JEE Advanced Challenge: Mastering Complex Number Proofs & Tough Locus Problems
Rigorous proofs, advanced locus problems, and problem-solving strategies for JEE Advanced Complex Numbers section.
📚 What You'll Master
Why This Challenge Matters for JEE Advanced
Complex Numbers contribute 12-15% weightage in JEE Advanced Mathematics. The proofs and locus problems separate top rankers from average performers. Mastering these will give you:
- Conceptual clarity to tackle any complex number variation
- Geometric intuition for complex plane problems
- Proof-writing skills for theoretical questions
- Time management in solving complex locus problems
- 8-12 marks secured in every JEE Advanced paper
1. Triangle Inequalities & Modulus Properties
Proof: Triangle Inequality for Complex Numbers
Prove that $|z_1 + z_2| \leq |z_1| + |z_2|$ for all complex numbers $z_1, z_2$
Proof Strategy:
Step 1: Let $z_1 = a + bi$, $z_2 = c + di$ where $a,b,c,d \in \mathbb{R}$
Step 2: Compute $|z_1 + z_2|^2 = (a+c)^2 + (b+d)^2$
Step 3: Expand: $a^2 + 2ac + c^2 + b^2 + 2bd + d^2$
Step 4: Compare with $(|z_1| + |z_2|)^2 = a^2+b^2+c^2+d^2 + 2\sqrt{(a^2+b^2)(c^2+d^2)}$
Step 5: Use Cauchy-Schwarz: $ac + bd \leq \sqrt{(a^2+b^2)(c^2+d^2)}$
Step 6: Therefore, $|z_1 + z_2|^2 \leq (|z_1| + |z_2|)^2$
Step 7: Taking square roots (all terms non-negative): $|z_1 + z_2| \leq |z_1| + |z_2|$
Geometric Interpretation:
In the complex plane, this inequality states that the length of one side of a triangle is less than or equal to the sum of the lengths of the other two sides.
Proof: Reverse Triangle Inequality
Prove that $| |z_1| - |z_2| | \leq |z_1 - z_2|$ for all complex numbers $z_1, z_2$
Proof Approach:
Method 1: Use $|z_1| = |(z_1 - z_2) + z_2| \leq |z_1 - z_2| + |z_2|$
Thus, $|z_1| - |z_2| \leq |z_1 - z_2|$
Similarly, $|z_2| - |z_1| \leq |z_2 - z_1| = |z_1 - z_2|$
Combining: $| |z_1| - |z_2| | \leq |z_1 - z_2|$
2. Advanced Locus Problems: Circles
Locus Problem: Apollonius Circle
If $|z - 3 - 4i| = 2|z + 2 - i|$, find the locus of $z$ and its center and radius.
Solution Strategy:
Step 1: Let $z = x + iy$
Step 2: Substitute: $|(x-3) + i(y-4)| = 2|(x+2) + i(y-1)|$
Step 3: Square both sides: $(x-3)^2 + (y-4)^2 = 4[(x+2)^2 + (y-1)^2]$
Step 4: Expand and simplify:
$x^2 - 6x + 9 + y^2 - 8y + 16 = 4(x^2 + 4x + 4 + y^2 - 2y + 1)$
$x^2 + y^2 - 6x - 8y + 25 = 4x^2 + 16x + 16 + 4y^2 - 8y + 4$
Step 5: Rearrange: $3x^2 + 3y^2 + 22x + 5 = 0$
Step 6: Complete the square:
$x^2 + y^2 + \frac{22}{3}x + \frac{5}{3} = 0$
$(x + \frac{11}{3})^2 + y^2 = \frac{121}{9} - \frac{5}{3} = \frac{106}{9}$
Step 7: Center: $(-\frac{11}{3}, 0)$, Radius: $\frac{\sqrt{106}}{3}$
🚀 Advanced Problem-Solving Strategies
For Proofs:
- Always start with algebraic form $z = x + iy$
- Use geometric interpretations for intuition
- Apply triangle inequalities systematically
- Remember $|z|^2 = z\overline{z}$ is powerful
For Locus Problems:
- Identify the type: circle, line, ellipse, etc.
- Use geometric distance interpretations
- Complete the square for circle equations
- Check special cases and boundaries
6 More Advanced Proofs & 5 Locus Problems Available
Includes De Moivre's Theorem proofs, argument properties, ellipse locus, and mixed advanced problems
📝 Quick Self-Test
Try these JEE Advanced level problems:
1. Prove that $|z_1 + z_2|^2 + |z_1 - z_2|^2 = 2(|z_1|^2 + |z_2|^2)$
2. Find the locus of $z$ if $\arg\left(\frac{z-1}{z+1}\right) = \frac{\pi}{4}$
3. If $|z| = 1$ and $z \neq \pm1$, prove that $\frac{z-1}{z+1}$ is purely imaginary
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