3. konjugat va modul
Oldingida biz murakkab sonlar ustidagi algebraik amallarni ko'rib chiqdik. Biz ko'rib chiqishimiz kerak bo'lgan yana bir nechta amallar bor, chunki ular vaqti-vaqti bilan paydo bo'ladi. Shuningdek, biz ushbu operatsiyalar haqida bir nechta yoqimli faktlarni ko'rib chiqamiz.
Murakkab konjugat
Biz ko'rib chiqadigan birinchi narsa murakkab konjugat, (yoki shunchaki konjugat). \(z = a + bi\) kompleks raqamini hisobga olgan holda, kompleks konjugat \(\overline z\) bilan belgilanadi va quyidagicha aniqlanadi:
\begin{equation}\overline z = a - bi\end{tenglama}Boshqacha qilib aytganda, biz raqamning xayoliy qismidagi belgini almashtiramiz.
Bu erda konjugatlar haqida ba'zi asosiy faktlar mavjud.
\begin{align} \overline{\overline{z}} & = z \\ \overline {{z_1} \pm {z_2}} & = {\overline z_1} \pm {\overline z_2} \label{eq:conjsum} \\ \overline {{z_1}{z_2}} &= {\overline z_1}\,{\overline z_2} \label{eq:conjprod} \\ \overline {\left( {\frac{{{z_1}}}{{{z_2}}}} \right)} & = \frac{{{{\overline z}_1}}}{{{{\overline z}_2}}} \end{tekislash}Birinchisi shunchaki aytadiki, agar biz ikki marta konjugatsiya qilsak, biz dastlab boshlagan narsamizga qaytamiz va umid qilamanki, bu qandaydir ma'noga ega. Qolgan uchtasi shunchaki yig'indi, farqlar, mahsulot va bo'laklarni alohida qismlarga ajratib, keyin konjugatsiya qilishimiz mumkinligini aytadi.
Shunday qilib, biz bir yoki ikkita misol ishladik deb aytishimiz mumkin, keling, yuqoridagi faktlarni ko'rsatadigan bir nechta misol keltiramiz.
- \(z = 3 - 15i\) uchun \(\overline{\overline{z}}\)
- \({z_1} = 5 + i\) va \({z_2} =- 8 + 3i\) uchun \(\overline {{z_1} - {z_2}} \)
- \({z_1} = 5 + i\) va \({z_2} =- 8 + 3i\) uchun \({\overline{z_1}} - {\overline{z_2}}\)
Yechim
a \(\overline{z} = 3 + 15i \hspace{0.5in} \Rightarrow \hspace{0.5in} \overline{\overline{z}} = \overline{3 + 15i}= 3 - 15i = z\)
Albatta, biz ikki marta konjugatsiyadan so'ng asl raqamimizga qaytishimizni ko'rishimiz mumkin.
b \({z_1} - {z_2} = 13 - 2i \hspace{0.5in} \Rightarrow \hspace{0.5in} \overline {{z_1} - {z_2}}= \overline {13 - 2i}= 13 + 2i\)
c \({\overline z_1} - {\overline z_2} = \overline {5 + i}- \left( {\overline { - 8 + 3i} } \right) = 5 - i - \left( { - 8 - 3i} \right) = 13 + 2i\)
Buning natijasini ko'rishimiz mumkin (b) va (c) ular nazarda tutilgan haqiqat bilan bir xil.
Konjugatlardan foydalanadigan yana bir yoqimli fakt borki, biz uni ko‘rib chiqishimiz kerak. Biroq, shunchaki faktni aytib berish o‘rniga, keling, uni chiqaramiz. Biz \(z = a + bi\) kompleks raqamidan boshlaymiz va keyin quyidagi amallarning har birini bajaramiz.
\[\begin{array}{rlcrl}z + \overline{z} & = a + bi + \left( {a - bi} \right) & \hspace{0.5in} & z - \overline{z} & = a + bi - \left( {a - bi} \right)\\ & = 2a & \hspace{0.5in} & & = 2bi\end{array}\]Endi \({\mathop{\rm Re}\nolimits}\, z = a\) va \({\mathop{\rm Im}\nolimits}\, z = b\) ni eslab, bizda borligini ko'ramiz,
\begin{equation}{\mathop{\rm Re}\nolimits} z = \frac{{z + \overline z}}{2} \hspace{0,75in} {\mathop{\rm Im}\nolimits} z = \frac{{z - \overline z}}{{2i}} \labelm{eq:} \labelm{eq:}Modul
Ushbu bo'limda biz ko'rib chiqmoqchi bo'lgan boshqa operatsiya modul kompleks son. \(z = a + bi\) kompleks son berilganda modul \(\left| z \right|\) bilan belgilanadi va quyidagicha aniqlanadi.
\begin{tenglama}\left| z \right| = \sqrt {{a^2} + {b^2}} \label{eq:ModDefn} \end{tenglama}E'tibor bering, kompleks sonning moduli har doim haqiqiy sondir va aslida u hech qachon manfiy bo'lmaydi, chunki kvadrat ildizlar radikal ostidagi narsaga qarab har doim ijobiy son yoki nolni qaytaradi.
E'tibor bering, agar \(z\) haqiqiy raqam bo'lsa (i.e. \(z = a + 0i\)) keyin,
\[\left| z \right| = \sqrt {{a^2}}= \left| a \right|\]bu yerda \(z\) da \(\left| {\,\, \cdot \,} \right|\) kompleks sonning moduli va \(a\) da \(\left| {\,\, \cdot \,} \right|\) haqiqiy sonning mutlaq qiymatidir (esda tutingki, umuman har qanday haqiqiy son uchun \(\sqrt {{a^2}}= \left| a \right|\) bizda mavjud). Demak, bundan biz haqiqiy va mutlaq qiymatlar uchun asosiy modul ekanligini ko‘rishimiz mumkin.
Kompleks sonning moduli bilan uning haqiqiy va xayoliy qismlari o‘rtasidagi munosabat haqida yaxshi ma’lumotga ega bo‘lishimiz mumkin. Buni ko‘rish uchun \(\eqref{eq:ModDefn}\) ning ikkala tomonini kvadrat qilib, \({\mathop{\rm Re}\nolimits}\, z = a\) va \({\mathop{\rm Im}\nolimits}\, z = b\) faktlaridan foydalanamiz.
\[{\left| z \right|^2} = {a^2} + {b^2} = {\left( {{\mathop{\rm Re}\nolimits}\, z} \right)^2} + {\left( {{\mathop{\rm Im}\nolimits}\, z} \right)^2}\]Ushbu uchta atamaning barchasi ijobiy bo'lganligi sababli biz \({\mathop{\rm Im}\nolimits}\, z\) qismini chapga tushirishimiz mumkin, bu esa quyidagi tengsizlikni beradi,
\[{\left| z \right|^2} = {\left( {{\mathop{\rm Re}\nolimits}\, z} \right)^2} + {\left( {{\mathop{\rm Im}\nolimits}\, z} \right)^2} \ge {\left( {{\mathop{\rm Re}\nolimits}\, z} \right)^2}\]Agar buning ikkala tomonining kvadrat ildizi bo'lsa, biz,
\[\left| z \right| \ge \left| {{\mathop{\rm Re}\nolimits}\, z} \right|\]bu erda \(z\)-dagi \(\left| {\,\, \cdot \,} \right|\) kompleks sonning moduli va \({\mathop{\rm Re}\nolimits}\, z\)-dagi \(\left| {\,\, \cdot \,} \right|\) mutlaq qiymatlar satridir. Va nihoyat, har qanday \(a\) haqiqiy soni uchun biz \(a \le \left| a \right|\) (mutlaq qiymat…) ekanligini ham bilamiz va shuning uchun biz quyidagilarni olamiz:
\begin{tenglama}\left| z \right| \ge \left| {{\mathop{\rm Re}\nolimits}\, z} \right| \ge {\mathop{\rm Re}\nolimits}\, z \label{eq:zRez} \end{tenglama}Biz shunga o'xshash argumentdan foydalanishimiz mumkin,
\begin{tenglama}\left| z \right| \ge \left| {{\mathop{\rm Im}\nolimits}\, z} \right| \ge {\mathop{\rm Im}\nolimits}\, z \label{eq:zImz} \end{tenglama}Kompleks sonning moduli va uning konjugati o'rtasida juda yaxshi bog'liqlik mavjud. Keling, \(z = a + bi\) kompleks raqamidan boshlaymiz va quyidagi ko'paytmani ko'rib chiqamiz.
\[z\,\overline z = \left( {a + bi} \right)\left( {a - bi} \right) = {a^2} + {b^2}\]Ushbu mahsulotdan biz buni ko'rishimiz mumkin
\begin{tenglama}z\,\overline z = {\left| z \right|^2}\label{eq:zConjz} \end{tenglama}Bu ba'zida yoqimli va qulay fakt.
E'tibor bering, modulni hisoblashda kompleks sonning haqiqiy va xayoliy qismidagi belgi modul qiymatiga ta'sir qilmaydi va shuning uchun biz buni ham ko'rishimiz mumkin,
\begin{tenglama}\left| z \right| = \left| {\overline z} \right|\label{eq:MzMzbar}\end{tenglama}va
\begin{tenglama}\left| { - z} \right| = \left| z \right|\end{tenglama}Modul va konjugat belgilariga ega boʻlganimiz uchun biz avvalgi boʻlimdan boʻlinish jarayonini ham rasmiylashtirishimiz mumkin. Boʻlimning maxrajidan \(i\) ni olish uchun biz pay va maxrajni maxrajning konjugatiga koʻpaytirdik. Keyin \(\eqref{eq:zConjz}\) formulasidan foydalanib, quyidagi formulani biroz soddalashtira olmaymiz. bo'linish,
\[\frac{{{z_1}}}{{{z_2}}} = \frac{{{z_1}\,{{\overline z}_2}}}{{{z_2}\,{{\overline z}_2}}} = \frac{{{z_1}\,{{\overline z}_2}}}{{{{\left| {{z_2}} \right|}^2}}}\]Buni ko'rsatadigan tezkor misol,
Yechim
Bunda koeffitsient,
\[\frac{{6 + 3i}}{{10 + 8i}} = \frac{{\left( {6 + 3i} \right)\left( {10 - 8i} \right)}}{{164}} = \frac{{60 - 48i + 30i - 24{i^2}}}{{164}} = \frac{{21}}{{41}} - \frac{9}{{82}}i\]Mana, kompleks sonning moduli haqida yana bir qancha yoqimli faktlar.
\begin{align}{\rm{Agar }}\left bo'lsa| z \right| & = 0\,\,\,{\rm{keyin}}\, z = 0 \label{eq:Mzero} \\ \left| {{z_1}\,{z_2}} \right| & = \left| {{z_1}} \right|\,\left| {{z_2}} \right| \label{eq:MProd} \\ \left| {\frac{{{z_1}}}{{{z_2}}}} \right| & = \frac{{\left| {{z_1}} \right|}}{{\left| {{z_2}} \right|}} \label{eq:MQuot} \end{tekislash}\(\eqref{eq:Mzero}\) mulki siz uchun ma'noli bo'lishi kerak. Agar modul nolga teng bo'lsa, \({a^2} + {b^2} = 0\), lekin bu nolga teng bo'lishining yagona yo'li \(a\) va \(b\) nolga teng bo'lsa.
\(\eqref{eq:MProd}\) ni tekshirish uchun quyidagilarni ko'rib chiqing:
\begin{align*}{\left| {{z_1}\,{z_2}} \right|^2} & = \left( {{z_1}\,{z_2}} \right)\overline {\left( {{z_1}\,{z_2}} \right)} & \hspace{0,5 dyuym} va {\text{xususiyatdan foydalanish }}e \eq:j & = \left( {{z_1}\,{z_2}} \right)\left( {{{\overline z}_1}\,{{\overline z}_2}} \right) & \hspace{0,5in} & {\text{xususiyatdan foydalanish }} \eqref{eq:conjprod} \\ & = {z_1}{{\overline z}_1}\,{z_2}{{\overline z}_2} & \hspace{0,5in} & {\text{shartlarni qayta tartiblash}}\\ & = {\ chap | {{z_1}} \right|^2}\,{\left| {{z_2}} \right|^2} & \hspace{0,5in} & {\text{\eqref{eq:zConjz} xususiyatidan foydalanish (ikki marta)}}\end{align*}Demak, bundan biz buni ko'rishimiz mumkin
\[{\left| {{z_1}\,{z_2}} \right|^2} = {\left| {{z_1}} \right|^2}\,{\left| {{z_2}} \right|^2}\]Va nihoyat, modul har doim ijobiy ekanligini bilamiz, shuning uchun ikkala tomonning kvadrat ildizini oling.
\[\left| {{z_1}\,{z_2}} \right| = \left| {{z_1}} \right|\,\,\left| {{z_2}} \right|\]\(\eqref{eq:MQuot}\) mulki shunga o'xshash argument yordamida tekshirilishi mumkin.
Uchburchak tengsizligi va variantlari
\(\eqref{eq:MProd}\) va \(\eqref{eq:MQuot}\) xossalari ikkita kompleks sonning ko‘paytmasi/ko‘paytmasi modulini alohida sonlar modulining ko‘paytmasi/bo‘limi bilan bog‘laydi. Endi biz kompleks sonlar yig‘indisi uchun o‘xshash munosabatni ko‘rib chiqishimiz kerak. Bu munosabat deyiladi uchburchak tengsizligi va,
\begin{tenglama}\left| {{z_1} + {z_2}} \right| \le \left| {{z_1}} \right| + \left| {{z_2}} \right| \label{eq:uchburchak} \end{tenglama}Bundan murakkab sonlar farqiga munosabatni olish uchun ham foydalanishimiz mumkin.
Uchburchak tengsizligini isbotlash juda oddiy, shuning uchun buni qilaylik. Biz chap tomonni kvadrat bilan boshlaymiz va uni biroz qayta yozish uchun \(\eqref{eq:zConjz}\) va \(\eqref{eq:conjsum}\) dan foydalanamiz.
\[{\left| {{z_1} + {z_2}} \right|^2} = \left( {{z_1} + {z_2}} \right)\left( {\overline {{z_1} + {z_2}} } \right) = \left( {{z_1} + {z_2}} \right)\left( {\overline {{z_1}}+ \overline {{z_2}} } \right)\]Endi olish uchun o'ng tomonni ko'paytiring,
\begin{equation}\left| {{z_1} + {z_2}} \right|^2} = {z_1}\,{\overline z_1} + {z_1}\,{\overline z_2} + {z_2}\,{\overline z_1} + {z_2}\,{\overline z_2}\label{eq:tripquationKeyingi e'tibor bering,
\[\overline {{z_2}{{\overline z}_1}}= {\overline z_2}{\overline{\overline{z_1}}} = {\overline z_2}{z_1}\]va shuning uchun \(\eqref{eq:ReImDefn}\), \(\eqref{eq:zRez}\) va \(\eqref{eq:MzMzbar}\) dan foydalanib, biz \(\eqref{eq:tripfone}\) ning o'ng tomonining o'rta ikkita hadini quyidagicha yozishimiz mumkin.
\[{z_1}\,{\overline z_2} + {z_2}\,{\overline z_1} = {z_1}\,{\overline z_2} + \overline {{z_1}\,{{\overline z}_2}}= 2{\mathop{\rm Re}\nolimits} \left( {{z_1}\,{{\overline z}_2}} \right) \le 2\left| {{z_1}\,{{\overline z}_2}} \right| = 2\left| {{z_1}} \right|\,\left| {{{\overline z}_2}} \right| = 2\left| {{z_1}} \right|\,\left| {{z_2}} \right|\]Shuningdek, \(\eqref{eq:zConjz}\) dan \(\eqref{eq:tripfone}\) da birinchi va to‘rtinchi atamalarni shunday yozish uchun foydalaning:
\[{z_1}\,{\overline z_1} = {\left| {{z_1}} \right|^2} \hspace{0.75in} {z_2}\,{\overline z_2} = {\left| {{z_2}} \right|^2}\]O'rtadagi ikkita atamani qayta yozish bilan biz endi \(\eqref{eq:tripfone}\) ni shunday yozishimiz mumkin
\begin{align*}{\left| {{z_1} + {z_2}} \right|^2} & = {z_1}\,{{\overline z}_1} + {z_1}\,{{\overline z}_2} + {z_2}\,{{\overline z}_1} + {z_2}\,{{\overline z}_2}\le {{z_1}} \right|^2} + {z_1}\,{{\overline z}_2} + {z_2}\,{{\overline z}_1} + {\left| {{z_2}} \right|^2}\\ & \le {\left| {{z_1}} \right|^2} + 2\left| {{z_1}} \right|\left| {{z_2}} \right| + {\left| {{z_2}} \right|^2}\\ & = {\left( {\left| {{z_1}} \right| + \left| {{z_2}} \right|} \right)^2}\end{tizalang*}Shunday qilib, bularning barchasini birlashtirganda,
\[{\left| {{z_1} + {z_2}} \right|^2} \le {\left( {\left| {{z_1}} \right| + \left| {{z_2}} \right|} \right)^2}\]Endi modul har doim ijobiy ekanligini eslab, biz ikkala tomonning kvadrat ildizini olamiz va biz uchburchak tengsizligiga erishamiz.
\[\left| {{z_1} + {z_2}} \right| \le \left| {{z_1}} \right| + \left| {{z_2}} \right|\]Uchburchak tengsizligining bir nechta o'zgarishlari mavjud bo'lib, ularning barchasini osongina olish mumkin.
Avval \(\left| {{z_1}} \right| \ge \left| {{z_2}} \right|\) deb faraz qilishdan boshlaylik. Bu hosila olish uchun talab qilinmaydi, lekin biz bu yerda olinadigan narsaning umumiy versiyasini olishga yordam beradi. Shunday qilib, keling, \(\left| {{z_1}} \right|\) dan boshlaylik va u ustida biroz ishlaymiz.
\begin{align*}\left| {{z_1}} \right| & = \left| {{z_1} + {z_2} - {z_2}} \right| & \\ & \le \left| {{z_1} + {z_2}} \right| + \left| { - {z_2}} \right| & \hspace{0,25in} {\text{Uchburchak tengsizligidan foydalanish}}\\ & = \left| {{z_1} + {z_2}} \right| + \left| {{z_2}} \right| & \end{tekislash*}Endi narsalarni biroz qayta yozing va biz olamiz:
\begin{tenglama}\left| {{z_1} + {z_2}} \right| \ge \left| {{z_1}} \right| - \left| {{z_2}} \right| \ge 0\label{eq:revtrione} \end{tenglama}Agar biz \(\left| {{z_1}} \right| \le \left| {{z_2}} \right|\) ni endi yuqoridagiga o'xshash jarayondan o'tishimiz mumkin deb hisoblasak, \({z_1}\) va \({z_2}\) vaqtini almashtiramiz va biz quyidagilarni olamiz:
\begin{tenglama}\left| {{z_1} + {z_2}} \right| \ge \left| {{z_2}} \right| - \left| {{z_1}} \right| =- \left( {\left| {{z_1}} \right| - \left| {{z_2}} \right|} \right) \ge 0\label{eq:revtritwo}\end{tenglama}Endi, mutlaq qiymat ta'rifini eslab, biz \(\eqref{eq:revtrione}\) va \(\eqref{eq:revtritwo}\) ni uchburchak tengsizligining quyidagi o'zgarishiga birlashtira olamiz.
\begin{tenglama}\left| {{z_1} + {z_2}} \right| \ge \left| {{\kern 1pt} \left| {{z_1}} \right| - \left| {{z_2}} \right|{\ yadro 1pt} } \right|\yorliq{eq:revtrithree}\end{tenglama}Bundan tashqari, \({z_2}\) ni \(\eqref{eq:triangle} \) va \(\eqref{eq:revtrithree} \) da \( - {z_2}\) bilan almashtirsak, biz uchburchak tengsizligining yana ikkita variantiga erishamiz.
\begin{tenglama}\left| {{z_1} - {z_2}} \right| \le \left| {{z_1}} \right| + \left| {{z_2}} \right|\end{tenglama} \begin{tenglama}\left| {{z}_{1}}-{{z}_{2}} \right|\ge \left| \left| {{z}_{1}} \right|-\left| {{z}_{2}} \right| \right| \label{eq:revtri} \end{tenglama}Ba'zida siz \(\eqref{eq:revtri}\) ni ko'rasiz teskari uchburchak tengsizligi.