$\frac{2\left(\cos\frac{5\pi}{12}+\mathrm{i}\sin\frac{5\pi}{12}\right)}{-1-\mathrm{i}}$

Project ID: 
3000000017
SubArea: 
Fórmula binómica y trigonométrica de números complejos
Question: 

José tenía que encontrar la expresión binómica del número complejo: $$\frac{2\left(\cos\frac{5\pi}{12}+\mathrm{i}\sin\frac{5\pi}{12}\right)}{-1-\mathrm{i}}.$$

¿En qué paso ha cometido un error José?

La resolución de José:

$$ \begin{aligned} \frac{2\left(\cos\frac{5\pi}{12}+\mathrm{i}\sin\frac{5\pi}{12}\right)}{-1-\mathrm{i}} &\stackrel{(1)}= \frac{2\left(\cos\frac{5\pi}{12}+\mathrm{i}\sin\frac{5\pi}{12}\right)}{\sqrt2\left(\cos\frac{5\pi}{4}+\mathrm{i}\sin\frac{5\pi}{4}\right)} = \cr &\stackrel{(2)}= \frac{2}{\sqrt2}\left(\cos\frac{\frac{5}{12}}{\frac{5}{4}}\pi+\mathrm{i}\sin\frac{\frac{5}{12}}{\frac{5}{4}}\pi\right) =\cr &\stackrel{(3)}= \sqrt2\left(\cos\frac{\pi}{3}+\mathrm{i}\sin\frac{\pi}{3}\right) =\cr&\stackrel{(4)}= \frac{\sqrt2}{2}+\mathrm{i}\frac{\sqrt6}{2} \end{aligned} $$

Answer 1: 

En el paso (1). La forma trigonométrica de $-1-\mathrm{i}$ es $-\cos\frac{\pi}{4}-\mathrm{i}\sin\frac{\pi}{4}$ y por tanto $$ \frac{2\left(\cos\frac{5\pi}{12}+\mathrm{i}\sin\frac{5\pi}{12}\right)}{-1-\mathrm{i}} = \frac{2\left(\cos\frac{5\pi}{12}+\mathrm{i}\sin\frac{5\pi}{12}\right)}{-\left(\cos\frac{\pi}{4}+\mathrm{i}\sin\frac{\pi}{4}\right)} $$

Answer 2: 

En el paso (2). La simplificación correcta es:

$$ \frac{2\left(\cos\frac{5\pi}{12}+\mathrm{i}\sin\frac{5\pi}{12}\right)}{\sqrt2\left(\cos\frac{5\pi}{4}+\mathrm{i}\sin\frac{5\pi}{4}\right)}=\frac{2}{\sqrt2}\left(\cos\left(\frac{5\pi}{12}-\frac{5\pi}{4}\right)+\mathrm{i}\sin\left(\frac{5\pi}{12}-\frac{5\pi}{4}\right)\right) $$

Answer 3: 

En el paso (3). La simplificación correcta es:

$$ \frac{2}{\sqrt2}\left(\cos\frac{\frac{5}{12}}{\frac{5}{4}}\pi+\mathrm{i}\sin\frac{\frac{5}{12}}{\frac{5}{4}}\pi\right)=\sqrt2\left(\cos3\pi+\mathrm{i}\sin3\pi\right) $$

Answer 4: 

En el paso (4). La simplificación correcta es:

$$ \sqrt2\left(\cos\frac{\pi}{3}+\mathrm{i}\sin\frac{\pi}{3}\right)=\frac{\sqrt6}{2}+\mathrm{i}\frac{\sqrt6}{2} $$

Fixed Answer: 
All Fixed
Correct Answer: 
Answer 2
Hint: 

$$ \begin{aligned} \frac{2\left(\cos\frac{5\pi}{12}+\mathrm{i}\sin\frac{5\pi}{12}\right)}{-1-\mathrm{i}} &\stackrel{(1)}= \frac{2\left(\cos\frac{5\pi}{12}+\mathrm{i}\sin\frac{5\pi}{12}\right)}{\sqrt2\left(\cos\frac{5\pi}{4}+\mathrm{i}\sin\frac{5\pi}{4}\right)} = \cr &\stackrel{(2)}= \frac{2}{\sqrt2}\left(\cos\left(\frac{5\pi}{12}-\frac{5\pi}{4}\right)+\mathrm{i}\sin\left(\frac{5\pi}{12}-\frac{5\pi}{4}\right)\right) =\cr &\stackrel{(3)}= \sqrt2\left(\cos\left(-\frac{5\pi}{6}\right)+\mathrm{i}\sin\left(-\frac{5\pi}{6}\right)\right) =\cr&\stackrel{(4)}= \sqrt2\left(\cos\left(\frac{7\pi}{6}\right)+\mathrm{i}\sin\left(\frac{7\pi}{6}\right)\right) =\cr&\stackrel{(5)}= -\frac{\sqrt6}{2}-\mathrm{i}\frac{\sqrt2}{2} \end{aligned} $$