Binomske enačbe

$z^3=i$
Naj bo rešitev enačbe $z=|z|(\cos\varphi+i\sin\varphi)$. Zapišimo število $i$ v polarni obliki: $i=1(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2})$. Oba izraza vstavimo v enačbo in primerjamo levo in desno stran enačbe.
$(|z|(\cos\varphi+i\sin\varphi))^3=1(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2})$
$|z|^3(\cos3\varphi+i\sin3\varphi)=1(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2})$

Iz $|z|^3=1$ izračunamo $|z|=\sqrt[3]1=1$ in iz $3\varphi=\frac{\pi}{2}+2k\pi$ izračunamo $\varphi=\frac{\frac{\pi}{2}+2k\pi}{3}$, $k\in\mathbb{Z}$. Izračunajmo nekaj kotov:
$k=0: \varphi=\frac{\frac{\pi}{2}+2\cdot 0\cdot\pi}{3}=\frac{\pi}{6}$
$k=1: \varphi=\frac{\frac{\pi}{2}+2\cdot 1\cdot\pi}{3}=\frac{5\pi}{6}$
$k=2: \varphi=\frac{\frac{\pi}{2}+2\cdot 2\cdot\pi}{3}=\frac{9\pi}{6}=\frac{3\pi}{2}$
$k=3: \varphi=\frac{\frac{\pi}{2}+2\cdot 3\cdot\pi}{3}=\frac{13\pi}{6}=\frac{\pi}{6}+2\pi$

Vrednosti funkcije sinus sta zaradi periodičnosti pri kotih $\frac{\pi}{6}$ in $\frac{13\pi}{6}$ enaki, prav tako vrednosti funkcije kosinus. Podobno bi ugotovili, če bi računali kote za $k>3$.

Sklepamo, da ima enačba $z^3=i$ natanko tri različne rešitve. Izračunajmo jih:
$z_0=1(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6})=\frac{\sqrt 3}{2}+\frac{1}{2}i$
$z_1=1(\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6})=-\frac{\sqrt 3}{2}+\frac{1}{2}i$
$z_2=1(\cos\frac{9\pi}{6}+i\sin\frac{9\pi}{6})=-i$

Če je $a=0$, je rešitev enačbe $z^n=0$ le $z=0$. V drugih primerih zapišemo $z$ in $a$ v polarni obliki, vstavimo števili v enačbo in primerjamo levo in desno stran.

$z^n=a$
Naj bo rešitev enačbe $z=|z|(\cos\varphi+i\sin\varphi)$ in $a=|a|(\cos\alpha+i\sin\alpha)$. Vstavimo v enačbo:

$(|z|(\cos\varphi+i\sin\varphi))^n=|a|(\cos\alpha+i\sin\alpha)$
$|z|^n(\cos n\varphi+i\sin n\varphi)=|a|(\cos\alpha+i\sin\alpha)$

Iz $|z|^n=|a|$ izračunamo $|z|=\sqrt[n] |a|$ in iz $n\varphi=\alpha+2k\pi$ izračunamo $\varphi=\frac{\alpha+2k\pi}{n}$, $k\in\mathbb{Z}$. Izračunajmo nekaj kotov:
$k=0: \varphi=\frac{\alpha+2\cdot 0\cdot\pi}{n}=\frac{\alpha}{n}$
$k=1: \varphi=\frac{\alpha+2\cdot 1\cdot\pi}{n}=\frac{\alpha+2\pi}{n}$
$k=2: \varphi=\frac{\alpha+2\cdot 2\cdot\pi}{n}=\frac{\alpha+4\pi}{n}$

Pri $k=n$ je $\varphi=\frac{\alpha+2\cdot n\cdot\pi}{n}=\frac{\alpha}{n}+2\pi$, kar pomeni, da se vrednosti funkcij sinus in kosinus zopet začnejo ponavljati.

Sklepamo, da ima enačba $z^n=a$ natanko $n$ različnih rešitev. Zapišimo jih z izrazom:

$z_k=\sqrt[n]{|a|}(\cos\frac{\alpha+2k\pi}{n}+i\sin \frac{\alpha+2k\pi}{n})$, $k=0, 1, 2  ...  n-1$

$a=1-i\sqrt 3=  2 (\cos \frac{5\pi}{3}+i\sin\frac{5\pi}{3})$

$z_k=\sqrt[3]{2}(\cos\frac{\frac{5\pi}{3}+2k\pi}{3}+i\sin \frac{\frac{5\pi}{3}+2k\pi}{3})$, $k=0, 1, 2$

$z_0=\sqrt[3]{2}(\cos\frac{\frac{5\pi}{3}+0}{3}+i\sin \frac{\frac{5\pi}{3}+0}{3})=\sqrt[3]{2}(\cos \frac{5\pi}{9}  +i\sin \frac{5\pi}{9})$
$z_1=\sqrt[3]{2}(\cos\frac{\frac{5\pi}{3}+2\pi}{3}+i\sin \frac{\frac{5\pi}{3}+2\pi}{3})=$
$=\sqrt[3]{2}(\cos \frac{11\pi}{9}  +i\sin \frac{11\pi}{9})$
$z_2=\sqrt[3]{2}(\cos\frac{\frac{5\pi}{3}+4\pi}{3}+i\sin \frac{\frac{5\pi}{3}+4\pi}{3})=$
$=\sqrt[3]{2}(\cos \frac{17\pi}{9}  +i\sin \frac{17\pi}{9})$

$a=-1+i = \sqrt 2 (\cos \frac{3\pi}{4}+i\sin\frac{3\pi}{4})$

$z_k=\sqrt[3]{\sqrt 2}(\cos\frac{\frac{3\pi}{4}+2k\pi}{4}+i\sin \frac{\frac{3\pi}{4}+2k\pi}{4})$, $k=0, 1, 2, 3$

$z_0=\sqrt[6]{2}(\cos\frac{\frac{3\pi}{4}+0}{4}+i\sin \frac{\frac{3\pi}{4}+0}{4})=$
$=\sqrt[6]{2}(\cos \frac{3\pi}{16}  +i\sin \frac{3\pi}{16})$
$z_1=\sqrt[6]{2}(\cos\frac{\frac{3\pi}{4}+2\pi}{4}+i\sin \frac{\frac{3\pi}{4}+2\pi}{4})=$
$=\sqrt[6]{2}(\cos \frac{11\pi}{4}  +i\sin \frac{11\pi}{4})$
$z_2=\sqrt[6]{2}(\cos\frac{\frac{3\pi}{4}+4\pi}{4}+i\sin \frac{\frac{3\pi}{4}+4\pi}{4})=$
$=\sqrt[6]{2}(\cos \frac{19\pi}{16}  +i\sin \frac{19\pi}{16})$
$z_3=\sqrt[6]{2}(\cos\frac{\frac{3\pi}{4}+6\pi}{4}+i\sin \frac{\frac{3\pi}{4}+6\pi}{4})=$
$=\sqrt[6]{2}(\cos \frac{27\pi}{16}  +i\sin \frac{27\pi}{16})$

$a=-1= 1 (\cos \pi+i\sin\pi)$

$z_k=\sqrt[4]{1}(\cos\frac{\pi+2k\pi}{4}+i\sin \frac{\pi+2k\pi}{4})$, $k=0, 1, 2, 3$

$z_0= \cos\frac{\pi+0}{4}+i\sin \frac{\pi+0}{4}= \frac{\sqrt 2}{2}+i \frac{\sqrt 2}{2}$
$z_1= \cos\frac{\pi+2\pi}{4}+i\sin \frac{\pi+2\pi}{4}=\cos\frac{3\pi }{4}+i\sin \frac{3\pi }{4}=$
$=- \frac{\sqrt 2}{2}+i \frac{\sqrt 2}{2}$
$z_2= \cos\frac{\pi+4\pi}{4}+i\sin \frac{\pi+4\pi}{4}=\cos\frac{5\pi }{4}+i\sin \frac{5\pi }{4}=$
$=- \frac{\sqrt 2}{2}-i \frac{\sqrt 2}{2}$
$z_3= \cos\frac{\pi+6\pi}{4}+i\sin \frac{\pi+6\pi}{4}=\cos\frac{7\pi }{4}+i\sin \frac{7\pi }{4}= $
$=\frac{\sqrt 2}{2}-i \frac{\sqrt 2}{2}$

$a=-i= 1 (\cos \frac{3\pi}{2}+i\sin\frac{3\pi}{2})$

$z_k=\sqrt[6]{1}(\cos\frac{\frac{3\pi}{2}+2k\pi}{6}+i\sin \frac{\frac{3\pi}{2}+2k\pi}{6})$, $k=0, 1, 2, ... 5$

$z_0= \cos\frac{\frac{3\pi}{2}+0}{6}+i\sin \frac{\frac{3\pi}{2}+0}{6}= \cos \frac{\pi} {4}+i\sin \frac{\pi} {4}=$
$=\frac{\sqrt 2}{2}+i \frac{\sqrt 2}{2}$
$z_1= \cos\frac{\frac{3\pi}{2}+2\pi}{6}+i\sin \frac{\frac{3\pi}{2}+2\pi}{6}= \cos \frac{7\pi} {12}+i\sin \frac{7\pi} {12}$
$z_2= \cos\frac{\frac{3\pi}{2}+4\pi}{6}+i\sin \frac{\frac{3\pi}{2}+4\pi}{6}= \cos \frac{11\pi} {12}+i\sin \frac{11\pi} {12}$
$z_3= \cos\frac{\frac{3\pi}{2}+6\pi}{6}+i\sin \frac{\frac{3\pi}{2}+6\pi}{6}= \cos \frac{15\pi} {12}+i\sin \frac{15\pi} {12}=$
$=-\frac{\sqrt 3}{2}+i\frac{1}{2}i$
$z_4= \cos\frac{\frac{3\pi}{2}+8\pi}{6}+i\sin \frac{\frac{3\pi}{2}+8\pi}{6}= \cos \frac{19\pi} {12}+i\sin \frac{19\pi} {12}$
$z_5= \cos\frac{\frac{3\pi}{2}+10\pi}{6}+i\sin \frac{\frac{3\pi}{2}+10\pi}{6}= \cos \frac{23\pi} {12}+i\sin \frac{23\pi} {12}$