সৃজনশীলঃ ‘Z1=10+6i,Z2=4+6i,p=a+bω+cω2,q=a+bω2+cω চারটি জটিল…’ ২০২৫ – জটিল সংখ্যা, উচ্চতর গণিত, উচ্চতর গণিত ২য় পত্র (এইচএসসি)

দেওয়া আছে, $x+y+z=R$ 
আবার, $R=0\therefore x+y+z=0$ 
এখানে, $\omega$ হলো একককর একটি কাল্পনিক ঘনমূল। 
$\therefore$ অপর কাল্পনিক ঘনমূলটি  $={\omega }^2$ 
ধরি, $x+y\omega +z{\omega }^2=a$ এবং $x+y{\omega }^2+z\omega =b$ 

$\begin{array}{cc}& \therefore {(x+y\omega +z{\omega }^2)}^3+{(x+y{\omega }^2+z\omega )}^3=a^3+b^3\\ & =(a+b)(a^2-ab+b^2)\\ & =(a+b)(a^2+b^2-ab)\\ & =(a+b)\left\{\right(a+b{)}^2-2ab-ab\}\\ & =(a+b)\left\{\right(a+b{)}^2-3ab\}\end{array}$ 

এখন, $a+b=(x+y\omega +z{\omega }^2)+(x+y{\omega }^2+z\omega )$ 

$\begin{array}{cc}& =2x+y(\omega +{\omega }^2)+z(\omega +{\omega }^2)\\ & =2x+y(-1)+z(-1)[\because 1+\omega +{\omega }^2=0\therefore \omega +{\omega }^2=-1]\\ & =2x-y-z=3x-(x+y+z)\\ & =3x-0[\because x+y+z=0]\\ & =3x\end{array}$ 

এবং $ab=(x+y\omega +z{\omega }^2)(x+y{\omega }^2+z\omega )$ 

$\begin{array}{cc}& =x^2+xy{\omega }^2+zx\omega +xy\omega +y^2{\omega }^3+yz{\omega }^2+zx{\omega }^2+yz{\omega }^4+z^2{\omega }^3\\ & =x^2+xy({\omega }^2+\omega )+zx(\omega +{\omega }^2)+yz({\omega }^2+{\omega }^2)+y^2+z^2[\because {\omega }^3=1]\\ & =x^2+y^2+z^2-xy-yz-zx[1+\omega +{\omega }^2=0]\end{array}$ 

$=x^2+(y+z{)}^2-2yz-yz-x(y+z)$ 

$=x^2+(-x{)}^2-3yz-x(-x\left)\right[\because x+y+z=0]$ 

$=x^2+x^2-3yz+x^2=3x^2-3yz$ 

বামপক্ষ $={(x+y\omega +z{\omega }^2)}^3+{(x+y{\omega }^2+z\omega )}^3$ 

$\begin{array}{cc}& =(a+b)\left\{\right(a+b{)}^2-3ab\}\\ & =3x\left\{\right(3x{)}^2-3(3x^2-3yz)\}\\ & =3x(9x^2-9x^2+9yz)\end{array}$ 

$=3x\times 9\mathrm{yz}=27\mathrm{xyz}=L$ ডানপক্ষ

$\therefore {(x+y\omega +z{\omega }^2)}^3+{(x+y{\omega }^2+z\omega )}^3=27xyz$. (প্রমাণিত) 

[বি: দ্র: x, y ও z এর পরিবর্তে যথাক্রমে a, b ও c হবে]

দেওয়া আছে, $z_1=10+6i$ 

$\begin{array}{cc}& z_2=4+6i\\ & z=x+iy\end{array}$ 

এবং $\arg \left(\frac{z-z_1}{z-z_2}\right)=\frac{\pi }{4}$
বা, $\arg \left(\frac{x+iy-10-6i}{x+iy-4-6i}\right)=\frac{\pi }{4}$
বা, $\arg \left\{\frac{x-10+i(y-6)}{x-4+i(y-6)}\right\}=\frac{\pi }{4}$
বা, $\arg \left\{\right(x-10)+i(y-6\left)\right\}-\arg \left\{\right(x-4)+i(y-6\left)\right\}=\frac{\pi }{4}$
বা, ${\tan }^{-1}\frac{y-6}{x-10}-{\tan }^{-1}\frac{y-6}{x-4}=\frac{\pi }{4}$ 

বা, ${\tan }^{-1}\frac{\frac{y-6}{x-10}-\frac{y-6}{x-4}}{1+\frac{y-6}{x-10}\times \frac{y-6}{x-4}}=\frac{\pi }{4}$
বা, ${\tan }^{-1}\frac{\frac{(y-6)(x-4)-(y-6)(x-10)}{(x-10)(x-4)}}{1+\frac{(y-6)(y-6)}{(x-10)(x-4)}}=\frac{\pi }{4}$ 

বা, $\frac{\frac{(xy-4y-6x+24)-(xy-10y-6x+60)}{(x-10)(x-4)}}{\frac{(x-10)(x-4)+(y-6{)}^2}{(x-10)(x-4)}}=\tan \frac{\pi }{4}$ 

বা, $\frac{\frac{xy-4y-6x+24-xy+10y+6x-60}{(x-10)(x-4)}}{\frac{x^2-4x-10x+40+y^2-12y+36}{(x-10)(x-4)}}=1$

বা, $\frac{\frac{6y-36}{(x-10)(x-4)}}{\frac{x^2+y^2-14x-12y+76}{(x-10)(x-4)}}=1$
বা, $\frac{6y-36}{(x-10)(x-4)}\times \frac{(x-10)(x-4)}{x^2+y^2-14x-12y+76}=1$
বা, $\frac{6y-36}{x^2+y^2-14x-12y+76}=1$
বা, $x^2+y^2-14x-12y+76=6y-36$
বা, $x^2+y^2-14x-12y+76-6y+36=0$
$\therefore x^2+y^2-14x-18y+112=0$. (দেখানো হলো)