Kompleksā skaitļa saknes aprēķināšana
Par n-tās pakāpes sakni no kompleksa skaitļa z sauc tādu kompleksu skaitli, kura n-tā pakāpe ir vienāda ar z:
$\sqrt[n]{z}=w$, ja ${w}^{n}=z$.

Pieņemsim, ka kompleksais skaitlis ir izteikts trigonometriskā formā $z=r\left(\mathit{cos}\phantom{\rule{0.147em}{0ex}}\mathrm{\varphi }+i\phantom{\rule{0.147em}{0ex}}\mathit{sin}\phantom{\rule{0.147em}{0ex}}\mathrm{\varphi }\right)$, un meklēsim tā n-tās pakāpes sakni $w=\mathrm{\rho }\left(\mathit{cos}\phantom{\rule{0.147em}{0ex}}\mathrm{\alpha }+i\phantom{\rule{0.147em}{0ex}}\mathit{sin}\phantom{\rule{0.147em}{0ex}}\mathrm{\alpha }\right)$. No vienādības ${w}^{n}=z$ sanāk, ka ${\mathrm{\rho }}^{n}\left(\mathit{cos}\phantom{\rule{0.147em}{0ex}}\left(n\mathrm{\alpha }\right)+i\phantom{\rule{0.147em}{0ex}}\mathit{sin}\phantom{\rule{0.147em}{0ex}}\left(n\mathrm{\alpha }\right)\right)=r\left(\mathit{cos}\phantom{\rule{0.147em}{0ex}}\mathrm{\varphi }+i\phantom{\rule{0.147em}{0ex}}\mathit{sin}\phantom{\rule{0.147em}{0ex}}\mathrm{\varphi }\right)$.
No vienādības izriet, ka ${\mathrm{\rho }}^{n}=r$ un $n\mathrm{\alpha }=\mathrm{\varphi }+2\mathrm{\pi }k,\phantom{\rule{0.147em}{0ex}}k\in \mathrm{ℤ}$. Jeb $\mathrm{\rho }=\sqrt[n]{r}$ un $\mathrm{\alpha }=\frac{\mathrm{\varphi }}{n}+\frac{2\mathrm{\pi }k}{n}$. Un šeit ir iespējamas n dažādas saknes vērtības.

Svarīgi!
Saknes aprēķināšanas formula:
$\sqrt[n]{r\left(\mathit{cos}\phantom{\rule{0.147em}{0ex}}\mathrm{\varphi }+i\phantom{\rule{0.147em}{0ex}}\mathit{sin}\phantom{\rule{0.147em}{0ex}}\mathrm{\varphi }\right)}=\sqrt[n]{r}\left(\mathit{cos}\phantom{\rule{0.147em}{0ex}}\frac{\mathrm{\varphi }+2\mathrm{\pi }k}{n}+i\phantom{\rule{0.147em}{0ex}}\mathit{sin}\phantom{\rule{0.147em}{0ex}}\frac{\mathrm{\varphi }+2\mathrm{\pi }k}{n}\right)$, kur $k=0,\phantom{\rule{0.147em}{0ex}}1,\phantom{\rule{0.147em}{0ex}}...,\phantom{\rule{0.147em}{0ex}}n-1$.
Piemērs:
$\begin{array}{l}\sqrt[4]{1}=\sqrt[4]{1+0i}=\\ =\sqrt[4]{1\left(\mathit{cos}\phantom{\rule{0.147em}{0ex}}0+i\phantom{\rule{0.147em}{0ex}}\mathit{sin}\phantom{\rule{0.147em}{0ex}}0\right)}=\\ =\mathit{cos}\phantom{\rule{0.147em}{0ex}}\frac{0+2\mathrm{\pi }k}{4}+i\phantom{\rule{0.147em}{0ex}}\mathit{sin}\phantom{\rule{0.147em}{0ex}}\frac{0+2\mathrm{\pi }k}{4},\phantom{\rule{0.147em}{0ex}}\mathit{kur}\phantom{\rule{0.147em}{0ex}}k=0,\phantom{\rule{0.147em}{0ex}}1,\phantom{\rule{0.147em}{0ex}}2,\phantom{\rule{0.147em}{0ex}}3\\ \\ \mathit{ja}\phantom{\rule{0.147em}{0ex}}k=0,\phantom{\rule{0.294em}{0ex}}\sqrt[4]{1}=\mathit{cos}\phantom{\rule{0.147em}{0ex}}0+i\phantom{\rule{0.147em}{0ex}}\mathit{sin}\phantom{\rule{0.147em}{0ex}}0=1+0i=1\\ \mathit{ja}\phantom{\rule{0.147em}{0ex}}k=1,\phantom{\rule{0.294em}{0ex}}\sqrt[4]{1}=\mathit{cos}\phantom{\rule{0.147em}{0ex}}\frac{\mathrm{\pi }}{2}+i\phantom{\rule{0.147em}{0ex}}\mathit{sin}\phantom{\rule{0.147em}{0ex}}\frac{\mathrm{\pi }}{2}=0+1i=i\\ \mathit{ja}\phantom{\rule{0.147em}{0ex}}k=2,\phantom{\rule{0.294em}{0ex}}\sqrt[4]{1}=\mathit{cos}\phantom{\rule{0.147em}{0ex}}\mathrm{\pi }+i\phantom{\rule{0.147em}{0ex}}\mathit{sin}\phantom{\rule{0.147em}{0ex}}\mathrm{\pi }=-1+0i=-1\\ \mathit{ja}\phantom{\rule{0.147em}{0ex}}k=3,\phantom{\rule{0.294em}{0ex}}\sqrt[4]{1}=\mathit{cos}\phantom{\rule{0.147em}{0ex}}\frac{3\mathrm{\pi }}{2}+i\phantom{\rule{0.147em}{0ex}}\mathit{sin}\phantom{\rule{0.147em}{0ex}}\frac{3\mathrm{\pi }}{2}=0-1i=-i\end{array}$