সৃজনশীলঃ ‘দৃশ্য-১ :f(x)=2x+3i+1, h(x)=2x−3i+1…’ সিলেট বোর্ড ২০২৫ – জটিল সংখ্যা, উচ্চতর গণিত, উচ্চতর গণিত ২য় পত্র (এইচএসসি)

 দেওয়া আছে, $f\left(x\right)=2x+\sqrt{3}i+1$ এবং $h\left(x\right)=2x-\sqrt{3}i+1$

$f\left(p\right)=0$ হলে, $2p+\sqrt{3}i+1=0$

বা, $2p=-1-\sqrt{3}i$

$\therefore p=\frac{1}{2}(-1-\sqrt{3}i)$

আবার, $h\left(q\right)=0$ হলে, $2q-\sqrt{3}i+1=0$
বা, $2q=-1+\sqrt{3}i$

$\therefore q=\frac{1}{2}(-1+\sqrt{3}i)$ 

$\begin{array}{cc}\therefore p^n+q^n& ={\left\{\frac{1}{2}\right(-1-\sqrt{3}i\left)\right\}}^n+{\left\{\frac{1}{2}\right(-1+\sqrt{3}i\left)\right\}}^n\\ & ={\left({\omega }^2\right)}^n+{\omega }^n={\omega }^{2n}+{\omega }^n={\omega }^n+{\omega }^{2n}\end{array}$  

এখানে, $n=3m$ হলে, 

$\begin{array}{cc}{p}^n+q^n& ={\omega }^n+{\omega }^{2n}={\omega }^{3m}+{\left({\omega }^2\right)}^{3m}\\ & ={\left({\omega }^3\right)}^m+{\left({\omega }^3\right)}^{2m}\\ & =1+1[\because {\omega }^3=1]\\ & =2\end{array}$ 

$n=3m+1$ হলে,

$\begin{array}{cc}{p}^n+q^n& ={\omega }^n+{\omega }^{2n}={\omega }^{3m+1}+{\left({\omega }^2\right)}^{3m+1}\\ & ={\omega }^{3m+1}+{\omega }^{6m+2}={\omega }^{3m}\cdot \omega +{\omega }^{6m}\cdot {\omega }^2\\ & ={\left({\omega }^3\right)}^m\cdot \omega +{\left({\omega }^3\right)}^{2m}\cdot {\omega }^2\\ & =1\cdot \omega +1\cdot {\omega }^2[\because {\omega }^3=1]\\ & =1+{\omega }^2=-1[\because 1+\omega +{\omega }^2=0]\end{array}$ 

$n=3m+2$ হলে, 

$\begin{array}{cc}{p}^n+q^n& ={\omega }^n+{\omega }^{2n}={\omega }^{3m+2}+{\left({\omega }^2\right)}^{3m+2}\\ & ={\omega }^{3m}\cdot {\omega }^2+{\omega }^{6m}\cdot {\omega }^4\\ & ={\left({\omega }^3\right)}^m\cdot {\omega }^2+{\left({\omega }^3\right)}^{2m}\cdot {\omega }^3\cdot \omega \\ & =1\cdot {\omega }^2+1\cdot 1\cdot \omega \cdot [\because {\omega }^3=1]\\ & ={\omega }^2+\omega \\ & =-1[\because 1+\omega +{\omega }^2=0]\end{array}$ 

অর্থাৎ, $p^n+q^n=2;$ যখন $n,3$ দ্বারা বিভাজ্য, অথবা - $1$ যখন $n$ অপর যেকোনো পূর্ণসংখ্যা। (দেখানো হলো) 

দেওয়া আছে, $g\left(x\right)=x+1$ এবং $z=p+\mathrm{iq}$

$\therefore g(z-1)=z-1+1=z$

এবং $g(z+1)=z+1+1=z+2$
এখন, $\left|g\right(z-1)|+|g(z+1)|=5$
বা, $|z|+|z+2|=5$
বা, $|p+iq|+|p+iq+2|=5$
বা, $\left|\right(p+2)+iq|=5-|p+iq|$
বা, $\sqrt{(p+2{)}^2+q^2}=5-\sqrt{p^2+q^2}$
বা, $(p+2{)}^2+q^2={(5-\sqrt{p^2+q^2})}^2$ [বর্গ করে]
বা, $p^2+2.p.2+2^2+q^2=5^2-2.5\sqrt{p^2+q^2}+(p^2+q^2)$
বা, $p^2+4p+4+q^2=25-10\sqrt{p^2+q^2}+p^2+q^2$
বা, $p^2+4p+4+q^2-25-p^2-q^2=-10\sqrt{p^2+q^2}$
বা, $4p-21=-10\sqrt{p^2+q^2}$
বা, $(4p-21{)}^2={(-10\sqrt{p^2+q^2})}^2$ [বর্গ করে]
বা, $(4p{)}^2-2.4p.21+(21{)}^2=100(p^2+q^2)$
বা, $16p^2-168p+441=100p^2+100q^2$
বা, $100p^2+100q^2-16p^2+168p=441$
বা, $84p^2+168p+100q^2=441$
বা, $84(p^2+2p)+100q^2=441$
বা, $84(p^2+2p+1)+100q^2=441+84$
বা, $84(p+1{)}^2+100q^2=525$
বা, $\frac{84(p+1{)}^2}{525}+\frac{100q^2}{525}=1$
বা, $\frac{4(p+1{)}^2}{25}+\frac{4q^2}{21}=1$
$\therefore \frac{(p+1{)}^2}{{\left(\frac{5}{2}\right)}^2}+\frac{q^2}{{\left(\frac{\sqrt{21}}{2}\right)}^2}=1$; যা একটি উপবৃত্ত নির্দেশ করে।

নির্ণেয় সঞ্চারপথের সমীকরণ : $\frac{(p+1{)}^2}{{\left(\frac{5}{2}\right)}^2}+\frac{q^2}{{\left(\frac{\sqrt{21}}{2}\right)}^2}=1$ এবং ইহা একটি উপবৃত্ত নির্দেশ করে।