Problem 15 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(z=\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^{5}+\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^{5}\), then (A) \(\operatorname{Re}(z)=0\) (B) \(I_{m}(z)=0\) (C) \(\operatorname{Re}(z)>0, I_{s}(z)>0\) (D) \(\operatorname{Re}(z)>0, I_{m}(z)<0\)

Step-by-Step Solution

Answer

(B) \(I_{m}(z)=0\)

Step 1: Define the Expression

The expression given is \(z = \left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^{5}+\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^{5}\). We need to find the real and imaginary parts of this expression.

Step 2: Identify the Complex Numbers

Identify the two complex numbers: \(a = \frac{\sqrt{3}}{2} + \frac{i}{2}\) and \(b = \frac{\sqrt{3}}{2} - \frac{i}{2}\). These are complex conjugates of each other.

Step 3: Use De Moivre's Theorem

Since \(a\) and \(b\) are on the unit circle (as they satisfy \(\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2 = 1\)), express them in polar form: \(\frac{\sqrt{3}}{2} + \frac{i}{2} = e^{i\frac{\pi}{6}}\) and \(\frac{\sqrt{3}}{2} - \frac{i}{2} = e^{-i\frac{\pi}{6}}\). Apply De Moivre's Theorem: \( \left(e^{i\frac{\pi}{6}}\right)^5 = e^{i\frac{5\pi}{6}}\) and \(\left(e^{-i\frac{\pi}{6}}\right)^5 = e^{-i\frac{5\pi}{6}}\).

Step 4: Simplify the Expression

Combine the results from De Moivre's theorem: \(z = e^{i\frac{5\pi}{6}} + e^{-i\frac{5\pi}{6}}\). Simplify using the identity \(e^{ix} + e^{-ix} = 2\cos(x)\), resulting in \(z = 2\cos\left(\frac{5\pi}{6}\right)\).

Step 5: Calculate \(\cos\left(\frac{5\pi}{6}\right)\)

Calculate \(\cos\left(\frac{5\pi}{6}\right)\): Since \(\cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}\) (as \(\frac{5\pi}{6}\) is in the second quadrant where cosine is negative), we find that \(z = 2(-\frac{\sqrt{3}}{2}) = -\sqrt{3}\).

Step 6: Determine the Real and Imaginary Parts of \(z\)

The expression \(z = -\sqrt{3}\) confirms that \(z\) is a purely real number with \(\operatorname{Re}(z) = -\sqrt{3}\) and \(\operatorname{Im}(z) = 0\).

Key Concepts

De Moivre's Theorem

De Moivre's Theorem is a powerful tool for solving complex number equations. It helps us find powers of complex numbers that are expressed in polar form.
This theorem states that if you have a complex number in polar form, say \( z = r e^{i\theta} \), then \( z^n = r^n e^{i n\theta} \). Let's break that down with an example. Suppose we have \( z = e^{i\frac{\pi}{6}} \), which describes a point on the unit circle (so \( r = 1 \)). If we wish to calculate \( z^5 \), we simply multiply the angle inside the expression by 5, getting \( e^{i\frac{5\pi}{6}} \).
This angle representation enlightens us about how complex numbers behave when raised to powers, which isn't as intuitive in the rectangular form. By applying De Moivre's Theorem, it becomes much simpler to calculate higher powers of complex numbers.

Complex Conjugates

Complex conjugates are a fascinating concept when working with complex numbers. If you have a complex number \( z = a + bi \), its complex conjugate is \( \bar{z} = a - bi \).
These pairs have a unique relationship: their product always results in a real number, specifically \( a^2 + b^2 \). This property is very useful when solving problems, as it simplifies computations.In the exercise, we see complex numbers \( a = \frac{\sqrt{3}}{2} + \frac{i}{2} \) and \( b = \frac{\sqrt{3}}{2} - \frac{i}{2} \), where one is the conjugate of the other.
When you sum their powers, the imaginary components cancel out due to their symmetrical nature on the complex plane, leaving only the real component. Knowing this can save significant time when identifying the real and imaginary parts of expressions involving complex conjugates.

Polar Form

The polar form of complex numbers is like translating the rectangular coordinates into a language that speaks better with angles and distances. In polar form, a complex number \( z = x + yi \) is expressed as \( z = r(\cos \theta + i\sin \theta) \) or \( z = r e^{i\theta} \).
Here, \( r \) is the modulus or magnitude of the complex number, and \( \theta \) is the argument or angle made with the positive real axis.Polar form simplifies the multiplication and division of complex numbers because angles can be added or subtracted, and magnitudes can be multiplied or divided.
In the context of the exercise, the expression \( \frac{\sqrt{3}}{2} + \frac{i}{2} \) can be converted into polar form as \( e^{i\frac{\pi}{6}} \) because it lies on the unit circle (\( r = 1 \), \( \theta = \frac{\pi}{6} \)). Understanding polar form will help you easily visualize and manipulate complex numbers, especially when using De Moivre's Theorem.