If z = 2 + 3i, then z⁵ + (z̄)⁵ is equal to :

Q: Why can't I just calculate $z^5$ and $(\bar{z})^5$ separately and add them?

You absolutely can, but it is a trap! Expanding $(2+3i)^5$ involves complex binomial coefficients and powers of $i$ that are prone to arithmetic errors. Using the binomial identity is not just faster; it is mathematically elegant and minimizes the chance of making a sign error.

Q: Why do the imaginary parts always cancel out in this type of problem?

Great observation! Because $z$ and $\bar{z}$ are conjugates, their sum $z + \bar{z}$ is $2Re(z)$, which is real. Similarly, any power $z^n + (\bar{z})^n$ will always result in a purely real number because the imaginary components, which arise from odd powers of $i$, will always appear in pairs with opposite signs, effectively neutralizing each other.

Q: Is there a general formula for $z^n + (\bar{z})^n$?

Yes! If $z = r(\cos \theta + i \sin \theta)$, then $z^n = r^n(\cos n\theta + i \sin n\theta)$ and $\bar{z}^n = r^n(\cos n\theta - i \sin n\theta)$. Adding them gives $2r^n \cos(n\theta)$. This is a powerful tool in complex analysis!

Question

If $z = 2 + 3i$, then $z^5 + (\bar{z})^5$ is equal to :

Options:

(A) 244

(B) 224

(C) 245

(D) 265

Sigma Percentile · Accepted Answer

### The Elegance of Symmetry: Unlocking Complex Powers

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a problem; we are uncovering a hidden symmetry in the world of complex numbers.

You are given $z = 2 + 3i$, and you are asked to find the value of $z^5 + (\bar{z})^5$. At first glance, you might be tempted to reach for your pen and start expanding $(2+3i)^5$ manually.

I urge you to pause. In the JEE, the path of least resistance is often the path of deepest insight.

### Phase 1: The Geometry of Conjugates

Imagine you are standing on the Argand plane. You walk two units to the right and three units up. That is your $z$.

Now, consider $\bar{z} = 2 - 3i$. Geometrically, this is a perfect mirror reflection of $z$ across the real axis.

This reflection is not just a visual curiosity; it is the key to the entire problem. When we look at the expression $z^5 + (\bar{z})^5$, we are essentially asking what happens when we combine these two reflected points raised to the fifth power. The symmetry here is profound.

### Phase 2: The Binomial Shortcut

Instead of brute force, let us invoke the Binomial Theorem. We know that $(a+b)^n + (a-b)^n$ is a classic structure.

When we expand these, the terms involving odd powers of $b$ (like $b^1, b^3, b^5$) will have opposite signs in the two expansions. Specifically, $(a+b)^n$ will have a positive term, and $(a-b)^n$ will have a negative term for those odd powers.

When we add them, they vanish! We are left with exactly twice the sum of the even-powered terms. This is the 'Aha!' moment. We don't need to calculate everything; we only need the terms where the power of $b$ is even.

### Phase 3: The Calculation

Let $a = 2$ and $b = 3i$. Our expression becomes:

$$2 \left[ ^{5}C_{0} a^5 + ^{5}C_{2} a^3 (3i)^2 + ^{5}C_{4} a^1 (3i)^4 \right]$$

Let us break this down with precision:

1. **The First Term:** $^{5}C_{0} \cdot 2^5 = 1 \cdot 32 = 32$. Simple, clean, and correct.

2. **The Second Term:** $^{5}C_{2} \cdot 2^3 \cdot (3i)^2$. Here, $^{5}C_{2} = 10$, $2^3 = 8$, and $(3i)^2 = 9i^2 = -9$. So, $10 \cdot 8 \cdot (-9) = -720$. Watch that negative sign! It is the most common place to stumble.

3. **The Third Term:** $^{5}C_{4} \cdot 2^1 \cdot (3i)^4$. $^{5}C_{4} = 5$, $2^1 = 2$, and $(3i)^4 = 81i^4 = 81(1) = 81$. So, $5 \cdot 2 \cdot 81 = 810$.

### Phase 4: The Final Reveal

Now, we sum these components inside our bracket: $32 - 720 + 810$.

Combining these, $810 - 720 = 90$, and $90 + 32 = 122$. But wait! Do not forget the factor of $2$ we pulled out at the beginning.

The final result is $2 \cdot 122 = \mathbf{244}$.

Look at that result. It is a purely real number. The imaginary parts have completely vanished, just as the symmetry of the conjugate promised.

This is the beauty of complex numbers—they often resolve into simple, elegant, real-world values. Keep this binomial trick in your toolkit; it will serve you well in many JEE problems to come. You have mastered the symmetry, and that is the true victory.

Animated Solution for Mathematics - Complex Numbers: If $z = 2 + 3 i$ , then $z^{5} + (\overset{z}{ˉ})^{5}$ is equal to :

Select Answer:

Visualized Solution

Identify the Complex Number $z$

Find the Conjugate $\overset{z}{ˉ}$

Define the Target Expression

Apply the Binomial Identity

Set up the Expansion

Calculate the First Term

Calculate the Second Term

Calculate the Third Term

Sum the Terms Inside Brackets

Final Result

Key Takeaway

The Sigma Insight: Algebraic Operations on Complex Numbers

The Elegance of Symmetry

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Complex Numbers: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If z=2+3i, then z5+(zˉ)5 is equal to :