সৃজনশীলঃ ‘f(x)=a+bx+cx2;z=r(cos⁡θ+isin⁡θ).…’ ২০২৫ – জটিল সংখ্যা, উচ্চতর গণিত, উচ্চতর গণিত ২য় পত্র (এইচএসসি)

দেওয়া আছে, $z=r(\cos \theta +i\sin \theta )=r\cos \theta +ir\sin \theta$ 

$\therefore |z|=\sqrt{r^2{\cos }^2\theta +r^2{\sin }^2\theta }=\sqrt{r^2({\sin }^2\theta +{\cos }^2\theta )}=\sqrt{r^2\times 1}=r$ 

এখানে, $|z|=1\therefore r=1$ 

$r=1,\theta =\frac{2\pi }{3}$ হলে. $z=1(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3})=-\frac{1}{2}+i\times \frac{\sqrt{3}}{2}$ 

$\begin{array}{cc}{z}^2& ={(-\frac{1}{2}+i\frac{\sqrt{3}}{2})}^2={(-\frac{1}{2})}^2+2(-\frac{1}{2})i\frac{\sqrt{3}}{2}+{\left(i\frac{\sqrt{3}}{2}\right)}^2\\ & =\frac{1}{4}-i\frac{\sqrt{3}}{2}+i^2\frac{3}{4}=\frac{1}{4}-i\frac{\sqrt{3}}{2}-\frac{3}{4}[\because i^2=-1]\\ & =-\frac{1}{2}-\frac{\sqrt{3}}{2}i\end{array}$ 

$\begin{array}{cc}{z}^3=z^2\cdot z& =(-\frac{1}{2}-\frac{\sqrt{3}}{2}i)(-\frac{1}{2}+\frac{\sqrt{3}}{2}i)\\ & ={(-\frac{1}{2})}^2-{\left(\frac{\sqrt{3}}{2}i\right)}^2=\frac{1}{4}-\frac{3}{4}{i}^2=\frac{1}{4}+\frac{3}{4}[\because i^2=-1]\\ & =\frac{1+3}{4}=\frac{4}{4}=1\\ z^4=z^3\cdot z& =1(-\frac{1}{2}+\frac{\sqrt{3}}{2}i)=-\frac{1}{2}+\frac{\sqrt{3}}{2}i\\ \therefore 1+z^2+z^4& =1+(-\frac{1}{2}-\frac{\sqrt{3}}{2}i)+(-\frac{1}{2}+\frac{\sqrt{3}}{2}i)\\ & =1-\frac{1}{2}-\frac{1}{2}=1-2\times \frac{1}{2}=1-1=0\end{array}$ 

বামপক্ষ $=(a+b{)}^2+{(az+b{z}^2)}^2+{(a{z}^2+bz)}^2$ 

$\begin{array}{cc}& =a^2+2ab+b^2+a^2{z}^2+2ab{z}^3+b^2{z}^4+a^2{z}^4+2ab{z}^3+b^2{z}^2\\ & =a^2(1+z^2+z^4)+b^2(1+z^2+z^4)+2ab+4ab{z}^3.\\ & =a^2\times 0+b^2\times 0+2ab+4ab\times 1\end{array}$ 

$=0+0+2\mathrm{ab}+4\mathrm{ab}=6\mathrm{ab}=$ ডানপক্ষ 

$\therefore (a+b{)}^2+{(az+b{z}^2)}^2+{(a{z}^2+bz)}^2=6ab$. (প্রমাণিত) 

দেওয়া আছে, $f\left(x\right)=a+bx+c{x}^2$ 

$\begin{array}{cc}f\left(\omega \right)& =a+b\omega +c{\omega }^2\\ f\left(\frac{1}{\omega }\right)& =f\left(\frac{{\omega }^2}{{\omega }^3}\right)=f\left(\frac{{\omega }^2}{1}\right)[\because {\omega }^3=1]\\ & =f\left({\omega }^2\right)=a+b{\omega }^2+c{\left({\omega }^2\right)}^2\\ & =a+b{\omega }^2+c{\omega }^3\cdot \omega \\ & =a+b{\omega }^2+c\omega [\because {\omega }^3=1]\end{array}$ 

এখন, $\left\{f\right(\omega ){\}}^3+{\left\{f\left(\frac{1}{\omega }\right)\right\}}^3=0$ হলে , 

${(a+b\omega +c{\omega }^2)}^3+{(a+b{\omega }^2+c\omega )}^3=0$ 

বা, ${(a+b\omega +c{\omega }^2)}^3=-{(a+b{\omega }^2+c\omega )}^3$ 

বा, ${\left(\frac{a+b\omega +c{\omega }^2}{a+b{\omega }^2+c\omega }\right)}^3=-1$ 

বा, $\frac{a+b\omega +c{\omega }^2}{a+b{\omega }^2+c\omega }=\sqrt[3]{-1}$ 

বा, $\frac{a+b\omega +c{\omega }^2}{a+b{\omega }^2+c\omega }=-1,-\omega$, ${\omega }^2$ 

অর্থাৎ, $\frac{a+b\omega +c{\omega }^2}{a+b{\omega }^2+c\omega }=-1$ 

বা, $a+b\omega +c{\omega }^2=-a-b{\omega }^2-c\omega$ 

বা, $2a+b(\omega +{\omega }^2)+c({\omega }^2+\omega )=0$ 

বा, $2a-b-c=0[\because 1+\omega +{\omega }^2=0]$ 

বা, $2a=b+c$ 

$\therefore a=\frac{1}{2}(b+c)$

আবার, $\frac{a+b\omega +c{\omega }^2}{a+b{\omega }^2+c\omega }=-\omega$
বা, $a+b\omega +c{\omega }^2=-a\omega -b{\omega }^3-c{\omega }^2$
বা, $a(1+\omega )+b(\omega +{\omega }^3)+c({\omega }^2+{\omega }^2)=0$ 

বা, $-{\omega }^{2}a+b(\omega +1)+2c{\omega }^2=0[1+\omega +{\omega }^2=0$ এবং  ${\omega }^3=1]$
বা, $-{\omega }^{2}a-b{\omega }^2+2c{\omega }^2=0$
বা, $-a-b+2c=0[{\omega }^2$ দ্বারা ভাগ করে। ]
বা, $2c=a+b$ 

$\therefore c=\frac{1}{2}(a+b)$ 

আবার, $\frac{a+b(\omega +c{\omega }^2}{a+b{\omega }^2+c\omega }=-{\omega }^2$
বা, $a+b\omega +c{\omega }^2=-a{\omega }^2-b{\omega }^4-c{\omega }^3$
বা, $a(1+{\omega }^2)+b(\omega +{\omega }^4)+c({\omega }^2+{\omega }^3)=0$
বা, $a(1+{\omega }^2)+b(\omega +\omega )+c({\omega }^2+1)=0[\because {\omega }^3=1]$
বা, $-a\omega +2b\omega -c\omega =0[\because 1+\omega +{\omega }^2=0]$
বা, $-a+2b-c=0$ [ $\omega$ দ্বারা ভাগ করে]
বা, $\mathrm{ab}=c+a\therefore b=\frac{1}{2}(c+a)$

বা,  $2b=c+a\therefore b=\frac{1}{2}(c+a)$$\therefore a=\frac{1}{2}(b+c)$ অথবা, $b=\frac{1}{2}(c+a)$ অথবা, $c=\frac{1}{2}(a+b)$. 

(দেখানো হলো)