5. kuchlar va ildizlar
Ushbu bo'limda biz murakkab sonlarning butun sonlari va ildizlarini tezda hisoblashning ajoyib usulini ko'rib chiqamiz.
Biz \(z = r{{\bf{e}}^{i\theta }}\) ning butun sonidan boshlaymiz, chunki ular etarlicha oson. Agar \(n\) butun son bo'lsa,
\begin{tengation}{z^n} = {\left( {r{{\bf{e}}^{i\theta }}} \right)^n} = {r^n}{{\bf{e}}^{i\,\,n\theta }}\label{eq:eq1}\end{tenglama}Haqiqatan ham tez misol qilishdan boshqa kuchlar bilan bog'liq juda ko'p narsa yo'q.
Yechim
Albatta, biz buni raqamni ko'paytirish orqali amalga oshirishimiz mumkin edi, lekin bu ko'p vaqt talab qiladi va xatolarga moyil bo'ladi. Buning o'rniga biz eksponensial shaklga aylantira olamiz va keyin tezda javob olish uchun \(\eqref{eq:eq1}\) dan foydalanamiz.
Mana \(3 + 3i\) ning eksponensial shakli.
\[r = \sqrt {9 + 9} = 3\sqrt 2 \hspace{0.5in} \tan \theta = \frac{3}{3} \hspace{0.25in} \Rightarrow \hspace{0.25in} {\mathop{\rm Arg}\nolimits} \,z = \frac{\pi }{4}\] \[3 + 3i = 3\sqrt 2 {{\bf{e}}^{i\frac{\pi }{4}}}\]E'tibor bering, biz eksponensial shakl uchun argumentning asosiy qiymatidan foydalanganmiz, garchi bunga majbur bo'lmasak ham.
Endi hisobni tezda bajarish uchun \(\eqref{eq:eq1}\) dan foydalaning.
\begin{align*}{\left( {3 + 3i} \right)^5} & = {\left( {3\sqrt 2 } \right)^5}{{\bf{e}}^{i\,\frac{{5\pi }}{4}}}\\ & = 972\sqrt 2 \,\sqrt 2\,\sqrt ({\pi) }}{4}} \right) + i\sin \left( {\frac{{5\pi }}{4}} \right)} \right)\\ & = 972\sqrt 2 \left( { - \frac{{\sqrt 2 }}{2} - \frac{{\sqrt 2 }}{2}i} \u003d -7) 972i\end{align*}Shunday qilib, murakkab sonning butun soni uchun juda ko'p narsa yo'q.
E'tibor bering, agar \(r = 1\), bizda,
\({z^n} = {\left( {{{\bf{e}}^{i\theta }}} \right)^n} = {{\bf{e}}^{i\,\,n\theta }}\)va agar biz oxirgi ikki atamani olib, qutbli shaklga aylantirsak, biz deyilgan formulaga erishamiz de Moivre formulasi.
\[{\left( {\cos \theta + i\sin \theta } \right)^n} = \cos \left( {n\,\theta } \right) + i\sin \left( {n\,\theta } \right) \hspace{0.25in} n = 0, \pm 1, \pm 2, \ldots \]Endi biz murakkab sonlarning ildizlarini hisoblashga o'tishimiz kerak. Biz buni "oddiy" ni topishdan boshlaymiz nth birlik ildizlari. nth \(n = 2,3, \ldots \) uchun birlik ildizlari tenglamaning aniq echimlari,
\[{z^n} = 1\]Shubhasiz (umid qilamanki) \(z = 1\) yechimlardan biridir. Biz boshqa echimlar mavjudligini aniqlamoqchimiz. Buni amalga oshirish uchun biz \({z_1} = {z_2}\), agar va faqat agar bo'lsa, deb ko'rsatadigan oldingi bo'limlardagi faktdan foydalanamiz.
Shunday qilib, keling, tenglamaning ikkala tomonini murakkab shaklga aylantirib, keyin chap tomonda quvvatni hisoblashdan boshlaylik. Buni qilish,
\[{\left( {r{{\bf{e}}^{i\theta }}} \right)^n} = 1\,{{\bf{e}}^{i\left( 0 \right)}} \hspace{0.5in} \Rightarrow \hspace{0.5in} {r^n}{{\bf{e}}^{i\,\,n\theta }} = 1\,{{\bf{e}}^{i\left( 0 \right)}}\]Shunday qilib, ular teng bo'ladi,
\[{r^n} = 1 \hspace{0.5in} n\theta = 0 + 2\pi k\, \hspace{0.25in} k = 0, \pm 1, \pm 2, \ldots \]Endi, \(r\) musbat butun son (eksponensial/qutbli shaklni qabul qilish orqali) va shuning uchun yechish shuni beradi:
\[r = 1 \hspace{0.5in} \theta = \frac{{2\pi k}}{n}\, \hspace{0.25in} k = 0, \pm 1, \pm 2, \ldots \]Tenglamaning yechimlari u holda,
\[z = \exp \left( {i\,\,\frac{{2\pi k}}{n}} \right) \hspace{0.25in} k = 0, \pm 1, \pm 2, \ldots \]Eslab qoling qutb shakli (va demak, eksponensial shakl) haqidagi bahsimizdan bu nuqtalar \(r\) radiusi aylanasida yotadi. Shunday qilib, bizning nuqtalarimiz birlik aylanasida yotadi va ular har bir \(\frac{{2\pi }}{n}\) radianida birlik aylanasida teng masofada joylashgan bo'ladi. E'tibor bering, bu bizga \(k = 0,1,2, \ldots ,n - 1\) ga mos keladigan \(n\) alohida ildizlari mavjudligini ham bildiradi, chunki biz \(k = n\) ga yetganimizdan so'ng biz boshlagan joyga qaytamiz.
Shuning uchun \(n\) n mavjudth birlikning ildizlari va ular tomonidan beriladi,
\begin{equation}\exp \left( {i\,\,\frac{{2\pi k}}{n}} \right) = \cos \left( {\frac{{2\pi k}}{n}} \right) + i\sin \left( {\frac{{2\pi k}}{n}} \right) \hspace = 5 dyuym, \hspace = 0,ld \u003d ,n - 1\label{eq:eq2}\end{tenglama}Ko'pincha n ni belgilash uchun ishlatiladigan oddiyroq belgi mavjudth birlik ildizlari. Avval belgilang,
\begin{equation}{\omega _n} = \exp \left( {i\,\,\frac{{2\pi }}{n}} \right)\label{eq:eq3}\end{tenglama}keyin nth birlikning ildizlari,
\[\omega _n^k = {\left( {\exp \left( {i\,\,\frac{{2\pi }}{n}} \right)} \right)^k} = \exp \left( {i\,\,\frac{{2\pi k}}{n}} \right) \hspace{0.5in} k = 0,1,2, \ldots n - 1\]Yoki oddiyroq qilib aytganda nth birlikning ildizlari,
\begin{tengation}1,{\omega _n},\omega _n^2, \ldots ,\omega _n^{n - 1}\label{eq:eq4}\end{tenglama}bu erda \({\omega _n}\) \(\eqref{eq:eq3}\) da aniqlangan.
Yechim
Biz \(n\) = 2 dan boshlaymiz.
\[{\omega _2} = \exp \left( {i\,\,\frac{{2\pi }}{2}} \right) = {{\bf{e}}^{i\,\pi }}\]Bu beradi,
\begin{align*}1 = 1 \hspace{0,5in} {\rm{va}} \hspace{0,5in} {\omega _2} & = {{\bf{e}}^{i\,\pi }}\\ & = \cos \left(\pi \right) + i\sin \left(\pi \right)\\ & = - 1\end{tekislash*}Shunday qilib, \(n\) = 2 uchun bizda -1 va n sifatida 1 borth birlik ildizlari. Bu juda ajablanarli bo'lmasa kerak, chunki biz tenglamani yechish edi
\[{z^2} = 1\]va biz hammamiz bilamizki -1 va 1 ikkita echimdir.
\(n\) = 2 uchun natija ajablanarli bo'lmasa-da, \(n\) = 3 uchun biroz hayratlanarli bo'lishi mumkin. Bu holatda biz haqiqatan ham hal qilamiz
\[{z^3} = 1\]va real sonlar dunyosida buning yechimi \(z\) = 1 ekanligini bilamiz. Biroq, yuqoridagi ishdan biz bilamizki, 3 n.th bu holatda birlikning ildizlari. Bu erda muammo shundaki, qolgan ikkitasi murakkab echimlardir va shuning uchun bu tenglamaning haqiqiy yechimini hal qilishda odatda o'ylab topilmaydi, bu biz hozirgacha xohlagan narsadir.
Shunday qilib, keling, n ni topamizth \(n\) = 3 uchun birlik ildizlari.
\[{\omega _3} = \exp \left( {i\,\,\frac{{2\pi }}{3}} \right)\]Bu beradi,
\begin{align*}1 = 1 \hspace{0,5in} {\omega _3} & = \exp \left( {i\,\,\frac{{2\pi }}{3}} \right) & \omega _3^2 & = \exp \left( {i\,\,\frac{{3\pi}}} {4\pi} & = \cos \left( {\frac{{2\pi }}{3}} \right) + i\sin \left( {\frac{{2\pi }}{3}} \right) & \hspace{0,5 dyuym} & = \cos \left( {\frac{{4\pi }}{3}} \right) + {ft {4\}le {i\f \right)\\ & = - \frac{1}{2} + \frac{{\sqrt 3 }}{2}i & & = - \frac{1}{2} - \frac{{\sqrt 3 }}{2}i\end{align*}Agar oxirgi ikkita qiymatni kubga aylantirsangiz, aslida 1 ga ega bo'lishingizni tekshirishni sizga qoldiraman.
Nihoyat, \(n=4\) orqali o'tamiz. Biz buni oldingi holatlarga qaraganda tezroq bajaramiz.
\[{\omega _4} = \exp \left( {i\,\,\frac{{2\pi }}{4}} \right) = \exp \left( {i\,\,\frac{\pi }{2}} \right)\]Bu beradi,
\begin{align*}1 = 1 \hspace{0,25 dyuym} {\omega _4} & = \exp \left( {i\,\,\frac{\pi }{2}} \right) \hspace{0,5 dyuym} & \omega _4^2 & = \exp \left( {i\s,\pace \{0,0 dyuym})\ \omega _4^3 & = \exp \left( {i\,\,\frac{{3\pi }}{2}} \right)\\ & = i & & = - 1 & & = - i\end{tekislash*}Keling, umumiy ildizlarga o'tamiz. Avval ba'zi bir eslatmani yo'ldan olib chiqaylik. Biz \(z_{0}^{{1}/{n}\;}\) ni tenglamani qanoatlantiradigan har qanday raqam sifatida aniqlaymiz
\begin{tenglama}{z^n} = {z_0}\label{eq:eq5}\end{tenglama}\(z_{0}^{{1}/{n}\;}\) qiymatlarini topish uchun biz ushbu tenglamani echishimiz kerak va biz buni xuddi n ni topgandek qilishimiz mumkin.th birlik ildizlari. Shunday qilib, agar \({r_0} = \left| {{z_0}} \right|\) va \({\theta _0} = \arg {z_0}\) (eslatma \({\theta _0}\) argumentning istalgan qiymati bo'lishi mumkin, lekin biz odatda asosiy qiymatdan foydalanamiz) bizda,
\[{\left( {r{{\bf{e}}^{i\theta }}} \right)^n} = {r_0}\,{{\bf{e}}^{i\,\,{\theta _0}}} \hspace{0.5in} \Rightarrow \hspace{0.5in} {r^n}{{\bf{e}}^{i\,\,n\theta }} = {r_0}\,{{\bf{e}}^{i\,\,{\theta _0}}}\]Demak, bu bizga shuni aytadiki,
\[r = \sqrt[n]{{{r_0}}} \hspace{0.5in} \theta = \frac{{{\theta _0}}}{n} + \frac{{2\pi k}}{n} \hspace{0.25in} k = 0, \pm 1, \pm 2, \ldots \]\(\eqref{eq:eq5}\) uchun aniq echimlar quyidagilardir:
\begin{equation}{a_k} = \sqrt[n]{{{r_0}}}\exp \left( {i\left( {\frac{{{\theta _0}}}{n} + \frac{{2\pi k}}{n}} \right)} \right) \hspace{0,25in}, k =, \ld,s 1\label{eq:eq6}\end{tenglama}Shunday qilib, biz buni avvalgidek ko'rishimiz mumkin n nth birlikning ildizlari ham bor n nth \({z_0}\) ildizlari.
Va nihoyat, biz yozuvni yana biroz soddalashtira olamiz. Agar \(a\) n dan birortasi bo'lsath \({z_0}\) ildizlari, keyin barcha ildizlar quyidagicha yozilishi mumkin:
\[a,\,\,a{\omega _n},\,\,a\omega _n^2,\, \ldots ,,\,\,a\omega _n^{n - 1}\]bu erda \({\omega _n}\) \(\eqref{eq:eq3}\) da aniqlangan.
- \({\left( {2i} \right)^{\frac{1}{2}}}\)
- \({\left( {\sqrt 3 \, - i} \right)^{\frac{1}{3}}}\)
a Yechim
Birinchi narsa, biz ildiz olayotgan kompleks sonning eksponensial shaklini yozishdir.
\[2i = 2\exp \left( {i\,\frac{\pi }{2}} \right)\]Shunday qilib, agar biz \({\theta _0} = \frac{\pi }{2}\) dan foydalansak, ildizlarni yozish uchun \(\eqref{eq:eq6}\) dan foydalanishimiz mumkin.
\[{a_k} = \sqrt 2 \exp \left( {i\left( {\frac{\pi }{4} + \pi k} \right)} \right) \hspace{0.25in} k = 0,1\]\(k\)-ga ulanish,
\begin{align*}{a_0} & = \sqrt 2 \exp \left( {i\frac{\pi }{4}} \right) \hspace{0,25in} & {a_1} & = \sqrt 2 \exp \left( {i\left( {\frac{{5\pi})}}}\right & = \sqrt 2 \left( {\cos \left( {\frac{\pi }{4}} \right) + i\sin \left( {\frac{\pi }{4}} \right)} \right) \hspace{0,25 dyuym} & & = \sqrt 2 \left( {\cos f \left {{4)} ({4)} i\sin \left( {\frac{{5\pi }}{4}} \right)} \right)\\ & = 1 + i & & = - 1 - i\end{tekislash*}Agar ikkalasini ham kvadratga aylantirsangiz, 2\(i\) olishini tekshirishni sizga qoldiraman.
b Yechim
Mana sonning eksponensial shakli,
\[\sqrt 3 - i = 2\exp \left( {i\,\left( { - \frac{\pi }{6}} \right)} \right)\]\(\eqref{eq:eq6}\) dan foydalanib, ildizlar,
\[{a_k} = \sqrt[3]{2}\exp \left( {i\left( { - \frac{\pi }{{18}} + \frac{{2\pi k}}{3}} \right)} \right) \hspace{0.25in} k = 0,1,2\]\(k\) uchun ulanish, beradi,
\begin{align*}{a_0} & = \sqrt[3]{2}\exp \left( {i\left( { - \frac{\pi }{{18}}} \right)} \right) = \sqrt[3]{2}\left( {\cos \left( { - \frac{\pi } {{18}} \frac{\pi }{{18}}} \right)} \right) = 1,24078 - 0,21878\,i\\ {a_1} & = \sqrt[3]{2}\exp \left( {i\frac{{11\pi }}{{18}}} \right) = \sqrt[3]{2}\left( {\cos \left( {\frac{{11\pi }}{{18}}} \right) + i\sin \left }}{{18}}} \right)} \right) = - 0,43092 + 1,18394\,i\\ {a_2} & = \,\sqrt[3]{2}\exp \left( {i\frac{{23\pi }}{{18}}} \right) = \sqrt[3]{2}\left( {\cos \left( {\frac{{23\pi }}{{18}}} \right) {2\ft {2\pi }}{{18}}} \right)} \right) = - 0,80986 - 0,96516\,i\end{align*}Oldingi qismda bo'lgani kabi, agar siz ularning har birini kub qilsangiz, \(\sqrt 3 - i\) ga ega bo'lishingizni tekshirishni sizga qoldiraman.