ID#6178 HSC Higher Math 2nd CQ (Mymensingh 2025)

ক-এর উত্তর:
প্রদত্ত রাশি: $\frac{3+4i}{3-i} + 1$
$\implies \frac{(3+4i)(3+i)}{(3-i)(3+i)} + 1$
$\implies \frac{9+3i+12i+4i^2}{3^2+1^2} + 1$
$\implies \frac{9+15i-4}{9+1} + 1$
$\implies \frac{5+15i}{10} + 1$
$\implies \frac{5}{10} + \frac{15i}{10} + 1$
$\implies \frac{1}{2} + \frac{3}{2}i + 1$
$\implies \frac{3}{2} + i\frac{3}{2}$
এটি $A+iB$ আকার যেখানে $A = \frac{3}{2}$ এবং $B = \frac{3}{2}$।

খ-এর উত্তর:
দেওয়া আছে, $f(x) = 1-x+x^2$
আমরা জানি, $\omega^3 = 1$ এবং $1+\omega+\omega^2 = 0 \implies 1+\omega^2 = -\omega$
$\therefore f(\omega) = 1-\omega+\omega^2 = -\omega-\omega = -2\omega$
$f(\omega^2) = 1-\omega^2+(\omega^2)^2 = 1-\omega^2+\omega = -\omega^2-\omega^2 = -2\omega^2$
$f(\omega^4) = f(\omega) = -2\omega$ এবং $f(\omega^8) = f(\omega^2) = -2\omega^2$
$\therefore$ বামপক্ষ $= f(\omega) \cdot f(\omega^2) \cdot f(\omega^4) \dots 2n$ সংখ্যক পদ
$\implies \{f(\omega) \cdot f(\omega^2)\}^n$
$\implies \{(-2\omega) \cdot (-2\omega^2)\}^n$
$\implies \{4\omega^3\}^n = 4^n = 2^{2n}$
ডানপক্ষ $= 2^{2n} (1+\omega^8)(1+\omega^{10})$
$\implies 2^{2n} (1+\omega^2)(1+\omega)$
$\implies 2^{2n} (-\omega)(-\omega^2) = 2^{2n} \cdot \omega^3 = 2^{2n}$
$\therefore$ বামপক্ষ = ডানপক্ষ (দেখানো হলো)।

গ-এর উত্তর:
দেওয়া আছে, $\{f(-x)\}^n = \{1-(-x)+(-x)^2\}^n = (1+x+x^2)^n$
$\therefore (1+x+x^2)^n = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4 + a_5 x^5 + a_6 x^6 + \dots + a_{2n} x^{2n}$ --- (i)
(i) নং সমীকরণে $x = 1$ বসিয়ে পাই,
$(1+1+1)^n = a_0 + a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + \dots$
$\implies 3^n = a_0 + a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + \dots$ --- (ii)
(i) নং সমীকরণে $x = \omega$ বসিয়ে পাই,
$(1+\omega+\omega^2)^n = a_0 + a_1 \omega + a_2 \omega^2 + a_3 \omega^3 + a_4 \omega^4 + a_5 \omega^5 + a_6 \omega^6 + \dots$
$\implies 0 = a_0 + a_1 \omega + a_2 \omega^2 + a_3 + a_4 \omega + a_5 \omega^2 + a_6 + \dots$ --- (iii) [$\because \omega^3=1$]
(i) নং সমীকরণে $x = \omega^2$ বসিয়ে পাই,
$(1+\omega^2+\omega^4)^n = (1+\omega^2+\omega)^n = 0$
$\implies 0 = a_0 + a_1 \omega^2 + a_2 \omega^4 + a_3 \omega^6 + a_4 \omega^8 + \dots$
$\implies 0 = a_0 + a_1 \omega^2 + a_2 \omega + a_3 + a_4 \omega^2 + a_5 \omega + a_6 + \dots$ --- (iv)
এখন, (ii) + (iii) + (iv) করে পাই,
$3^n + 0 + 0 = 3a_0 + a_1(1+\omega+\omega^2) + a_2(1+\omega^2+\omega) + 3a_3 + a_4(1+\omega+\omega^2) + \dots$
$\implies 3^n = 3a_0 + 0 + 0 + 3a_3 + 0 + 0 + 3a_6 + \dots$
$\implies 3^n = 3(a_0 + a_3 + a_6 + \dots)$
$\implies a_0 + a_3 + a_6 + \dots = \frac{3^n}{3}$
$\therefore a_0 + a_3 + a_6 + \dots = 3^{n-1}$ (প্রমাণিত)।