সৃজনশীলঃ ‘z1=−1−i3,z2=3−i.…’ বরিশাল বোর্ড ২০২৩ – জটিল সংখ্যা, উচ্চতর গণিত, উচ্চতর গণিত ২য় পত্র (এইচএসসি)

দেওয়া আছে, $z_1=-1-i\sqrt{3}$ এবং $z_2=\sqrt{3}-i$

$Arg{z}_1=-\pi +{\tan }^{-1}\left|\frac{y}{x}\right|[\because$ বিন্দুতে ৩য় চতুর্ভাগে]

$=-\pi +{\tan }^{-1}\left|\frac{-\sqrt{3}}{-1}\right|$

$=-\pi +{\tan }^{-1}\left(\sqrt{3}\right)=-\pi +\frac{\pi }{3}=-\frac{2\pi }{3}$

এবং $\mathrm{Arg}{z}_2=-{\tan }^{-1}\left|\frac{y}{x}\right|[\because$ বিন্দুতে ৪র্থ চতুর্ভাগে]

$=-{\tan }^{-1}\left|\frac{-1}{\sqrt{3}}\right|=-{\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)=-\frac{\pi }{6}$

$\therefore \frac{z_1}{z_2}=\frac{-1-\sqrt{3}i}{\sqrt{3}-i}$

$=\frac{(-1-\sqrt{3}i)(\sqrt{3}+i)}{(\sqrt{3}-i)(\sqrt{3}+i)}$ [লব ও হরকে হরের অনুবদ্ধী জটিল সংখ্যা $(\sqrt{3}-i)$ দ্বারা গুণ করে]

$=\frac{-\sqrt{3}-i-3i-\sqrt{3}{i}^2}{(\sqrt{3}{)}^2-i^2}=\frac{-\sqrt{3}-4i+\sqrt{3}}{3+1}[\because i^2=-1]$

$=\frac{-4i}{4}=-i=0-i$

বামপক্ষ $=\mathrm{Arg}\left(\frac{z_1}{z_2}\right)=-{\tan }^{-1}\left|\frac{-1}{0}\right|=-{\tan }^{-1}\left(\infty \right)=-\frac{\pi }{2}$

ডানপক্ষ $={\mathrm{Argz}}_1-{\mathrm{Argz}}_2$

$=-\frac{2\pi }{3}-(-\frac{\pi }{6})=-\frac{2\pi }{3}+\frac{\pi }{6}=\frac{-4\pi +\pi }{6}=\frac{-3\pi }{6}=-\frac{\pi }{2}$

$\therefore \mathrm{Arg}\left(\frac{z_1}{z_2}\right)=\mathrm{Arg}{z}_1-\mathrm{Arg}{z}_2$. (দেখানো হলো)

দেওয়া আছে, $z_1=-1-i\sqrt{3}$

$\therefore {\overline{z}}_1=-1+i\sqrt{3}$

$\therefore {\left(\frac{1}{2}{\overline{z}}_1\right)}^n+{\left(\frac{1}{2}{z}_1\right)}^n={\left\{\frac{1}{2}\right(-1+\sqrt{3}i\left)\right\}}^n+{\left\{\frac{1}{2}\right(-1-\sqrt{3}i\left)\right\}}^n$

$={\omega }^n+{\left({\omega }^2\right)}^n={\omega }^n+{\omega }^{2n}$

এখানে, $n=3m$ হলে, ${\left(\frac{1}{2}{\overline{z}}_1\right)}^n+{\left(\frac{1}{2}{z}_1\right)}^n={\omega }^n+{\omega }^{2n}$

$={\omega }^{3m}+{\left({\omega }^2\right)}^{3m}$

$={\left({\omega }^3\right)}^m+{\left({\omega }^3\right)}^{2m}$

$=1+1[\because {\omega }^3=1]$

$=2$

$n=3m+1$ হলে, ${\left(\frac{1}{2}{\overline{z}}_1\right)}^n+{\left(\frac{1}{2}{z}_1\right)}^n={\omega }^n+{\omega }^{2n}$

$={\omega }^{3m+1}+{\left({\omega }^2\right)}^{3m+1}$

$={\left({\omega }^3\right)}^m\cdot \omega +{\left({\omega }^3\right)}^{2m}\cdot {\omega }^2$

$=\omega +{\omega }^2[\because {\omega }^3=1]$

$n=3m+2$ হলে, ${\left(\frac{1}{2}{\overline{z}}_1\right)}^n+{\left(\frac{1}{2}{z}_1\right)}^n={\omega }^n+{\omega }^{2n}$

$=-1[\because 1+\omega +{\omega }^2=0]$

$={\omega }^{3m+2}+{\left({\omega }^2\right)}^{3m+2}$

$={\left({\omega }^3\right)}^m\cdot {\omega }^2+{\left({\omega }^3\right)}^{2m}\cdot {\left({\omega }^2\right)}^2$

$=1\cdot {\omega }^2+1\cdot {\omega }^3\cdot \omega ={\omega }^2+\omega$

$=-1[\because 1+\omega +{\omega }^2=0]$

অতত্রব, ${\left(\frac{1}{2}{\overline{z}}_1\right)}^n+{\left(\frac{1}{2}{z}_1\right)}^n=2$, যখন $n$ এর মান 3 দ্বারা বিভাজ্য

অথবা, ${\left(\frac{1}{2}{\overline{z}}_1\right)}^n+{\left(\frac{1}{2}{z}_1\right)}^n=-1$ যখন $n$ এর মান অন্য যেকোনো পূর্ণসংখ্যা হয়। (প্রমাণিত)