Let i = √-1. If ((-1+i√3)²¹)/((1-i)²⁴) + ((1+i√3)²¹)/((1+i)²⁴) = k, and n = [|k|] be the greatest in

Q: Why can't we just expand $(-1+i\sqrt{3})^{21}$ using the Binomial Theorem?

You could, but it would be a nightmare! The Binomial Theorem is great for small powers, but for power 21, it would result in 22 terms, most of which are complex. Polar form turns exponentiation into simple multiplication of the argument, which is the elegant path.

Q: How do I know which quadrant the argument belongs to?

Always look at the signs of the real and imaginary parts. For $-1+i\sqrt{3}$, the real part is negative and the imaginary part is positive, placing it firmly in the second quadrant. The argument is $\pi - \tan^{-1}(\sqrt{3}/1) = 2\pi/3$. Never guess; always sketch the point!

Q: Is there a trick to the summation?

Since $n=0$, the summation is just a finite series of squares and linear terms. Don't overcomplicate it with summation formulas if the range is small. Just write out the terms and add them up—it's often faster and less prone to error than complex algebraic identities.

Question

Let $i = \sqrt{-1}$. If $\frac{(-1+i\sqrt{3})^{21}}{(1-i)^{24}} + \frac{(1+i\sqrt{3})^{21}}{(1+i)^{24}} = k$, and $n = [|k|]$ be the greatest integral part of $|k|$. Then $\sum_{j=0}^{n+5} (j + 5)^2 – \sum_{j=0}^{n} (j + 5)$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The expression provided is:

$$k = \frac{(-1+i\sqrt{3})^{21}}{(1-i)^{24}} + \frac{(1+i\sqrt{3})^{21}}{(1+i)^{24}}$$

In the world of JEE Advanced, intimidation is just a mask for elegance. We will avoid brute-force calculation and instead utilize the geometry of the complex plane.

### The Polar Transformation

Polar coordinates $(r, 	heta)$ are built for rotation and power. When you see a complex number raised to a high power, apply **De Moivre's Theorem**.

For the first numerator, $z_1 = -1 + i\sqrt{3}$:
The modulus is $r = \sqrt{(-1)^2 + (\sqrt{3})^2} = 2$.
Since it lies in the second quadrant, the argument is $	heta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.
Thus, $z_1 = 2e^{i\frac{2\pi}{3}}$.

For the first denominator, $z_2 = 1 - i$:
The modulus is $\sqrt{1^2 + (-1)^2} = \sqrt{2}$.
The argument is $-\frac{\pi}{4}$.
Thus, $z_2 = \sqrt{2}e^{-i\frac{\pi}{4}}$.

For the second term, we follow the same logic:
$z_3 = 1 + i\sqrt{3} = 2e^{i\frac{\pi}{3}}$
$z_4 = 1 + i = \sqrt{2}e^{i\frac{\pi}{4}}$

### The De Moivre Magic

Now, we substitute these into the expression for $k$.

For the first term:

$$\frac{(2e^{i\frac{2\pi}{3}})^{21}}{(\sqrt{2}e^{-i\frac{\pi}{4}})^{24}} = \frac{2^{21} \cdot e^{i14\pi}}{2^{12} \cdot e^{-i6\pi}} = 2^9 \cdot e^{i(14\pi - (-6\pi))} = 2^9 \cdot e^{i20\pi}$$

Since $e^{i20\pi} = 1$, the first term simplifies to $2^9$.

For the second term:

$$\frac{(2e^{i\frac{\pi}{3}})^{21}}{(\sqrt{2}e^{i\frac{\pi}{4}})^{24}} = \frac{2^{21} \cdot e^{i7\pi}}{2^{12} \cdot e^{i6\pi}} = 2^9 \cdot e^{i(7\pi - 6\pi)} = 2^9 \cdot e^{i\pi}$$

Since $e^{i\pi} = -1$, the second term simplifies to $-2^9$.

### The Final Calculation

Combining these results, we find:

$$k = 2^9 - 2^9 = 0$$

Given $n = |0| = 0$, we proceed to the final summation:

$$\sum_{j=0}^{5} (j + 5)^2 - \sum_{j=0}^{5} (j + 5)$$

This expands to:

$$(5^2 + 6^2 + 7^2 + 8^2 + 9^2 + 10^2) - (5 + 6 + 7 + 8 + 9 + 10)$$

Calculating the values:

$$355 - 45 = 310$$

The final result is **310**.

Enter Numerical Value:

Visualized Solution

Introduction to the Problem

Polar Form of $z_{1} = - 1 + i 3$

Polar Form of $z_{2} = 1 - i$

Polar Form of $z_{3} = 1 + i 3$

Polar Form of $z_{4} = 1 + i$

Substituting into $k$

Simplifying the First Term

Simplifying the Second Term

Calculating $k$ and $n$

Setting up the Summation

Calculating the First Sum

Calculating the Second Sum

Final Result

The Sigma Insight: Argand Plane and Polar Representation

Analyzing the Setup

The Polar Transformation

The De Moivre Magic

The Final Calculation

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Enter Numerical Value:

Visualized Solution

Introduction to the Problem

Polar Form of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z1​=−1+i3​

Polar Form of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z2​=1−i

Polar Form of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z3​=1+i3​

Polar Form of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z4​=1+i

Substituting into .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k