সৃজনশীলঃ ‘দৃশ্যকল্প-১ : z=−2−23i…’ যশোর বোর্ড ২০২৫ – জটিল সংখ্যা, উচ্চতর গণিত, উচ্চতর গণিত ২য় পত্র (এইচএসসি)

দেওয়া আছে, $z_1=1-\mathrm{ix};z_2=a-\mathrm{ib};a,b,x\in R$

এবং $i=\sqrt{-1}$ বা, $i^2=-1$ 

$\begin{array}{cc}& a^2+b^2=1\\ \therefore & {\overline{z}}_1=1+\mathrm{ix}\end{array}$ 

এখন, $z_1={\bar{z}}_1.z_2$
বা, $1-ix=(1+ix)(a-ib)$
বা, $1-ix=a-ib+iax-i^{2}bx$
বা, $1-ix=a-ib+iax+bx[\because i^2=-1]$
বা, $iax+bx+ix=1-a+ib$
বা, $x(ia+b+i)=1-a+ib$ 

$\begin{array}{cc}\therefore x& =\frac{1-a+ib}{b+i(1+a)}=\frac{(1-a+ib)\{b-i(1+a\left)\right\}}{\{b+i(1+a\left)\right\}\{b-i(1+a\left)\right\}}\\ & =\frac{b-i(1+a)-ab+ia(1+a)+i{b}^2-i^{2}b(1+a)}{b^2-\left\{i\right(1+a){\}}^2}\\ & =\frac{b-i-ia-ab+ia+i{a}^2+i{b}^2+b+ab}{b^2+(1+a{)}^2}[\because i^2=-1]\\ & =\frac{2b-i+i(a^2+b^2)}{b^2+1+2a+a^2}=\frac{2b-i+i(a^2+b^2)}{a^2+b^2+2a+1}\\ & =\frac{2b-i+i}{1+2a+1}[\because a^2+b^2=1]\\ & =\frac{2b}{2a+2}=\frac{2b}{2(a+1)}=\frac{b}{a+1}\end{array}$ 

$\therefore x=\frac{b}{a+1};$ যা x এর একটি বাস্তব মান।
: $z_1={\bar{z}}_1{z}_2$ সমীকরণে x এর একটি বাস্তব মান বিদ্যমান । 

(দেখানো হল) 

দেওয়া আছে, $z=-2-2\sqrt{3}i$ 

$\therefore \overline{z}=-2+2\sqrt{3}i$ 
$\begin{array}{cc}& \therefore {\left(\frac{z}{4}\right)}^m+{\left(\frac{\overline{z}}{4}\right)}^m={\left(\frac{-2-2\sqrt{3}i}{4}\right)}^m+{\left(\frac{-2+2\sqrt{3}i}{4}\right)}^m\\ & ={\left\{\frac{2(-1-\sqrt{3}i)}{4}\right\}}^m+{\left\{\frac{2(-1+\sqrt{3}i)}{4}\right\}}^m\\ & ={\left\{\frac{1}{2}\right(-1-\sqrt{3}i\left)\right\}}^m+{\left\{\frac{1}{2}\right(-1+\sqrt{3}i\left)\right\}}^m\\ & ={\left({\omega }^2\right)}^m+(\omega {)}^m={\omega }^{2m}+{\omega }^m\end{array}$ 

এখানে, m=3n হলে, 

$\begin{array}{cc}{\left(\frac{z}{4}\right)}^m+{\left(\frac{\overline{z}}{4}\right)}^m& ={\omega }^{2m}+{\omega }^m={\left({\omega }^2\right)}^{3n}+{\omega }^{3n}\\ & ={\left({\omega }^3\right)}^{2n}+{\left({\omega }^3\right)}^n\\ & =1+1[\because {\omega }^3=1]\\ & =2\end{array}$ 

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

$m=3n+2$ হলে, 

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

(প্রমাণিত)