pierwiastki z liczby zespolonej

[rafu007]

Witam,
Zadanko: Wylicz \(\displaystyle{ \sqrt[4]{-5+5 \sqrt{3}i }}\)

\(\displaystyle{ \left|z\right|= \sqrt{-5^{2}+ \left( 5 \sqrt{3} \right)^2 }=10 \\[1ex]
\sin \phi= \frac{5 \sqrt{3}}{10}= \frac{ \sqrt{3} }{2} \\[1ex]
\cos \phi= \frac{-5}{10}= \frac{ \sqrt{-1} }{2} \\[1ex]
\phi= \frac{2}{3}\pi}\)

\(\displaystyle{ k_{0} = \sqrt[4]{10} \left( \cos \frac{ \frac{2}{3}\pi }{4} +i\sin\frac{ \frac{2}{3}\pi }{4} \right) =
\sqrt[4]{10} \left( \cos \frac{\pi}{6} + i\sin\frac{\pi}{6} \right) = \sqrt[4]{10} \left( \frac{ \sqrt{3} }{2} = \frac{1}{2}i \right)}\)

\(\displaystyle{ k_{1} = \sqrt[4]{10} \left( \cos \frac{ \frac{2}{3}\pi+2\pi}{4} +i\sin\frac{ \frac{2}{3}\pi+2\pi }{4} \right) = \sqrt[4]{10} \left( \cos \frac{8\pi}{12} +i\sin \frac{8\pi}{12} \right) =
\sqrt[4]{10} \left( - \frac{1}{2}+ \frac{ \sqrt{3} }{2}i \right)}\)

\(\displaystyle{ k_{2} = \sqrt[4]{10} \left( \cos \frac{ \frac{2}{3}\pi+4\pi}{4} +i\sin\frac{ \frac{2}{3}\pi+4\pi }{4} \right) = \sqrt[4]{10} \left( \cos \frac{14\pi}{12} +i\sin \frac{14\pi}{12} \right) = \sqrt[4]{10} \left( \cos \frac{7\pi}{6} +i\sin \frac{7\pi}{6} \right) =\sqrt[4]{10} \left( - \frac{ \sqrt{3} }{2} - \frac{1}{2}i \right)}\)

\(\displaystyle{ k_{3} = \sqrt[4]{10} \left( \cos \frac{ \frac{2}{3}\pi+6\pi}{4} +i\sin\frac{ \frac{2}{3}\pi+6\pi }{4} \right) = \sqrt[4]{10} \left( \cos \frac{20\pi}{12} +i\sin \frac{20\pi}{12} \right) = \sqrt[4]{10} \left( \cos \frac{5\pi}{3} +i\sin \frac{5\pi}{3} \right) =\sqrt[4]{10} \left( - \frac{ 1}{2} - \frac{ \sqrt{3} }{2}i \right)}\)

Czy k2 i k3 są dobrze policzone?

[Poszukujaca]

Według mnie jest dobrze

[SlotaWoj]

Gdyby nie te znaki „\(\displaystyle{ =}\)” wewnątrz nawiasów we wzorach na \(\displaystyle{ k_0}\), byłoby dobrze.
Popraw!