Let z = (√3/2 + i/2)⁵ + (√3/2 - i/2)⁵. If R(z) and I(z) respectively denote the real and imaginary p

Q: Why can't I just use the Binomial Theorem to expand these terms?

You could, but it is a trap! JEE examiners love to test your ability to recognize patterns. Expanding $(\frac{\sqrt{3}}{2} + \frac{i}{2})^5$ is a recipe for calculation errors and wasted time. Always look for the geometric shortcut first; complex numbers are meant to be handled with rotation, not just algebra.

Q: Why does the imaginary part vanish at the end?

Because the two terms are complex conjugates. When you add a complex number $z$ to its conjugate $\bar{z}$, the imaginary parts are $i\sin(\theta)$ and $-i\sin(\theta)$, which cancel out perfectly. It is a beautiful property of symmetry on the complex plane.

Q: How do I handle the angle $\frac{5\pi}{6}$?

Think of the unit circle. $\frac{5\pi}{6}$ is in the second quadrant. In the second quadrant, cosine is negative. Use the reference angle $\frac{\pi}{6}$ and apply the ASTC rule (All Students Take Calculus) to get $-\cos(\frac{\pi}{6})$. It is all about visualizing the position of the vector.

Question

Let $z = \left(\frac{\sqrt{3}}{2} + \frac{i}{2}\right)^5 + \left(\frac{\sqrt{3}}{2} - \frac{i}{2}\right)^5$. If $R(z)$ and $I(z)$ respectively denote the real and imaginary parts of $z$, then :

Options:

(A) R(z) > 0 and I(z) > 0

(B) R(z) < 0 and I(z) > 0

(C) R(z) = -3

(D) I(z) = 0

Sigma Percentile · Accepted Answer

### The Art of Avoiding the Grind: A Complex Number Odyssey

Welcome, future engineer! Today we are tackling a problem that, at first glance, looks like a nightmare of arithmetic. You see that power of 5, and your brain immediately screams, "Binomial expansion!"

But wait—stop. In the world of JEE Advanced, brute force is rarely the intended path. Let's look at the beauty hidden in the structure of:

$$z = \left( \frac{\sqrt{3}}{2} + \frac{i}{2} ight)^5 + \left( \frac{\sqrt{3}}{2} - \frac{i}{2} ight)^5$$

### Phase 1: The Geometric Insight

Before we touch a pen to paper, let's observe. We have two complex numbers, which we can define as:

$$z_1 = \frac{\sqrt{3}}{2} + \frac{i}{2} \quad 	ext{and} \quad z_2 = \frac{\sqrt{3}}{2} - \frac{i}{2}$$

Look at them. They are mirror images across the real axis, which means $z_2$ is the complex conjugate of $z_1$, or $z_2 = \bar{z}_1$.

This is not a coincidence; it is a gift from the problem setter. By recognizing this symmetry, we have already saved ourselves from a mountain of tedious algebra.

### Phase 2: The Power of Euler

Now, how do we handle the power of 5? We choose the path of the master: Euler's form. We know that any complex number can be written as $z = r e^{i	heta}$.

For $z_1$, the modulus $r$ is:

$$r = \sqrt{\left( \frac{\sqrt{3}}{2} ight)^2 + \left( \frac{1}{2} ight)^2} = 1$$

The angle $	heta$ is found by $	an(	heta) = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}}$, which gives us $	heta = \frac{\pi}{6}$. So, $z_1 = e^{i\frac{\pi}{6}}$.

Because $z_2$ is the conjugate, its angle is simply the negative of $z_1$'s angle. Thus, $z_2 = e^{-i\frac{\pi}{6}}$.

Suddenly, the problem has transformed from a messy binomial expansion into a clean, elegant exponential expression:

$$z = \left( e^{i\frac{\pi}{6}} ight)^5 + \left( e^{-i\frac{\pi}{6}} ight)^5$$

### Phase 3: The Elegant Cancellation

Here comes the magic of De Moivre's Theorem. When we raise an exponential to a power, we simply multiply the exponent by that power.

Our expression becomes:

$$z = e^{i\frac{5\pi}{6}} + e^{-i\frac{5\pi}{6}}$$

Look at this! We have the sum of a complex number and its conjugate. We know the identity $e^{i	heta} + e^{-i	heta} = 2\cos(	heta)$.

The imaginary parts, the sine components, are destined to cancel out. We are left with:

$$z = 2\cos\left( \frac{5\pi}{6} ight)$$

### The Final Act

Now, we just need to evaluate $\cos\left( \frac{5\pi}{6} ight)$. We know $\frac{5\pi}{6}$ is in the second quadrant, where cosine is negative.

The reference angle is $\frac{\pi}{6}$, so:

$$\cos\left( \frac{5\pi}{6} ight) = -\cos\left( \frac{\pi}{6} ight) = -\frac{\sqrt{3}}{2}$$

Substituting this back, we get:

$$z = 2 \cdot \left( -\frac{\sqrt{3}}{2} ight) = -\sqrt{3}$$

The imaginary part is zero, and the real part is negative. We have arrived at the solution not by grinding through calculations, but by understanding the geometric soul of the complex number.

The final answer is **$-\sqrt{3}$**. Keep this mindset, and no problem will ever be too daunting.

Animated Solution for Mathematics - Complex Numbers: Let $z = (\frac{3}{2} + \frac{i}{2})^{5} + (\frac{3}{2} - \frac{i}{2})^{5}$ . If $R (z)$ and $I (z)$ respectively denote the real and imaginary parts of $z$ , then :

Select Answer:

Visualized Solution

Identify the Complex Numbers

Conjugate Relationship

Euler's Form

Modulus of $z_{1}$

Argument of $z_{1}$

Euler Form of $z_{1}$ and $z_{2}$

Substitute into Original Expression

De Moivre's Theorem

Apply the Power of 5

Sum of Conjugate Exponentials

Simplify the Expression

Evaluate $cos (\frac{5 π}{6})$

Final Value of $z$

Analyze Real and Imaginary Parts

The Sigma Insight: Argand Plane and Polar Representation

The Art of Avoiding the Grind

Phase 1

Phase 2

Phase 3

The Final Act

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