Olga and Eugene solved the same problem. They had to find the algebraic form of the complex number $$\frac{\cos\frac{2\pi}{3}+\mathrm{i}\sin\frac{2\pi}{3}}{\sqrt{3}-\mathrm{i}}.$$
Who made a mistake in their solution and in which step?
Olga's solution: $$ \begin{aligned} \frac{\cos\frac{2\pi}{3}+\mathrm{i}\sin\frac{2\pi}{3}}{\sqrt{3}-\mathrm{i}} &\stackrel{(1)}= \frac{\cos\frac{2\pi}{3}+\mathrm{i}\sin\frac{2\pi}{3}}{2\left(\cos\frac{11\pi}{6}+\mathrm{i}\sin\frac{11\pi}{6}\right)}= \cr &\stackrel{(2)}= \frac{1}{2}\left[\cos\left(\frac{2\pi}{3}-\frac{11\pi}{6}\right)+ \mathrm{i}\sin\left(\frac{2\pi}{3}-\frac{11\pi}{6}\right)\right]=\cr &\stackrel{(3)}= \frac{1}{2}\left[\cos\left(-\frac{7\pi}{6}\right)+ \mathrm{i}\sin\left(-\frac{7\pi}{6}\right)\right] =\cr&\stackrel{(4)}= \frac{1}{2}\left(-\frac{\sqrt{3}}{2}-\frac{1}{2}\mathrm{i}\right)=\cr& \stackrel{(5)}= -\frac{\sqrt{3}}{4}-\frac14\mathrm{i} \end{aligned} $$
Eugene's solution: $$ \begin{aligned} \frac{\cos\frac{2\pi}{3}+\mathrm{i}\sin\frac{2\pi}{3}}{\sqrt{3}-\mathrm{i}} &\stackrel{(1)}= \frac{-\frac12+\frac{\sqrt{3}}{2}\mathrm{i}}{\sqrt{3}-\mathrm{i}}= \cr &\stackrel{(2)}= \frac{\left(-\frac12+\frac{\sqrt{3}}{2}\mathrm{i}\right)(\sqrt3 +\mathrm{i})}{(\sqrt{3}-\mathrm{i})(\sqrt{3}+\mathrm{i})}=\cr &\stackrel{(3)}= \frac{-\frac{\sqrt{3}}{2}-\frac12\mathrm{i} + \frac32\mathrm{i}-\frac{\sqrt{3}}{2}}{4}=\cr&\stackrel{(4)}= \frac{-\sqrt{3}+\mathrm{i}}{4}=\cr& \stackrel{(5)}= -\frac{\sqrt{3}}{4}+\frac14\mathrm{i} \end{aligned} $$
Olga, in step (1). It is a mistake to convert the denominator into trigonometric form. On the contrary, everything should be converted into algebraic form as Eugene did.
Eugene, in step (1). It is a mistake to convert the numerator into algebraic form. On the contrary, everything should be converted into trigonometric form as Olga did.
Olga, in step (4). The correct simplification is: $$\frac{1}{2}\left[\cos\left(-\frac{7\pi}{6}\right)+ \mathrm{i}\sin\left(-\frac{7\pi}{6}\right)\right]=\frac{1}{2}\left(-\frac{\sqrt{3}}{2}+\frac{1}{2}\mathrm{i}\right)$$
Eugene, in step (4). The correct simplification is: $$\frac{-\frac{\sqrt{3}}{2}-\frac12\mathrm{i} + \frac32\mathrm{i}-\frac{\sqrt{3}}{2}}{4}=-\frac{\sqrt{3}+\mathrm{i}}{4}$$
Olga's solution with the corrected mistake:
$$ \begin{aligned} \frac{\cos\frac{2\pi}{3}+\mathrm{i}\sin\frac{2\pi}{3}}{\sqrt{3}-\mathrm{i}} &\stackrel{(1)}= \frac{\cos\frac{2\pi}{3}+\mathrm{i}\sin\frac{2\pi}{3}}{2\left(\cos\frac{11\pi}{6}+\mathrm{i}\sin\frac{11\pi}{6}\right)}= \cr &\stackrel{(2)}= \frac{1}{2}\left[\cos\left(\frac{2\pi}{3}-\frac{11\pi}{6}\right)+ \mathrm{i}\sin\left(\frac{2\pi}{3}-\frac{11\pi}{6}\right)\right]=\cr &\stackrel{(3)}= \frac{1}{2}\left[\cos\left(-\frac{7\pi}{6}\right)+ \mathrm{i}\sin\left(-\frac{7\pi}{6}\right)\right] =\cr&\stackrel{(4)}= \frac{1}{2}\left(-\frac{\sqrt{3}}{2}+\frac{1}{2}\mathrm{i}\right)=\cr& \stackrel{(5)}= -\frac{\sqrt{3}}{4}+\frac14\mathrm{i} \end{aligned} $$