zespolone

[zadanko]

oblicz pierwiastki stopnia 3 z liczby \(\frac{(1+i)^{7}}{(1-i)^{3}}\)

[alexx17]

\((1+i)^7=(sqrt2 (\cos \frac{1}{\sqrt2} + i \sin \frac{1}{\sqrt2}))^7=8 sqrt2 (\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})^7= 8 sqrt2(\cos \frac{7\pi}{4} + i \sin \frac{7\pi}{4})=\\=2 sqrt2 ( \sin 45^o - \cos 45^o)= 8 sqrt2 ( \frac{\sqrt2}{2} + i \frac{\sqrt2}{2} )=8-8i\)

\((1-i)^3=(sqrt2 (\cos \frac{1}{\sqrt2} + i \sin \frac{1}{\sqrt2}))^7=2 sqrt2 (\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})^7= 2 sqrt2(\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4})=\\=2 sqrt2 (- \sin 45^o - \cos 45^o)= 2 sqrt2 (- \frac{\sqrt2}{2} - i \frac{\sqrt2}{2} )=-2-2i\)

\(\frac{8-8i}{-2-2i} \cdot \frac{-2+2i}{-2+2i}= \frac{-16+16i+16i-16i^2}{4-4i^2}= \frac{32i}{8}=4i\)

\(w_k=\sqrt[n]{|z|}(\cos \frac{ \alpha }{n} + i \sin \frac{ \alpha }{n})\\w_k=\sqrt[n]{4}\(\cos \frac{ \frac{\pi}{2}+2 k \pi }{3} + i \sin \frac{ \frac{\pi}{2}+2 k \pi }{3}\)\\w_0=\sqrt[3]{4}( \cos \frac{\pi}{6}+ i \sin \frac{\pi}{6})=\sqrt[3]{4}( \frac{1}{2} + i \frac{1}{2} )= \frac{\sqrt[3]{4}}{2} + i \frac{\sqrt[3]{4}}{2}\\w_1=...\\w_2=...\)

[Pol]

można też tak:

\((1+i)^2 = 2i
(1+i)^7 = \( (1+i)^2 \) ^3(1+i) = -8i(1+i)\)

\((1-i)^2 = -2i
(1-i)^3 = -2i(1-i)\)

\(\frac{(1+i)^7}{(1-i)^3} = \frac{-8i(1+i)}{-2i(1-i)} = 4\frac{(1+i)(1+i)}{(1-i)(1+i)} =4\frac{2i}{2}=4i\)