Maximum/Minimum Values Using Triangle Inequality in JEE Main – Ultimate Guide

Introduction

JEE Main frequently tests optimization problems with complex numbers. While calculus approaches exist, the Triangle Inequality method provides faster, more elegant solutions with geometric insight.

---

Core Inequalities – Memorize These!

Triangle Inequality (Two Forms)

1. Upper Bound (For Maximum): $|z_1 + z_2| \leq |z_1| + |z_2|$ Equality when: $z_1$ and $z_2$ have same direction ($\arg(z_1) = \arg(z_2)$)

2. Lower Bound (For Minimum): $|z_1 + z_2| \geq \big||z_1| - |z_2|\big|$ Equality when: $z_1$ and $z_2$ have opposite directions ($\arg(z_1) = \arg(z_2) + \pi$)

---

Geometric Picture – Visualize This!

For $|z| = r$ (circle centered at origin):

• $|z + a|$ = distance from $z$ to $-a$
• Maximum: When $O$, $z$, and $-a$ collinear with $z$ farthest from $-a$
• Minimum: When $O$, $z$, and $-a$ collinear with $z$ closest to $-a$

---

Standard Problems – Quick Formulas

Case 1: $|z| = r$ fixed

$| |a| - r | \ \leq\ |z + a|\ \leq\ |a| + r$

Example: If $|z| = 3$, max of $|z + 4i| = 3 + 4 = 7$, min = $|4 - 3| = 1$

---

Case 2: $|z - z_0| \leq r$ (disk)

$\big||z_0 - a| - r\big| \ \leq\ |z - a|\ \leq\ |z_0 - a| + r$

Example: If $|z - (2+3i)| \leq 4$, range of $|z - i|$: Distance between centers: $|(2+3i) - i| = |2+2i| = 2\sqrt{2}$ Range: $[|2\sqrt{2} - 4|, \ 2\sqrt{2} + 4] = [4 - 2\sqrt{2}, \ 4 + 2\sqrt{2}]$

---

Case 3: $|z - z_1| = |z - z_2|$ (perpendicular bisector)

No general formula – solve geometrically by finding point on bisector closest/farthest from target.

---

JEE Main PYQs Solved Strategically

PYQ 1: JEE Main 2020

If $|z + 4| \leq 3$, maximum of $|z + 1|$ is?

30-second solution:

• $z$ lies in disk: center $-4$, radius $3$
• We want distance from $z$ to $-1$
• Distance between centers: $|-4 - (-1)| = 3$
• Max distance = center distance + radius = $3 + 3 = 6$

Answer: $6$

---

PYQ 2: JEE Main 2014

If $\left|z - \frac{4}{z}\right| = 2$, maximum $|z|$ is?

Triangle inequality approach: Let $|z| = r$. Then: $\left|z - \frac{4}{z}\right| \geq \left|r - \frac{4}{r}\right|$ So $\left|r - \frac{4}{r}\right| \leq 2$

Solve: $-2 \leq r - \frac{4}{r} \leq 2$

From right: $r - \frac{4}{r} \leq 2$ $r^2 - 2r - 4 \leq 0$ $r \leq 1 + \sqrt{5}$

From left: $r - \frac{4}{r} \geq -2$
$r^2 + 2r - 4 \geq 0$ $r \geq \sqrt{5} - 1$

Maximum: $r_{\text{max}} = 1 + \sqrt{5}$

Answer: $\sqrt{5} + 1$

---

PYQ 3: JEE Main 2021

For $|z - 2 - 2i| \leq 1$, maximum of $|3iz + 6|$ attained at $a+ib$. Find $a+b$.

Strategic solution: $|3iz + 6| = 3|iz + 2| = 3|i||z - 2i| = 3|z - 2i|$

So maximize $|z - 2i|$ with $|z - (2+2i)| \leq 1$

Geometrically:

• Disk center: $2+2i$, radius $1$
• Target point: $2i = (0,2)$
• Distance between centers: $|(2,2) - (0,2)| = 2$
• Max distance = $2 + 1 = 3$

Achieved at: $(2,2) + 1 \cdot \text{unit vector toward } (0,2)$ Vector = $(0,2) - (2,2) = (-2,0)$ Unit vector = $(-1,0)$ Point = $(2,2) + (-1,0) = (1,2)$

So $a=1, b=2$, $a+b=3$

Wait — check: $|(1,2) - (0,2)| = 1$, not 3! Mistake.

Actually: We want point on circle farthest from $(0,2)$. Line through centers: $(2,2)$ to $(0,2)$ is horizontal. Farthest point = $(2,2) + (1,0) = (3,2)$ $|(3,2) - (0,2)| = 3$ ✓

So $a=3, b=2$, $a+b=5$

Answer: $5$

---

Advanced Pattern: $\left|z + \frac{k}{z}\right| = a$

Special Formula (Memorize!):

For $\left|z + \frac{k}{z}\right| = a$: $|z|_{\text{max}} = \frac{a + \sqrt{a^2 + 4|k|}}{2}$ $|z|_{\text{min}} = \frac{\sqrt{a^2 + 4|k|} - a}{2}$

Derivation hint: Use $|z + k/z| \geq \left||z| - |k|/|z|\right|$ and $|z + k/z| \leq |z| + |k|/|z|$

---

Common Mistake Alert!

Wrong: Assuming $\left|z - \frac{k}{z}\right|$ gives same formula as $\left|z + \frac{k}{z}\right|$

Correct: For $\left|z - \frac{k}{z}\right| = a$: $|z|_{\text{max/min}} = \frac{\sqrt{a^2 + 4|k|} \pm a}{2}$

Note the ± order is swapped!

---

Optimization Framework

Step-by-Step Approach:

1. Identify constraint: Circle ($|z-z_0|=r$) or disk ($|z-z_0|\leq r$)
2. Identify objective: Expression to maximize/minimize
3. Geometrize: Interpret as distance between points
4. Find extremes: Use collinearity principle
5. Verify: Check if extreme points satisfy constraint

---

Special Cases & Tricks

1. Linear Combinations

For $|\alpha z + \beta|$, factor: $|\alpha|\left|z + \frac{\beta}{\alpha}\right|$

2. Reciprocal Relations

If $|z| = 1$, then $\bar{z} = 1/z$ Useful for expressions like $|z + 1/z|$

3. Rotation Effects

Multiplying by $i$ rotates by 90°: $|iz + a| = |z - ai|$

---

Practice Problems

Problem 1: $|z| = 2$, find max and min of $|z - 3 + 4i|$

Solution: $|3-4i| = 5$ Max = $5 + 2 = 7$ Min = $|5 - 2| = 3$

---

Problem 2: $|z - i| \leq 2$, find max of $|z + 3|$

Solution: Center distance: $|i - (-3)| = |3+i| = \sqrt{10}$ Max = $\sqrt{10} + 2$

---

Problem 3: If $\left|z + \frac{9}{z}\right| = 6$, find $|z|_{\text{max}}$

Solution: Using formula: $|z|_{\text{max}} = \frac{6 + \sqrt{36 + 36}}{2} = \frac{6 + \sqrt{72}}{2} = 3 + 3\sqrt{2}$

---

Quick Reference Table

---

Exam Strategy

1. Sketch mentally: Visualize circles and points
2. Check extremes: Aligned vs anti-aligned cases
3. Use symmetry: Often simplifies calculations
4. Time target: 1-2 minutes max for these problems
5. Verify: Plug back if unsure

---

Pro Tip: For $|z + a| + |z + b|$ type problems, think ellipse definitions rather than triangle inequality.

---

Related: Complex Geometry | Rotation Methods | Quick Formula Sheet