{"id":8071,"date":"2026-08-30T01:19:18","date_gmt":"2026-08-29T22:19:18","guid":{"rendered":"https:\/\/www.schooler.org.ua\/rozuminnja-matematichnogo-korinnja-vid-prostih-rishen-do-kompleksnih\/"},"modified":"2026-08-30T01:19:18","modified_gmt":"2026-08-29T22:19:18","slug":"rozuminnja-matematichnogo-korinnja-vid-prostih-rishen-do-kompleksnih","status":"publish","type":"post","link":"https:\/\/www.schooler.org.ua\/cs\/pochopeni-matematickych-korenu-od-jednoduchych-reseni-po-komplexni\/","title":{"rendered":"Pochopen\u00ed matematick\u00fdch ko\u0159en\u016f: od jednoduch\u00fdch \u0159e\u0161en\u00ed po komplexn\u00ed \u010d\u00edsla"},"content":{"rendered":"

Slovo \u201eko\u0159en\u201c mo\u017en\u00e1 zn\u00e1te jako m\u00edsto, kde roste rostlina, ale v matematice znamen\u00e1 n\u011bco \u00fapln\u011b jin\u00e9ho. Je to jen \u0159e\u0161en\u00ed rovnice. Toto rozhodnut\u00ed je obvykle \u010d\u00edslo. N\u011bkdy je to algebraick\u00fd vzorec. Koncept poch\u00e1z\u00ed z 9. stolet\u00ed. Arab\u0161t\u00ed spisovatel\u00e9 nazvali jeden ze stejn\u00fdch faktor\u016f \u010d\u00edsla jadhr<\/em>, co\u017e se p\u0159ekl\u00e1d\u00e1 jako \u201eko\u0159en\u201c. Pozd\u011bj\u0161\u00ed st\u0159edov\u011bc\u00ed evrop\u0161t\u00ed p\u0159ekladatel\u00e9 pou\u017e\u00edvali latinsk\u00e9 slovo radix<\/em>. Odtud poch\u00e1z\u00ed term\u00edn \u201eradik\u00e1ln\u00ed\u201c. <\/p>\n

Z\u00e1kladn\u00ed pravidlo je jednoduch\u00e9. Jestli\u017ee a<\/em> je kladn\u00e9 re\u00e1ln\u00e9 \u010d\u00edslo a n<\/em> je kladn\u00e9 cel\u00e9 \u010d\u00edslo, pak existuje jedine\u010dn\u00e9 kladn\u00e9 re\u00e1ln\u00e9 \u010d\u00edslo x<\/em> takov\u00e9, \u017ee x<\/em> n<\/em> = a<\/em>. Toto \u010d\u00edslo je (hlavn\u00ed) n<\/em> -tou odmocninou a<\/em>. Zapisuje se jako \u221aa<\/em> nebo a<\/em> 1\/n<\/em>. Cel\u00e9 \u010d\u00edslo n<\/em> se naz\u00fdv\u00e1 ko\u0159enov\u00fd index. <\/p>\n

Kdy\u017e n<\/em> = 2, naz\u00fdv\u00e1me to odmocnina. P\u00ed\u0161e se jako \u221aa<\/em>. Kdy\u017e n<\/em> = 3, m\u00e1me krychlovou odmocninu zapsanou jako \u221ba<\/em>. Co kdy\u017e je a<\/em> z\u00e1porn\u00e9? Pokud je n<\/em> lich\u00e9, jedin\u00e1 z\u00e1porn\u00e1 n<\/em> t\u00e1 odmocnina z a<\/em> se naz\u00fdv\u00e1 hlavn\u00ed ko\u0159en. Pod\u00edvejme se na tento p\u0159\u00edklad. Hlavn\u00ed odmocnina z \u201327 je \u20133. <\/p>\n

Kdy\u017e jsou ko\u0159eny racion\u00e1ln\u00ed<\/h3>\n

Zde je pravidlo, na kter\u00e9 se m\u016f\u017eete spolehnout. Pokud m\u00e1 cel\u00e9 \u010d\u00edslo racion\u00e1ln\u00ed n<\/em> -tou odmocninu, to znamen\u00e1, \u017ee jej lze zapsat jako oby\u010dejn\u00fd zlomek, pak tento ko\u0159en mus\u00ed b\u00fdt cel\u00e9 \u010d\u00edslo. Vezm\u011bme 5. Nem\u00e1 racion\u00e1ln\u00ed druhou odmocninu. Pro\u010d? Proto\u017ee 2\u00b2 je men\u0161\u00ed ne\u017e 5 a 3\u00b2 je v\u00edce ne\u017e 5. Mezi 2 a 3 nen\u00ed cel\u00e9 \u010d\u00edslo. Odmocnina z 5 proto nem\u016f\u017ee b\u00fdt jednoduch\u00fd zlomek. <\/p>\n

Komplexn\u00ed ko\u0159eny a jednotka<\/h3>\n

Nyn\u00ed se obr\u00e1zek st\u00e1v\u00e1 zaj\u00edmav\u011bj\u0161\u00edm. P\u0159esn\u011b n<\/em> komplexn\u00edch \u010d\u00edsel vyhovuje rovnici x<\/em> n<\/em> = 1. \u0158\u00edk\u00e1 se jim komplexn\u00ed n<\/em> ko\u0159eny jednoty. P\u0159edstavte si jednotkovou kru\u017enici se st\u0159edem v po\u010d\u00e1tku. Vlo\u017ete do n\u011bj pravideln\u00fd n<\/em> -\u00faheln\u00edk. Jeden z vrchol\u016f le\u017e\u00ed na kladn\u00e9 poloose x<\/em>. Polom\u011bry nakreslen\u00e9 k vrchol\u016fm jsou vektory. Tyto vektory p\u0159edstavuj\u00ed n<\/em> komplexn\u00ed n<\/em> -t\u00fd ko\u0159en jednoty. <\/p>\n

Jestli\u017ee ko\u0159en, jeho\u017e vektor sv\u00edr\u00e1 nejmen\u0161\u00ed kladn\u00fd \u00fahel s kladn\u00fdm sm\u011brem osy x<\/em>, ozna\u010d\u00edme \u0159eck\u00fdm p\u00edsmenem omega (\u03c9), pak \u03c9, \u03c9\u00b2, \u03c9\u00b3, \u2026, \u03c9n<\/em> = 1 tvo\u0159\u00ed v\u0161echny n<\/em> -t\u00e9 ko\u0159eny jednoty. <\/p>\n

Vezm\u011bme si jako p\u0159\u00edklad krychlov\u00e9 ko\u0159eny jednoty.
\n\u03c9 = \u22121\/2 + \u221a\u22123 \/2
\n\u03c9\u00b2 = \u22121\/2 \u2212 \u221a\u22123 \/2
\n\u03c9\u00b3 = 1<\/p>\n

Jak\u00fdkoli ko\u0159en, ozna\u010den\u00fd \u0159eck\u00fdm p\u00edsmenem epsilon (\u03b5), kter\u00fd m\u00e1 vlastnost, \u017ee \u03b5, \u03b5\u00b2, \u2026, \u03b5n<\/em> = 1 d\u00e1v\u00e1 v\u0161em n<\/em> -t\u00fdm ko\u0159en\u016fm jednoty, se naz\u00fdv\u00e1 primitivn\u00ed. Nalezen\u00ed t\u011bchto ko\u0159en\u016f je ekvivalentn\u00ed veps\u00e1n\u00ed pravideln\u00e9ho n<\/em> -\u00faheln\u00edku do kruhu. <\/p>\n

Je mo\u017en\u00e9 sestrojit tyto ko\u0159eny pouze pomoc\u00ed prav\u00edtka a kru\u017e\u00edtka? Ano, pro ur\u010dit\u00e1 \u010d\u00edsla. Toti\u017e, pokud n<\/em> je sou\u010din r\u016fzn\u00fdch prvo\u010d\u00edsel ve tvaru 2h<\/em> + 1. Nebo 2k<\/em> kr\u00e1t takov\u00fd sou\u010din. Nebo m\u00e1 tvar 2k<\/em>. Pro \u017e\u00e1dn\u00e9 jin\u00e9 cel\u00e9 \u010d\u00edslo n<\/em> je nebudete moci vykreslit t\u00edmto zp\u016fsobem. M\u016f\u017eete je definovat pomoc\u00ed racion\u00e1ln\u00edch operac\u00ed a radik\u00e1l\u016f. Ale ne s jednoduch\u00fdmi aritmetiky a odmocniny. <\/p>\n

Ko\u0159eny v polynomi\u00e1ln\u00edch rovnic\u00edch<\/h3>\n

Term\u00edn \u201eko\u0159en\u201c p\u0159es\u00e1hl r\u00e1mec jednoduch\u00fdch rovnic. Nyn\u00ed plat\u00ed pro v\u0161echny polynomick\u00e9 rovnice. \u0158e\u0161en\u00ed rovnice f<\/em> (x<\/em> ) = a<\/em> 0x<\/em> n<\/em> + a<\/em> 1x<\/em> n<\/em> \u2212 1 + \u2026 + a<\/em> n<\/em> \u2212 1x<\/em> + a<\/em> n<\/em> = 0 se naz\u00fdv\u00e1 ko\u0159en. Zde a<\/em> 0 \u2260 0.<\/p>\n

Jestli\u017ee koeficienty le\u017e\u00ed v komplexn\u00edm poli, pak rovnice n<\/em> t\u00e9ho stupn\u011b m\u00e1 p\u0159esn\u011b n<\/em> ko\u0159en\u016f. Nemus\u00ed se nutn\u011b li\u0161it. Pokud jsou koeficienty re\u00e1ln\u00e9 a n<\/em> je lich\u00e9, pak existuje alespo\u0148 jeden skute\u010dn\u00fd ko\u0159en. <\/p>\n

Ale rovnice nem\u00e1 v\u017edy ko\u0159en ve sv\u00e9m koeficientov\u00e9m poli. Uva\u017eujme x<\/em> \u00b2 \u2212 5 = 0. Nem\u00e1 \u017e\u00e1dn\u00fd racion\u00e1ln\u00ed ko\u0159en. Jeho koeficienty (1 a \u20135) jsou racion\u00e1ln\u00ed \u010d\u00edsla. \u0158e\u0161en\u00ed v\u0161ak zahrnuje iracion\u00e1ln\u00ed \u010d\u00edslo. <\/p>\n

\u0160irok\u00e1 definice ko\u0159ene<\/h3>\n

Obecn\u011bji m\u016f\u017ee b\u00fdt term\u00edn “ko\u0159en” aplikov\u00e1n na jak\u00e9koli \u010d\u00edslo, kter\u00e9 spl\u0148uje jakoukoli danou rovnici. Pat\u0159\u00ed sem i nepolynomi\u00e1ln\u00ed rovnice. Vezm\u011bme si nap\u0159\u00edklad \u03c0. Toto je ko\u0159en rovnice x<\/em> sin (x<\/em> ) = 0.<\/p>\n

Pro\u010d je to pro v\u00e1s d\u016fle\u017eit\u00e9? Proto\u017ee porozum\u011bn\u00ed ko\u0159en\u016fm pom\u00e1h\u00e1 \u0159e\u0161it probl\u00e9my od z\u00e1kladn\u00ed algebry po pokro\u010dilou geometrii. To spojuje jednoduchou aritmetiku se slo\u017eit\u00fdmi strukturami. A\u017e p\u0159\u00ed\u0161t\u011b uvid\u00edte symbol jako \u221a nebo \u221b, nezapome\u0148te, \u017ee p\u0159edstavuje konkr\u00e9tn\u00ed \u0159e\u0161en\u00ed. Jeden z potenci\u00e1ln\u011b mnoha. Most mezi rovnic\u00ed a jej\u00ed odpov\u011bd\u00ed. <\/p>\n

Historie samotn\u00e9ho slova vypr\u00e1v\u00ed p\u0159\u00edb\u011bh p\u0159ekladu a adaptace. Od jadhr<\/em> k radix<\/em> a d\u00e1le k modern\u00ed angli\u010dtin\u011b. Matematika se vyv\u00edj\u00ed spolu s jazykem. Ale z\u00e1kladn\u00ed logika z\u016fst\u00e1v\u00e1 stejn\u00e1. Odmocnina je hodnota, d\u00edky kter\u00e9 je rovnice pravdiv\u00e1. A\u0165 u\u017e je to cel\u00e9 \u010d\u00edslo. Zlomek. Nebo komplexn\u00ed \u010d\u00edslo s imagin\u00e1rn\u00edmi \u010d\u00e1stmi. <\/p>\n

Znamen\u00e1 to, \u017ee ka\u017ed\u00e1 rovnice m\u00e1 ko\u0159en? Ve slo\u017eit\u00e9m oboru \u2013 ano. V racion\u00e1ln\u00edm poli – ne. Toto rozli\u0161en\u00ed je kl\u00ed\u010dov\u00e9. Formuje to, jak k probl\u00e9m\u016fm p\u0159istupujeme. Ur\u010duje, jak\u00e9 n\u00e1stroje pou\u017e\u00edv\u00e1me. A definuje hranice toho, co m\u016f\u017eeme postavit.<\/p>\n","protected":false},"excerpt":{"rendered":"

Slovo \u201eko\u0159en\u201c mo\u017en\u00e1 zn\u00e1te jako m\u00edsto, kde roste rostlina, ale v matematice znamen\u00e1 n\u011bco \u00fapln\u011b jin\u00e9ho. Je to jen \u0159e\u0161en\u00ed rovnice. Toto rozhodnut\u00ed je obvykle \u010d\u00edslo. N\u011bkdy je to algebraick\u00fd vzorec. Koncept poch\u00e1z\u00ed z 9. stolet\u00ed. Arab\u0161t\u00ed spisovatel\u00e9 nazvali jeden ze stejn\u00fdch faktor\u016f \u010d\u00edsla jadhr, co\u017e se p\u0159ekl\u00e1d\u00e1 jako \u201eko\u0159en\u201c. Pozd\u011bj\u0161\u00ed st\u0159edov\u011bc\u00ed evrop\u0161t\u00ed p\u0159ekladatel\u00e9 pou\u017e\u00edvali […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"tdm_status":"","tdm_grid_status":""},"categories":[50],"tags":[],"wpm_language_slugs":{"cs":"pochopeni-matematickych-korenu-od-jednoduchych-reseni-po-komplexni","de":"mathematische-wurzeln-verstehen-von-einfachen-losungen-zu-komplexen","en":"understanding-mathematical-roots-from-simple-solutions-to-complex","es":"comprension-de-las-raices-matematicas-de-soluciones-simples-a","fr":"comprendre-les-racines-mathematiques-des-solutions-simples-aux-nombres","id":"pengertian-akar-matematika-dari-penyelesaian-sederhana-hingga-bilangan","it":"comprendere-le-radici-matematiche-dalle-soluzioni-semplici-ai-numeri","nl":"wiskundige-wortels-begrijpen-van-eenvoudige-oplossingen-tot-complexe","pl":"zrozumienie-pierwiastkow-matematycznych-od-prostych-rozwiazan-po","pt":"compreendendo-as-raizes-matematicas-de-solucoes-simples-a-numeros","ru-ru":"ponimanie-matematicheskih-kornej-ot-prostyh-reshenij-k-kompleksnym","uk-ua":"rozuminnja-matematichnogo-korinnja-vid-prostih-rishen-do-kompleksnih"},"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.schooler.org.ua\/cs\/wp-json\/wp\/v2\/posts\/8071"}],"collection":[{"href":"https:\/\/www.schooler.org.ua\/cs\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.schooler.org.ua\/cs\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.schooler.org.ua\/cs\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.schooler.org.ua\/cs\/wp-json\/wp\/v2\/comments?post=8071"}],"version-history":[{"count":0,"href":"https:\/\/www.schooler.org.ua\/cs\/wp-json\/wp\/v2\/posts\/8071\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.schooler.org.ua\/cs\/wp-json\/wp\/v2\/media?parent=8071"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.schooler.org.ua\/cs\/wp-json\/wp\/v2\/categories?post=8071"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.schooler.org.ua\/cs\/wp-json\/wp\/v2\/tags?post=8071"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}