Moivre formula

Moivre Formula:

The power n-th entry of a complex number r_\alpha it is another complex module r^n and argument n times the argument of the first

In consequence, we have that

z^n={(r_\alpha)}^n=\overbrace{r_\alpha \cdots r_\alpha}^{n\;\rm times}={(r^n)}_{n\cdot\alpha}

Example of Moivre's formula

\tiny\begin{cases}z^4={((-3)+4\cdot{i})}^4 \\ \|z\|=\|-3+4\cdot{i}\|=\sqrt{(-3)^2+4^2}=\sqrt{9+16}=\sqrt{25}=5 \\ \alpha=\frac{atan2(4, -3)\cdot{360}}{2\cdot\pi}=\frac{2.2143\cdot{360}}{2\cdot\pi}=\frac{797.148}{2\cdot\pi}=126.8701^{\circ}\approx\frac{2\cdot\pi}{3} \end{cases}

z^4={(5_{\frac{2\cdot\pi}{3}})}^4=(5^4)_{4\cdot{\frac{2\cdot\pi}{3}}}=625_{\frac{8\cdot\pi}{3}}

So we have to:

z^4={(r_\alpha)}^n=625_{\frac{8\cdot\pi}{3}}

Roots n-ésimas of a complex number

Given a complex number z=r_{\alpha}\text{ si }w=s_{\beta} it is a root n-th we have that

\begin{cases} z={(s_\beta)}^n={(r^n)}_{n\cdot\beta}=r_{\alpha} \\ s^n=r \\ n\cdot{\beta}=\alpha+2\cdot{k}\cdot\pi \\ s=\sqrt[n]{r} \\ \beta=\frac{(\alpha+2\cdot{k}\cdot\pi)}{n} \\ z_k=\sqrt[n]{r}\cdot{e}^{{\frac{(\alpha+2\cdot{k}\cdot\pi)}{n}}\cdot{i}} \end{cases}

with k=0,1,2,\cdots,(n-1) since for k=n gives the same value as for k=0

There are thus n roots of the n-ésimas different if z\not=0

Example of cube roots

\begin{cases} z^3+2=0\rightarrow{z^3=-2}\rightarrow{z=(-2)^{1/3}} \\ \|z\|=\|(-2)+0\cdot{i}\|=\sqrt{(-2)^2+0^2}=\sqrt{4+0}=\sqrt{4}=2 \\ \alpha=\frac{atan2(0, -2)\cdot{360}}{2\cdot\pi}=\frac{\pi\cdot{360}}{2\cdot\pi}=\frac{360}{2}=180^{\circ}\approx\pi \\ \beta=\frac{\pi+2\cdot{k}\cdot{\pi}}{3} \end{cases}

\begin{cases} z_1={\sqrt[3]{2}} \cdot{e}^{{\frac{\pi+2\cdot{0}\cdot{\pi}}{3}}\cdot{i}}={\sqrt[3]{2}}\cdot{e}^{{\frac{\pi}{3}}\cdot{i}} \\ z_2={\sqrt[3]{2}} \cdot{e}^{{\frac{\pi+2\cdot{1}\cdot{\pi}}{3}}\cdot{i}}={\sqrt[3]{2}}\cdot{e}^{{\frac{\pi+2\cdot\pi}{3}}\cdot{i}}={\sqrt[3]{2}}\cdot{e}^{{\frac{3\cdot\pi}{3}}\cdot{i}}={\sqrt[3]{2}}\cdot{e}^{\pi\cdot{i}} \\ z_3={\sqrt[3]{2}} \cdot{e}^{{\frac{\pi+2\cdot{2}\cdot{\pi}}{3}}\cdot{i}}={\sqrt[3]{2}}\cdot{e}^{{\frac{\pi+4\cdot\pi}{3}}\cdot{i}}={\sqrt[3]{2}}\cdot{e}^{{\frac{5\cdot\pi}{3}}\cdot{i}} \end{cases}

So we have to:

z^3+2=0\rightarrow\begin{cases} z_1={\sqrt[3]{2}}\cdot{e}^{{\frac{\pi}{3}}\cdot{i}} \\ z_2={\sqrt[3]{2}}\cdot{e}^{\pi\cdot{i}} \\ z_3={\sqrt[3]{2}}\cdot{e}^{{\frac{5\cdot\pi}{3}}\cdot{i}} \end{cases}