Aptarimas:Kompleksiniai skaičiai

nesklandumai

• Ištraukti šaknį 5 laipsnio šaknį iš kompleksinio skaičiaus $\alpha =-16+i16{\sqrt {3}}.$

Sprendimas. Randame spindulį

$r={\sqrt {16^{2}+(16{\sqrt {3}})^{2}}}={\sqrt {256+256\cdot 3}}={\sqrt {1024}}=32.$

Toliau taikydami formule galime užrašyti

${\sqrt {\alpha }}=\beta _{k}={\sqrt[{5}]{r}}(\cos {\frac {\phi +2k\pi }{5}}+i\sin {\frac {\phi +2k\pi }{5}})\quad (k=0,\;1,\;2,\;3,\;4).$

Pažymėkime:

$\epsilon ={\frac {\alpha }{r}}={\frac {-16+i16{\sqrt {3}}}{32}}=-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}.$

Dabar žinome, kad $\phi =-60^{\circ }=360^{\circ }-60^{\circ }=300^{\circ }$ arba $\phi =-{\frac {\pi }{3}}=2\pi -{\pi \over 3}={\frac {5\pi }{3}},$ nes $\cos {\frac {5\pi }{3}}={\frac {1}{2}}$ (kosinusas lyginė funkcija todėl nėra minuso) ir $\sin {\frac {5\pi }{3}}=$

Nesklandumai 2

Antras būdas surasti visas šaknis yra toks:

$\beta _{0}={\sqrt[{5}]{r}}\epsilon _{0},\;\beta _{1}={\sqrt[{5}]{r}}\epsilon _{0}^{2},\;\beta _{2}={\sqrt[{5}]{r}}\epsilon _{0}^{3},\;\beta _{3}={\sqrt[{5}]{r}}\epsilon _{0}^{4},$

$\beta _{4}={\sqrt[{5}]{r}}\epsilon _{0}^{5}={\sqrt[{5}]{32}}(0.913545457+i0.406736643)(0.913545457+i0.406736643)(0.913545457+i0.406736643)^{3}=$

$=2(0.834565303-0.165434696+i0.743144825)(0.913545457+i0.406736643)^{3}=$

$=2(0.669130606+i0.743144825)(0.913545457+i0.406736643)(0.913545457+i0.406736643)^{2}=$

$=2(0.611281225+i0.272159936+i0.678896578-0.302264231)(0.913545457+i0.406736643)^{2}=$

$=2(0.309016994+i0.951056514)(0.913545457+i0.406736643)^{2}=2(0.309016994+i0.951056514)(0.669130606+i0.743144825)=$

$=2(0.206772728-0.706772726+i0.636381021+i0.229644379)(0.669130606+i0.743144825)=2(-0.5+i0.8660254).$

Antras būdas surasti visas šaknis yra toks:

$\beta _{0}={\sqrt[{5}]{r}}\epsilon _{0},\;\beta _{1}={\sqrt[{5}]{r}}\epsilon _{0}^{2},\;\beta _{2}={\sqrt[{5}]{r}}\epsilon _{0}^{3},\;\beta _{3}={\sqrt[{5}]{r}}\epsilon _{0}^{4},$

$\beta _{4}={\sqrt[{5}]{r}}\epsilon _{0}^{5}={\sqrt[{5}]{32}}(0.913545457+i0.406736643)(0.913545457+i0.406736643)(0.913545457+i0.406736643)^{3}=$

$=2(0.834565303-0.165434696+i0.743144825)(0.913545457+i0.406736643)^{3}=$

$=2(0.669130606+i0.743144825)(0.913545457+i0.406736643)(0.913545457+i0.406736643)^{2}=$

$=2(0.611281225+i0.272159936+i0.678896578-0.302264231)(0.913545457+i0.406736643)^{2}=$

$=2(0.309016994+i0.951056514)(0.913545457+i0.406736643)^{2}=$

$=2(0.282301071-0.386829533+i0.868833357+i0.125688534)(0.913545457+i0.406736643)=$

$=2(-0.104528462+i0.994521891)(0.913545457+i0.406736643)=2(-0.095491501-0.404508495+i0.908540955-i0.042515555)=$

$=2(-0.5+0.866025399)=-1+{\sqrt {3}}.$