Problem 133 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(z_{1}\) and \(\bar{z}_{1}\) represent adjacent vertices of a regular polygon of \(n\) sides and if \(\frac{\operatorname{Im}\left(z_{1}\right)}{\operatorname{Re}\left(z_{1}\right)}=\sqrt{2}-1\), then \(n\) is equal to (A) 4 (B) 8 (C) 16 (D) None of these

Step-by-Step Solution

Answer

None of these

Step 1: Express the Condition Geometrically

If \(z_1\) and \(\bar{z}_1\) represent adjacent vertices of a regular polygon, then \(z_1\) and \(\bar{z}_1\) are symmetric with respect to the real axis.This means the angle between these adjacent vertices is \(\frac{2\pi}{n}\). Since \(\bar{z}_1 = \overline{r e^{i\theta}} = r e^{-i\theta}\), which implies \(\bar{z}_1\) is the reflection of \(z_1\).

Step 2: Calculate the Argument Condition

For adjacent vertices, the argument of \(\bar{z}_1\) should be \(-\theta\) and for the polygon symmetry, the angle \(\theta\) between \(z_1\) and \(-\theta\) is \(\frac{2\pi}{n}\).Hence, \(2\theta = \frac{2\pi}{n}\), giving \(\theta = \frac{\pi}{n}\).

Step 3: Use Given Imaginary to Real Ratio

Given \(\frac{\operatorname{Im}(z_1)}{\operatorname{Re}(z_1)} = \sqrt{2} - 1\), use \(z_1 = re^{i\theta} = r(\cos \theta + i \sin \theta)\).This gives \(\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \sqrt{2} - 1\), where \(\theta = \frac{\pi}{n}\).

Step 4: Solve for \(n\) using Tangent Identity

The tangent of \(\theta = \frac{\pi}{n}\) must satisfy \(\tan\theta = \sqrt{2}-1\). Since \(2\theta = \frac{2\pi}{n}\), we must find \(n\) such that \(\theta = 15^\circ\) (since \(\tan(15^\circ) = \sqrt{2} - 1\)).

Step 5: Evaluate \(n\) Value

Find \(n\) such that \(\frac{\pi}{n} = 15^\circ\). This implies \(n = \frac{180}{15} = 12\). Since 12 is not a provided option, the answer must be 'None of these'.

Key Concepts

Regular Polygon

A regular polygon is a geometrical figure with all sides of equal length and all angles of equal measure. Understanding the properties of regular polygons is crucial in geometry and helps in visualizing complex problems, such as those involving complex numbers. In a regular polygon with \(n\) sides, each angle between adjacent vertices can be calculated as \(\frac{2\pi}{n}\). This symmetry makes the regular polygon a significant concept in complex numbers, where every vertex can be represented as a point on a circle in the complex plane.

Key properties of a regular polygon include:

• Equal-length sides and angles which ensure symmetry.
• Vertices lie on a common circle, creating uniform arcs between them.
• Each central angle is \(\frac{2\pi}{n}\), where \(n\) is the number of sides.

For example, if you have a regular hexagon, the angle between any two adjacent points (vertices) on the circle is \(\frac{2\pi}{6} = \frac{\pi}{3}\). This uniform structure is mirrored in complex numbers, where vertices can be expressed through angles and radii, aiding in more straightforward calculations.

In our exercise, we look at adjacent vertices \(z_1\) and \(\bar{z}_1\), which are reflections over the real axis. The characteristics of regular polygons help deduce properties of these vertices.

Imaginary to Real Ratio

The imaginary to real ratio in complex numbers provides insights into the angle that the number makes with the real axis. This ratio can be expressed as \(\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\), where \(z = x + yi\). This tells us the tangent of the angle formed.

In the given exercise, the value \(\sqrt{2} - 1\) provides a specific tangent for the angle \(\theta\). This means that our complex number lies at an angle \(\theta\) from the positive real axis. It's important to note that since this ratio defines the tangent:

• \(\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \sqrt{2} - 1\).

By understanding the ratio, students can convert this information into an angle measurement that they can work with, herding towards solving angular problems, like those involving rotations and reflections in the complex plane.

Such ratios are also pivotal when dealing with polygons, as they help dictate the geometrical arrangement of vertices on the complex plane.

Tangent Identity

The tangent identity refers to the trigonometric identity which relates the tangent function with angles. Knowing this identity allows the transition between angle measurements and simple trigonometric ratios, crucial for solving equations involving complex numbers.

In the exercise problem, we are given that \(\tan(\theta) = \sqrt{2} - 1\) for \(\theta = \frac{\pi}{n}\). This means that:

• The tangent identity allows us to explore relationships between different angle representations.
• Since \(\tan(15^\circ) = \sqrt{2} - 1\), \(\theta\) is thus 15 degrees.

With this insight, students use the tangent identity to establish equations based on angular measurements and solve them effectively.

This direct relationship between angles and their tangent identities allows for a better grasp of how trigonometric concepts simplify complex equations. By understanding the inherent geometric properties of tangents, solving for \(\theta\) becomes more straightforward.

Symmetry with Real Axis

Symmetry with the real axis is significant in understanding complex numbers. If a complex number and its conjugate are symmetric with respect to the real axis, it means when we reflect one vertex across the real axis, it mirrors the other.

For example, if you have \(z = a + bi\), its complex conjugate \(\bar{z} = a - bi\) lies directly opposite \(z\), mirrored over the real axis. This geometrical relationship:

• Helps in solving problems by breaking them into simpler symmetrical components.
• Infers properties of symmetrical shapes like regular polygons.

In reference to the exercise, \(z_1\) and \(\bar{z}_1\) being symmetrical over the real axis tells us they form a part of the regular polygon. Such symmetric properties reveal not only structural but also intrinsic aesthetic qualities of the numerical solutions in complex analysis.

Understanding this symmetry allows students to visualize complex numbers in a real-world geometric context, aiding in deeper comprehension and problems solving.