Problem 37 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(z_{1}\) and \(z_{2}\) be two complex numbers such that \(\frac{z_{1}}{z_{2}}+\frac{z_{2}}{z_{1}}=1\), then (A) \(z_{1}, z_{2}\) are collinear (B) \(z_{1}, z_{2}\) and the origin from a right angled triangle (C) \(z_{1}, z_{2}\) and the origin form an equilateral triangle (D) None of these

Step-by-Step Solution

Answer

The complex numbers form an equilateral triangle with the origin.

Step 1: Convert the equation using polar form

To solve this problem, let's assume \(z_1 = r_1 e^{i \theta_1}\) and \(z_2 = r_2 e^{i \theta_2}\), where \(r_1\) and \(r_2\) are the magnitudes of \(z_1\) and \(z_2\), respectively, and \(\theta_1\), \(\theta_2\) are their arguments. This can be written as \(\frac{r_1 e^{i \theta_1}}{r_2 e^{i \theta_2}} + \frac{r_2 e^{i \theta_2}}{r_1 e^{i \theta_1}} = 1\).

Step 2: Simplify the expression

Simplify the expression: \(\frac{r_1}{r_2} e^{i(\theta_1 - \theta_2)} + \frac{r_2}{r_1} e^{i(\theta_2 - \theta_1)}=1\). This can be rewritten using cosine and sine components: \(\frac{r_1}{r_2}(\cos(\theta_1 - \theta_2) + i \sin(\theta_1 - \theta_2)) + \frac{r_2}{r_1}(\cos(\theta_2 - \theta_1) + i \sin(\theta_2 - \theta_1)) = 1\).

Step 3: Solve for real and imaginary parts

For the equality to hold true, both the real part and the imaginary part must equal zero (since the other side is purely real). Focus on the real part first: \(\frac{r_1}{r_2}\cos(\theta_1 - \theta_2) + \frac{r_2}{r_1}\cos(\theta_2 - \theta_1) = 1\). This simplifies to \(\frac{r_1}{r_2} + \frac{r_2}{r_1} = 1\).

Step 4: Use trigonometric identities

Since \(\cos(\theta) = \cos(-\theta)\), it implies \(\cos(\theta) = 1/2\). This leads us to two angles, \(\theta = \pm \pi/3\). Thus, the complex numbers must have an angular separation of \(\pm \pi/3\), suggesting they lie symmetrically about the origin as vertices of an equilateral triangle.

Step 5: Analyze geometric interpretation

Given \(\theta_1 - \theta_2 = \pm \pi/3\), \(z_1, z_2\) and the origin form an equilateral triangle in the complex plane. Therefore, the numbers lie such that their angle difference aids in forming the same magnitude along equal angular distances.

Key Concepts

Polar Form

Complex numbers can be expressed in different forms, and the polar form is particularly useful when dealing with multiplication and division. In polar form, a complex number is represented as \( z = r e^{i\theta} \), where \( r \) is the magnitude (or modulus) of the complex number, and \( \theta \) is the argument (or angle) from the positive real axis.

• This form helps simplify products and quotients of complex numbers.
• The magnitude gives us the distance from the origin in the complex plane.
• The argument tells us the angle from the real axis, measured in the counterclockwise direction.

For example, if we have two complex numbers \( z_1 = r_1 e^{i\theta_1} \) and \( z_2 = r_2 e^{i\theta_2} \), dividing them in polar form becomes straightforward: \( \frac{z_1}{z_2} = \frac{r_1}{r_2} e^{i(\theta_1 - \theta_2)} \). This shows how both angles and magnitudes are handled separately, which can greatly simplify the operation.
Understanding polar form is key to tackling problems involving complex number operations, especially when converting them into easy-to-manage expressions like the one in our exercise.

Geometric Interpretation

In the complex plane, interpreting the equation through geometry provides insightful understanding. The problem around \( \frac{z_1}{z_2} + \frac{z_2}{z_1} = 1 \) becomes easier to visualize if we think about the positioning and shape created by \( z_1 \), \( z_2 \), and the origin.

• The solution indicates that \( z_1 \), \( z_2 \), and the origin form points of an equilateral triangle.
• This geometric interpretation arises from the symmetric properties of the complex numbers involved.

Since an equilateral triangle is defined by equal sides and equal angles of \( 60^\circ \) or \( \pi/3 \) radians, the condition \( \theta_1 - \theta_2 = \pm \pi/3 \) shows that the angle between the vectors representing these numbers is \( 60^\circ \).
This means that not only do the magnitudes of these complex numbers play a role in their equality, but also their positional relationship in the plane. Interpreting complex equations in this visual manner can often make them less abstract and more intuitive.

Trigonometry in Complex Numbers

Trigonometry and complex numbers combine elegantly in many mathematical problems, as seen in our exercise where it provides a method to solve and understand the equality requirement. The polar form helped transition from complex division to an expression that could be tackled with trigonometry.

• The cosine and sine components are crucial when converting between the algebraic and polar forms.
• In our problem, the real part of the equation involves \( \cos(\theta_1 - \theta_2) \) which leads to critical angle analysis.

Using trigonometric identities such as \( \cos(\theta) = \cos(-\theta) \) allows for simplification of complex expressions. In this particular example, it led to \( \cos(\theta) = \frac{1}{2} \), which pointed directly to angles \( \theta = \pm \pi/3 \).
This angle separation is characteristic of an equilateral triangle arrangement. Therefore, a foundational understanding of trigonometry is vital when dealing with complex number problems, especially when those problems involve angular relationships and magnitudes. By mastering this connection, students can more easily transition between purely algebraic solutions and geometric interpretations.