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\begin{document}
\title{Raíz cúbica de un número complejo}
\author{Eduardo Virueña Silva}
\maketitle

Sea $z$ un número complejo en el círculo unitario.
\[
	z= \cos(\alpha) + i \sin(\alpha)
\]
deseamos encontrar la raíz cúbica de $z$, es decir, deseamos
encontrar $r$ tal que:  $r^3=z$.   Por la fórmula de
De Moivre sabemos que:
\[
	r= \cos\left(\frac{\alpha}{3}\right) 
	+ i \sin\left(\frac{\alpha}{3}\right)
\]
Entonces, resolver este problema se reduce a encontrar 
$\cos\left(\frac{\alpha}{3}\right)$  y 
$\sin\left(\frac{\alpha}{3}\right)$ en términos de 
$\cos\left(\alpha\right)$ y $\sin\left(\alpha\right)$.


Consideremos entonces:
\begin{align*}
	e^{3\varphi i} =& \left(e^{\varphi i}\right)^3\\
	=& \left(\cos\left(\varphi\right) + i \sin\left(\varphi\right) \right)^3\\
	=& \cos\left(\varphi\right)^3   
		+ 3i  \cos\left(\varphi\right)^2 sin\left(\varphi\right) 
		- 3  \cos\left(\varphi\right) sin\left(\varphi\right)^2
		- i \sin\left(\varphi\right)^3\\
	=& \cos\left(\varphi\right)^3   
		- 3  \cos\left(\varphi\right) sin\left(\varphi\right)^2
		+ 3i  \cos\left(\varphi\right)^2 sin\left(\varphi\right) 
		- i \sin\left(\varphi\right)^3\\
	=& \cos\left(\varphi\right)^3   
		- 3  \cos\left(\varphi\right) 
				\left(1-cos\left(\varphi\right)^2\right)
		+ 3i  \left(1 - \sin\left(\varphi\right)^2\right)
				\sin\left(\varphi\right) 
		- i \sin\left(\varphi\right)^3\\
	=& \cos\left(\varphi\right)^3   
		- 3  \cos\left(\varphi\right)+3cos\left(\varphi\right)^3
		+ 3i \sin\left(\varphi\right)-3i\sin\left(\varphi\right)^3
		- i \sin\left(\varphi\right)^3\\
	=& 4\cos\left(\varphi\right)^3 - 3\cos\left(\varphi\right)
	+ i \left( 3 \sin\left(\varphi\right)-4\sin\left(\varphi\right)^3 \right)
\end{align*}
es decir:
\begin{align*}
	\cos(3\varphi)=&4\cos\left(\varphi\right)^3 - 3\cos\left(\varphi\right)\\
	\sin(3\varphi)=&3 \sin\left(\varphi\right)-4\sin\left(\varphi\right)^3.
\end{align*}
Si tomamos $\varphi=\frac{\alpha}{3}$
\begin{align*}
	\cos(\alpha)=&4\cos\left(\frac{\alpha}{3}\right)^3 
		- 3\cos\left(\frac{\alpha}{3}\right)\\
	\sin(\alpha)=&3 \sin\left(\frac{\alpha}{3}\right)
		-4\sin\left(\frac{\alpha}{3}\right)^3.
\end{align*}
que podemos reacomodar:
\begin{align*}
	4\cos\left(\frac{\alpha}{3}\right)^3 
		- 3\cos\left(\frac{\alpha}{3}\right) - \cos(\alpha) &= 0\\
	4\sin\left(\frac{\alpha}{3}\right)^3 
		-3 \sin\left(\frac{\alpha}{3}\right) + \sin(\alpha) &= 0.
\end{align*}
o bien:
\begin{align}
	\label{cos}
	\cos\left(\frac{\alpha}{3}\right)^3 
		-\frac{3}{4} \cos\left(\frac{\alpha}{3}\right) - 
		\frac{1}{4}\cos(\alpha) &= 0\\
	\label{sin}
	\sin\left(\frac{\alpha}{3}\right)^3 
		-\frac{3}{4} \sin\left(\frac{\alpha}{3}\right) + 
		\frac{1}{4}\sin(\alpha) &= 0.
\end{align}
que son ecuaciones cúbicas de la forma:
\[
	t^3 + p x + q = 0
\]
cuyas soluciones pueden darse como sigue.  Sean:
\begin{align*}
	d=& \sqrt{\left(\frac{q}{2}\right)^2+ \left(\frac{p}{3}\right)^3}\\
	u=& \sqrt[3]{-\frac{q}{2} + d} \\
	v=& \sqrt[3]{-\frac{q}{2} - d} \\
\intertext{y, suponiendo que se cumple:}
	uv=& -\frac{p}{3} 
\intertext{entonces, las soluciones son:}
	t_1 =& u + v\\
	t_2 =& u \omega^2+ v\omega\\
	t_3 =& u \omega+ v\omega^2
\intertext{donde}
	\omega =& - \frac{1}{2} + \frac{\sqrt{3}}{2} i\\
	\omega^2 =& - \frac{1}{2} - \frac{\sqrt{3}}{2} i
\end{align*}

Empecemos por resolver (\ref{cos}).  Tenemos:
\begin{align*}
	p_c=&-\frac{3}{4} \\
	q_c=&-\frac{1}{4} \cos(\alpha)
\end{align*}
Sustituyendo, tenemos
\begin{align*}
	d_c=&\sqrt{\left(\frac{q_c}{2}\right)^2+\left(\frac{p_c}{3}\right)^3}\\
	=&\sqrt{\left(\frac{-\frac{1}{4} \cos(\alpha)}{2}\right)^2
		+\left(\frac{-\frac{3}{4}}{3}\right)^3}\\
	=&\sqrt{\left(\frac{\cos(\alpha)}{8}\right)^2
		+\left(-\frac{1}{4}\right)^3}\\
	=&\sqrt{\frac{\cos\left(\alpha\right)^2}{64}
		-\frac{1}{64}}\\
	=&\frac{1}{8}\sqrt{\cos\left(\alpha\right)^2 -1}\\
	=& \pm\frac{i}{8}\sin\left(\alpha\right)\\
\end{align*}
de donde:
\begin{align*}
	u_c=& = \sqrt[3]{-\frac{q_c}{2} + d_c } \\
	  =&\sqrt[3]{-\frac{\left(-\frac{\cos\left(\alpha\right)}{4}\right)}{2} 
			+ \frac{i}{8}\sin\left(\alpha\right) }\\
	  =&\sqrt[3]{\frac{\cos\left(\alpha\right)}{8}
			+ \frac{i}{8}\sin\left(\alpha\right) }\\
	  =& \frac{1}{2} \sqrt[3]{\cos\left(\alpha\right)
			+ i\sin\left(\alpha\right) }\\
	u_{ck}=& \frac{1}{2}e^{\frac{2k\pi+\alpha}{3}i}, \quad k= 0, 1, 2
\end{align*}
\begin{align*}
	v_c=& = \sqrt[3]{-\frac{q_c}{2} - d_c } \\
	  =&\sqrt[3]{-\frac{\left(-\frac{\cos\left(\alpha\right)}{4}\right)}{2} 
			- \frac{i}{8}\sin\left(\alpha\right) }\\
	  =&\sqrt[3]{\frac{\cos\left(\alpha\right)}{8}
			- \frac{i}{8}\sin\left(\alpha\right) }\\
	  =& \frac{1}{2} \sqrt[3]{\cos\left(\alpha\right)
			- i\sin\left(\alpha\right) }\\
	  =& \frac{1}{2} \sqrt[3]{\cos\left(-\alpha\right)
			+ i\sin\left(-\alpha\right) }\\
	v_{ck}=& \frac{1}{2}e^{\frac{2k\pi-\alpha}{3}i}, \quad k= 0, 1, 2
\end{align*}
Debemos recordar que la solución debe satisfacer:
\[
	u  v  = -\frac{p}{3} = -\frac{\left(-\frac{3}{4}\right)}{3} = \frac{1}{4}
\]
Las parejas que cumplen con la condición anterior son:
\begin{align*}
	u_{c0} v_{c0}=& \left(\frac{1}{2}e^{\frac{\alpha}{3}i} \right)
					\left(\frac{1}{2}e^{\frac{-\alpha}{3}i}\right) 
				= \frac{1}{4}\\
	u_{c1} v_{c2}=& 
			\left(\frac{1}{2}e^{\left(\frac{2\pi+\alpha}{3}\right)i}\right)
			\left(\frac{1}{2}e^{\left(\frac{4\pi-\alpha}{3}\right)i}\right)
			= \frac{1}{4} e^{2\pi i} = \frac{1}{4}\\
	u_{c2} v_{c1}=& 
			\left(\frac{1}{2}e^{\left(\frac{4\pi+\alpha}{3}\right)i}\right)
			\left(\frac{1}{2}e^{\left(\frac{2\pi-\alpha}{3}\right)i}\right)
			= \frac{1}{4} e^{2\pi i} = \frac{1}{4}
\end{align*}
de manera que las soluciones de la ecuación son:
\begin{align*}
	t_{c0}=& u_{c0} v_{c0} 
	= \frac{1}{2}\left(e^{\frac{\alpha}{3}i}+e^{-\frac{\alpha}{3}i}\right)\\
\intertext{que, irónicamente:}
	=& \cos\left(\frac{\alpha}{3}\right).\\
	t_{c1}=& u_{c1} v_{c2}
	= \frac{1}{2}\left( e^{\frac{2\pi+\alpha}{3}i} 
					+   e^{\frac{4\pi-\alpha}{3}i} \right)\\
	=& \frac{1}{2}\left( e^{\frac{2\pi}{3}i} e^{ \frac{\alpha}{3}i}
					+    e^{\frac{4\pi}{3}i} e^{-\frac{\alpha}{3}i} \right)\\
	=& \frac{1}{2}\left( e^{ \frac{2\pi}{3}i} e^{ \frac{\alpha}{3}i}
					+    e^{-\frac{2\pi}{3}i} e^{-\frac{\alpha}{3}i} \right)\\
	=& \frac{1}{2}\left( e^{ \frac{2\pi + \alpha}{3}i}
					+    e^{-\frac{2\pi - \alpha}{3}i} \right)\\
	=& \cos \left(\frac{2\pi + \alpha}{3}\right) \\
    t_{c2}=& u_{c2} v_{c1}
    = \frac{1}{2}\left( e^{\frac{4\pi+\alpha}{3}i}
                    +   e^{\frac{2\pi-\alpha}{3}i} \right)\\
    =& \frac{1}{2}\left( e^{\frac{4\pi}{3}i} e^{ \frac{\alpha}{3}i}
                    +    e^{\frac{2\pi}{3}i} e^{-\frac{\alpha}{3}i} \right)\\
    =& \frac{1}{2}\left( e^{ \frac{4\pi}{3}i} e^{ \frac{\alpha}{3}i}
                    +    e^{-\frac{4\pi}{3}i} e^{-\frac{\alpha}{3}i} \right)\\
    =& \frac{1}{2}\left( e^{ \frac{4\pi + \alpha}{3}i}
                    +    e^{-\frac{4\pi - \alpha}{3}i} \right)\\
    =& \cos \left(\frac{4\pi + \alpha}{3}\right)
\end{align*}

Ahora resolveremos (\ref{sin}). 
\begin{align*}
	    \sin\left(\frac{\alpha}{3}\right)^3
        -\frac{3}{4} \sin\left(\frac{\alpha}{3}\right) +
        \frac{1}{4}\sin(\alpha) &= 0.
\end{align*}
La ecuación tiene la forma:
\begin{align*}
    t^3 -\frac{3}{4} t  + \frac{1}{4}\sin(\alpha) &= 0.
\end{align*}
Para esta ecuación:
\begin{align*}
	p_s=& -\frac{3}{4}\\
	q_s=&  \frac{1}{4} \sin(\alpha).
\end{align*}
Tenemos:
\begin{align*}
    d_s=&\sqrt{\left(\frac{q_s}{2}\right)^2+\left(\frac{p_s}{3}\right)^3}\\
    =&\sqrt{\left(\frac{ \frac{1}{4} \sin(\alpha)}{2}\right)^2
        +\left(\frac{-\frac{3}{4}}{3}\right)^3}\\
    =&\sqrt{\left(\frac{\sin(\alpha)}{8}\right)^2
        -\left( \frac{1}{4}\right)^3}\\
    =&\sqrt{\frac{\sin\left(\alpha\right)^2}{64}
        -\frac{1}{64}}\\
    =&\frac{1}{8}\sqrt{\sin\left(\alpha\right)^2 -1}\\
    =& \pm\frac{i}{8}\cos\left(\alpha\right)\\
\end{align*}

\begin{align*}
    u_s=& = \sqrt[3]{-\frac{q_s}{2} + d_s } \\
      =&\sqrt[3]{-\frac{\left( \frac{\sin\left(\alpha\right)}{4}\right)}{2}
            + \frac{i}{8}\cos\left(\alpha\right) }\\
      =&\sqrt[3]{-\frac{\sin\left(\alpha\right)}{8}
            + \frac{i}{8}\cos\left(\alpha\right) }\\
      =& \frac{1}{2} \sqrt[3]{-\sin\left(\alpha\right)
            + i\cos\left(\alpha\right) }\\
      =& \frac{1}{2} \sqrt[3]{\cos\left(\frac{\pi}{2}+\alpha\right)
            + i\sin\left(\frac{\pi}{2}+\alpha\right) }\\
    u_{sk}=& \frac{1}{2}e^{\left(\frac{2k\pi}{3}+\frac{\frac{\pi}{2}+\alpha}{3}\right)i}, \quad k= 0, 1, 2
\end{align*}
\begin{align*}
    v_s=& = \sqrt[3]{-\frac{q_s}{2} - d_s } \\
      =&\sqrt[3]{-\frac{\left( \frac{\sin\left(\alpha\right)}{4}\right)}{2}
            - \frac{i}{8}\cos\left(\alpha\right) }\\
      =&\sqrt[3]{-\frac{\sin\left(\alpha\right)}{8}
            - \frac{i}{8}\cos\left(\alpha\right) }\\
      =& -\frac{1}{2} \sqrt[3]{ \sin\left(\alpha\right)
            + i\cos\left(\alpha\right) }\\
      =& - \frac{1}{2} \sqrt[3]{\cos\left(\frac{\pi}{2}-\alpha\right)
            + i\sin\left(\frac{\pi}{2}-\alpha\right) }\\
    v_{sk}=& \frac{1}{2}e^{\left(\frac{2k\pi}{3}+\frac{\frac{\pi}{2}-\alpha}{3}\right)i}, \quad k= 0, 1, 2
\end{align*}
Las parejas $(u,v)$ deben satisfacer:
\[
    u  v  = -\frac{p}{3} = -\frac{-\frac{3}{4}}{3} = \frac{1}{4}
\]
Las que satisfacen son:
\begin{align*}
	u_{s0} v_{s0} =  
\frac{1}{2}e^{\left(\frac{2k\pi}{3}+\frac{\frac{\pi}{2}+\alpha}{3}\right)i}
\frac{1}{2}e^{\left(\frac{2k\pi}{3}+\frac{\frac{\pi}{2}-\alpha}{3}\right)i}
\end{align*}


Lo cual es una soberana locura, porque no queríamos usar la fórmula de De Moivre, pero hemos llegado a ella... 
\end{document}
