Convertir de la forma módulo-argumental a la forma cartesiana es más simple que hacer lo contrario. Convertir de módulo-argumental a forma cartesiana sólo consiste en evaluar el valor de las expresiones trigonométricas correspondientes según el argumento y multiplicarlo por el módulo correspondiente.
Veamos varios ejemplos expresados a tres cifras significativas o exactamente:
\[
\begin{align*}
8e^{\left (0i\right )} &= 8\left (\cos 0 + i \textrm{ sen } 0 \right )\\
&= \left (8\cos 0\right ) + \left (2 \textrm{ sen } 0\right )i\\
&= (8)(1) + (8)(0)i\\
&= 8
\end{align*}
\]
\[
\begin{align*}
4e^{\left (i\pi\right )} &= 4\left (\cos \pi + i \textrm{ sen } \pi\right )\\
&= \left (4\cos \pi\right ) + \left (4 \textrm{ sen } \pi\right )i\\
&= (4)(-1) + (4)(0)i\\
&= -4
\end{align*}
\]
\begin{align*}
4e^{\left (i\pi\right )} &= 4\left (\cos \pi + i \textrm{ sen } \pi\right )\\
&= \left (4\cos \pi\right ) + \left (4 \textrm{ sen } \pi\right )i\\
&= (4)(-1) + (4)(0)i\\
&= -4
\end{align*}
\]
\[
\begin{align*}
2e^{\left (i\frac{\pi}{2}\right )} &= 2\left (\cos \frac{\pi}{2} + i \textrm{ sen } \frac{\pi}{2}\right )\\
&= \left (2\cos \frac{\pi}{2}\right ) + \left (2 \textrm{ sen } \frac{\pi}{2}\right )i\\
&= (2)(0) + (2)(1)i\\
&= 2i
\end{align*}
\]
\begin{align*}
2e^{\left (i\frac{\pi}{2}\right )} &= 2\left (\cos \frac{\pi}{2} + i \textrm{ sen } \frac{\pi}{2}\right )\\
&= \left (2\cos \frac{\pi}{2}\right ) + \left (2 \textrm{ sen } \frac{\pi}{2}\right )i\\
&= (2)(0) + (2)(1)i\\
&= 2i
\end{align*}
\]
\begin{align*}
3e^{\left (i\frac{3\pi}{2}\right )} &= 3\left (\cos \frac{3\pi}{2} + i \textrm{ sen } \frac{3\pi}{2}\right )\\
&= \left (3\cos \frac{3\pi}{2}\right ) + \left (3 \textrm{ sen } \frac{3\pi}{2}\right )i\\
&= (3)(0) + (3)(-1)i\\
&= -3i
\end{align*}
\]
\[
\begin{align*}
\sqrt{29}e^{\left (0.381i\right )} &= \sqrt{29}\left (\cos 0.381 + i \textrm{ sen } 0.381\right )\\
&= \left (\sqrt{29}\cos 0.381\right ) + \left (\sqrt{29} \textrm{ sen } 0.381\right )i\\
&= (\sqrt{29})(0.928) + (\sqrt{29})(0.372)i\\
&= 5+2i
\end{align*}
\]
\begin{align*}
\sqrt{29}e^{\left (0.381i\right )} &= \sqrt{29}\left (\cos 0.381 + i \textrm{ sen } 0.381\right )\\
&= \left (\sqrt{29}\cos 0.381\right ) + \left (\sqrt{29} \textrm{ sen } 0.381\right )i\\
&= (\sqrt{29})(0.928) + (\sqrt{29})(0.372)i\\
&= 5+2i
\end{align*}
\]
\[
\begin{align*}
\sqrt{13}e^{\left (-0.588i\right )} &= \sqrt{13}\left (\cos -0.588 + i \textrm{ sen } -0.588\right )\\
&= \left (\sqrt{13}\cos -0.588\right ) + \left (sqrt{13} \textrm{ sen } -0.588\right )i\\
&= (\sqrt{13})(0.832) + (\sqrt{13})(-0.555)i\\
&= 3-2i
\end{align*}
\]
\begin{align*}
\sqrt{13}e^{\left (-0.588i\right )} &= \sqrt{13}\left (\cos -0.588 + i \textrm{ sen } -0.588\right )\\
&= \left (\sqrt{13}\cos -0.588\right ) + \left (sqrt{13} \textrm{ sen } -0.588\right )i\\
&= (\sqrt{13})(0.832) + (\sqrt{13})(-0.555)i\\
&= 3-2i
\end{align*}
\]
\[
\begin{align*}
\sqrt{2}e^{\left (i\frac{5\pi}{4}\right )} &= \sqrt{2}\left (\cos \frac{5\pi}{4} + i \textrm{ sen } \frac{5\pi}{4}\right )\\
&= \left (\sqrt{2}\cos \frac{5\pi}{4}\right ) + \left (\sqrt{2} \textrm{ sen } \frac{5\pi}{4}\right )i\\
&= (\sqrt{2})\left (-\frac{\sqrt{2}}{2}\right ) + (\sqrt{2})\left (-\frac{\sqrt{2}}{2}\right )i\\
&= -1-1i
\end{align*}
\]
\begin{align*}
\sqrt{2}e^{\left (i\frac{5\pi}{4}\right )} &= \sqrt{2}\left (\cos \frac{5\pi}{4} + i \textrm{ sen } \frac{5\pi}{4}\right )\\
&= \left (\sqrt{2}\cos \frac{5\pi}{4}\right ) + \left (\sqrt{2} \textrm{ sen } \frac{5\pi}{4}\right )i\\
&= (\sqrt{2})\left (-\frac{\sqrt{2}}{2}\right ) + (\sqrt{2})\left (-\frac{\sqrt{2}}{2}\right )i\\
&= -1-1i
\end{align*}
\]
\[
\begin{align*}
e^{\left (i\frac{11\pi}{6}\right )} &= \left (\cos \frac{11\pi}{6} + i \textrm{ sen } \frac{11\pi}{6}\right )\\
&= \left (\cos \frac{11\pi}{6}\right ) + \left (\textrm{sen } \frac{11\pi}{6}\right )i\\
&= (1)(-0.260) + (1)(0.966)i\\
&= -0.260 + 0.966i
\end{align*}
\]
\begin{align*}
e^{\left (i\frac{11\pi}{6}\right )} &= \left (\cos \frac{11\pi}{6} + i \textrm{ sen } \frac{11\pi}{6}\right )\\
&= \left (\cos \frac{11\pi}{6}\right ) + \left (\textrm{sen } \frac{11\pi}{6}\right )i\\
&= (1)(-0.260) + (1)(0.966)i\\
&= -0.260 + 0.966i
\end{align*}
\]
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