Key Concepts and Formulas

• Polar Form of Complex Numbers and De Moivre's Theorem: A complex number $z = x + iy$ can be written as $z = r(\cos \theta + i \sin \theta)$, where $r = |z| = \sqrt{x^2 + y^2}$ and $\theta = \arg(z)$. De Moivre's Theorem: $(r(\cos \theta + i \sin \theta))^n = r^n(\cos n\theta + i \sin n\theta)$.

• Cube Roots of Unity: $\omega = e^{i 2\pi/3} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$, $\omega^2 = e^{i 4\pi/3} = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$, $1 + \omega + \omega^2 = 0$, $\omega^3 = 1$.

• Summation Formulas: $\sum_{j=0}^{N} j = \frac{N(N+1)}{2}$ and $\sum_{j=0}^{N} j^2 = \frac{N(N+1)(2N+1)}{6}$.

Step-by-Step Solution

Step 1: Simplify $(-1 + i\sqrt{3})^{21}$ We want to express $-1 + i\sqrt{3}$ in terms of $\omega$. Recall that $\omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$. Multiplying by 2, we have $2\omega = -1 + i\sqrt{3}$. Therefore, $(-1 + i\sqrt{3})^{21} = (2\omega)^{21} = 2^{21}\omega^{21}$. Since $\omega^3 = 1$, $\omega^{21} = (\omega^3)^7 = 1^7 = 1$. So, $(-1 + i\sqrt{3})^{21} = 2^{21}$.

Step 2: Simplify $(1 + i\sqrt{3})^{21}$ Since $\omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$, then $-\omega - 1 = - (-\frac{1}{2} + i\frac{\sqrt{3}}{2}) - 1 = \frac{1}{2} - i\frac{\sqrt{3}}{2} - 1 = -\frac{1}{2} - i\frac{\sqrt{3}}{2} = \omega^2$. Then $-2\omega^2 = 1 + i\sqrt{3}$. Therefore, $(1 + i\sqrt{3})^{21} = (-2\omega^2)^{21} = (-2)^{21}(\omega^2)^{21} = -2^{21}\omega^{42}$. Since $\omega^3 = 1$, $\omega^{42} = (\omega^3)^{14} = 1^{14} = 1$. So, $(1 + i\sqrt{3})^{21} = -2^{21}$.

Step 3: Simplify $(1 - i)^{24}$ We have $(1 - i)^2 = 1 - 2i + i^2 = 1 - 2i - 1 = -2i$. Thus, $(1 - i)^{24} = ((1 - i)^2)^{12} = (-2i)^{12} = (-2)^{12}i^{12} = 2^{12}(i^4)^3 = 2^{12}(1)^3 = 2^{12}$.

Step 4: Simplify $(1 + i)^{24}$ We have $(1 + i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i$. Thus, $(1 + i)^{24} = ((1 + i)^2)^{12} = (2i)^{12} = 2^{12}i^{12} = 2^{12}(i^4)^3 = 2^{12}(1)^3 = 2^{12}$.

Step 5: Substitute the simplified terms into the expression for $k$ $k = \frac{(-1 + i\sqrt{3})^{21}}{(1 - i)^{24}} + \frac{(1 + i\sqrt{3})^{21}}{(1 + i)^{24}} = \frac{2^{21}}{2^{12}} + \frac{-2^{21}}{2^{12}} = 2^{21-12} - 2^{21-12} = 2^9 - 2^9 = 0$

Step 6: Find $n$ We are given $n = [|k|]$. Since $k = 0$, $|k| = |0| = 0$. Therefore, $n = [0] = 0$.

Step 7: Evaluate the summation We need to find $\sum\limits_{j = 0}^{n + 5} {{{(j + 5)}^2} - \sum\limits_{j = 0}^{n + 5} {(j + 5)} }$. Since $n = 0$, we have $\sum_{j=0}^{5} (j+5)^2 - \sum_{j=0}^{5} (j+5) = \sum_{j=0}^{5} [(j+5)^2 - (j+5)] = \sum_{j=0}^{5} [j^2 + 10j + 25 - j - 5] = \sum_{j=0}^{5} (j^2 + 9j + 20)$ We can break this summation into three parts: $\sum_{j=0}^{5} j^2 + 9\sum_{j=0}^{5} j + 20\sum_{j=0}^{5} 1$ Using the summation formulas, we have: $\sum_{j=0}^{5} j^2 = \frac{5(5+1)(2(5)+1)}{6} = \frac{5(6)(11)}{6} = 55$ $\sum_{j=0}^{5} j = \frac{5(5+1)}{2} = \frac{5(6)}{2} = 15$ $\sum_{j=0}^{5} 1 = 6$ Substituting these values, we get: $55 + 9(15) + 20(6) = 55 + 135 + 120 = 190 + 120 = 310$

Common Mistakes & Tips

• Failing to recognize the connection between the complex numbers and the cube roots of unity can complicate the calculations.
• Remember to simplify complex expressions before raising them to high powers.
• Carefully apply the summation formulas to avoid errors.

Summary

By simplifying the complex number expression using properties of cube roots of unity and De Moivre's Theorem, we found $k=0$. This led to $n=0$. Then, we evaluated the summation by combining terms, expanding, and applying standard summation formulas for powers of integers. The final result of the summation is 310.

The final answer is \boxed{310}.