# Tricks to Solve JEE Complex Numbers Problems

Last updated on August 1, 2024 by Rajesh Saharan

Complex numbers are a crucial part of the JEE syllabus, often seen as challenging due to their abstract nature. However, with the right tricks and strategies, solving complex numbers problems can become much more manageable in JEE preparation. In this article, we’ll explore some effective tricks and provide example questions to illustrate these techniques.

### Understanding the Basics

Before diving into the tricks, it’s essential to have a solid grasp of the basics of complex numbers. A complex number is typically represented as z = a + bi, where:

• a is the real part
• b is the imaginary part
• i is the imaginary unit, satisfying i2 = -1

## Tricks to Solve JEE Complex Numbers Problems

### 1. Conjugate Method

The conjugate of a complex number z = a + bi is z̅ = a – bi. This method is particularly useful for division and finding magnitudes.

#### Example :

Simplify (1 + i) / (1 – i).

#### Solution :

Multiply the numerator and the denominator by the conjugate of the denominator:

(1 + i) / (1 – i) × (1 + i) / (1 + i) = [(1 + i)2 / (1 – i)(1 + i)] = [1 + 2i + i2 / 1 + 1] = [2i / 2] = i

### 2. Euler’s Formula

Euler’s formula states that e = cosθ + i sinθ. This is particularly useful in converting between exponential and trigonometric forms.

#### Example :

Express z = 1 + i in polar form.

#### Solution :

First, find the magnitude r:

r = √(12 + 12) = √2

Next, find the argument θ:

θ = tan-1(1 / 1) = π/4

Thus, in polar form, z = √2 (cos π/4 + i sin π/4) or z = √2 eiπ/4.

### 3. De Moivre’s Theorem

De Moivre’s Theorem is useful for raising complex numbers to a power and finding roots. It states that (r e)n = rn einθ.

Find (1 + i)4.

#### Solution :

Convert to polar form:

1 + i = √2 eiπ/4

Raise to the power 4 using De Moivre’s Theorem:

(1 + i)4 = (√2 eiπ/4)4 = (√2)4 ei 4 π/4 = 4 e = 4(-1) = -4

### 4. Roots of Unity

Roots of unity are solutions to the equation zn = 1. The n-th roots of unity are given by zk = e2kπi / n for k = 0, 1, 2, …, n-1.

#### Example :

Find the cube roots of unity.

#### Solution :

The cube roots of unity satisfy z3 = 1. These are given by:

zk = e2kπi / 3 for k = 0, 1, 2

Thus, the cube roots are:

z0 = 1, z1 = e2πi / 3 = -1/2 + i√3/2, z2 = e4πi / 3 = -1/2 – i√3/2

### 5. Geometric Interpretation

Understanding the geometric representation of complex numbers on the Argand plane can provide insights into their properties and operations.

#### Example :

If z1 = 3 + 4i and z2 = 1 + 2i, find the geometric interpretation of z1 + z2 and z1 – z2.

#### Solution :

Calculate the sums and differences:

z1 + z2 = (3 + 1) + (4 + 2)i = 4 + 6i

z1 – z2 = (3 – 1) + (4 – 2)i = 2 + 2i

Geometrically, z1 + z2 represents a vector addition resulting in a new point, and z1 – z2 represents a vector subtraction. Plotting these on the Argand plane shows the resultant vectors from the origin to the points (4, 6) and (2, 2), respectively.

### 6. Modulus and Argument

The modulus of a complex number z = a + bi is |z| = √(a2 + b2), and the argument is arg(z) = tan-1(b/a).

#### Example :

Find the modulus and argument of z = -1 + i.

#### Solution :

Calculate the modulus:

|z| = √((-1)2 + 12) = √2

Calculate the argument:

arg(z) = tan-1(-1) = -π/4 (since z is in the second quadrant, we add π)

Thus, arg(z) = π – π/4 = 3π/4.

### Mastering Complex Numbers for JEE

Mastering complex numbers for the JEE requires a blend of theoretical knowledge and practical tricks. By understanding and applying these tricks, you can simplify complex number problems and solve them efficiently. Practice these techniques with various problems to gain confidence and improve your problem-solving skills.

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