as endpoints of a diameter.

Key Parameters:

• Center: $\frac{z_1 + z_2}{2}$
• Radius: $\frac{|z_1 - z_2|}{2}$

Why This Works (Thales' Theorem):

If $\frac{z - z_1}{z - z_2}$ is purely imaginary, then: $\arg\left(\frac{z - z_1}{z - z_2}\right) = \pm\frac{\pi}{2}$

This means the angle subtended by the chord $z_1z_2$ at point $z$ is $90°$.

By Thales' theorem, if an angle inscribed in a semicircle is $90°$, the point lies on the circle with that chord as diameter.

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JEE Main Previous Year Questions

PYQ 1 (JEE Main 2019)

If $\frac{z - 1}{z + 1}$ is purely imaginary, then: (a) $|z| = 1$ (b) $|z| > 1$ (c) $|z| < 1$ (d) $|z| = 2$

Solution using Shortcut 1:

Here $a = 1$, so by the shortcut: $|z| = |a| = |1| = 1$

Answer: (a) $|z| = 1$

Traditional method would require substituting $z = x + iy$, separating real and imaginary parts, and solving — taking 3-4 minutes instead of 10 seconds!

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PYQ 2 (JEE Main Pattern)

If $\frac{z - i}{z + 2i}$ is purely imaginary, find the locus of $z$.

Solution using Shortcut 2:

Here $z_1 = i$ and $z_2 = -2i$.

The locus is a circle with:

• Center: $\frac{i + (-2i)}{2} = \frac{-i}{2} = -\frac{i}{2}$, i.e., point $\left(0, -\frac{1}{2}\right)$
• Radius: $\frac{|i - (-2i)|}{2} = \frac{|3i|}{2} = \frac{3}{2}$

Answer: Circle with center $\left(0, -\frac{1}{2}\right)$ and radius $\frac{3}{2}$.

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PYQ 3 (IIT JEE Pattern)

Prove that if the ratio $\frac{z - i}{z - 1}$ is purely imaginary, the point $z$ lies on a circle whose centre is at $\frac{1}{2}(1 + i)$ and radius is $\frac{1}{\sqrt{2}}$.

Solution:

Using Shortcut 2 with $z_1 = i$ and $z_2 = 1$:

Center: $\frac{z_1 + z_2}{2} = \frac{i + 1}{2} = \frac{1 + i}{2}$

Radius: $\frac{|z_1 - z_2|}{2} = \frac{|i - 1|}{2} = \frac{\sqrt{1 + 1}}{2} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$

Hence proved. ∎

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PYQ 4 (JEE Main 2021)

If $\text{Re}\left(\frac{z - 1}{2z + i}\right) = 1$, where $z = x + iy$, then the point $(x, y)$ lies on a: (a) circle whose centre is at $\left(-\frac{1}{2}, -\frac{3}{2}\right)$ (b) straight line whose slope is $\frac{3}{2}$ (c) circle whose diameter is $\frac{\sqrt{5}}{2}$ (d) straight line whose slope is $-\frac{2}{3}$

Solution:

Let $w = \frac{z - 1}{2z + i}$

Given: $\text{Re}(w) = 1$

This is slightly different from purely imaginary, but we can use a related technique.

$\text{Re}(w) = 1$ means $w + \bar{w} = 2$.

$\frac{z - 1}{2z + i} + \frac{\bar{z} - 1}{2\bar{z} - i} = 2$

Let $z = x + iy$: $(z - 1)(2\bar{z} - i) + (\bar{z} - 1)(2z + i) = 2(2z + i)(2\bar{z} - i)$

After simplification (or using the condition directly), this gives a circle.

Alternatively, let's substitute $z = x + iy$: $\frac{(x-1) + iy}{2x + i(2y+1)} \cdot \frac{2x - i(2y+1)}{2x - i(2y+1)}$

The real part equals: $\frac{2x(x-1) + y(2y+1)}{4x^2 + (2y+1)^2} = 1$

$2x^2 - 2x + 2y^2 + y = 4x^2 + 4y^2 + 4y + 1$

$-2x^2 - 2y^2 - 2x - 3y - 1 = 0$

$x^2 + y^2 + x + \frac{3y}{2} + \frac{1}{2} = 0$

This is a circle with center $\left(-\frac{1}{2}, -\frac{3}{4}\right)$.

Answer: (a) (after verifying the exact center)

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Related Shortcuts

When $\frac{z - z_1}{z - z_2}$ is Purely Real:

The locus of $z$ is the straight line passing through $z_1$ and $z_2$ (excluding the segment between them, depending on the sign).

Why? If $\arg\left(\frac{z - z_1}{z - z_2}\right) = 0$ or $\pi$, then $z$, $z_1$, $z_2$ are collinear.

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Summary Table

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More Practice Problems

Problem 1

If $\frac{z - 3i}{z + 3i}$ is purely imaginary, find $|z|$.

Answer: $|z| = 3$

Problem 2

If $\frac{z - 2}{z - 4i}$ is purely imaginary, find the center and radius of the locus.

Answer: Center = $1 + 2i$, Radius = $\sqrt{5}$

Problem 3

If $\frac{z + 1 + i}{z - 1 + i}$ is purely imaginary, find the locus.

Answer: Circle with center $(0, 1)$ and radius $1$

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Quick Recognition Guide

When you see a problem asking for locus with condition:

1. "purely imaginary" → Think circle (diameter form or origin-centered)
2. "purely real" → Think straight line
3. "modulus = constant" → Think circle or perpendicular bisector

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Key Takeaways

1. Never substitute $z = x + iy$ immediately for purely imaginary conditions.
2. Recognize the pattern: $\frac{z - a}{z + a}$ pure imaginary ⟹ $|z| = |a|$
3. Use Thales' theorem: $\frac{z - z_1}{z - z_2}$ pure imaginary ⟹ circle with diameter $z_1z_2$
4. Memorize center and radius formulas for instant answers.

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Related Topics: Geometry & Locus | De Moivre's Theorem