Czy mogę prosić o przejrzenie rachunków i ewentualne wskazowki?
\(\displaystyle{ (1+i)^{27} \\ \\
\left|z \right|= \sqrt{2} \\
cos \phi= \frac{ \sqrt{2} }{2} \\
sin \phi = \frac{ \sqrt{2} }{2} \\
\phi = \alpha_{0} = \frac{\pi}{4} \\ \\
z^{27}= \sqrt{2}^{27}(cos(27* \frac{\pi}{4})+i sin(27* \frac{\pi}{4})) = \\
\sqrt{2}^{27}(cos(6 \frac{3}{4}\pi)+i sin(6 \frac{3}{4}\pi))= \\
\sqrt{2}^{27}(cos(\pi - \frac{\pi}{4})+i sin(\pi - \frac{\pi}{4}))= \\
\sqrt{2}^{27}(cos(- \frac{\pi}{4})+i sin(\frac{\pi}{4}))= \\
\sqrt{2}^{27}(- \frac{ \sqrt{2} }{2}+\frac{ \sqrt{2} }{2}i)= \\
(2^{ \frac{1}{2}})^{26} \sqrt{2}(- \frac{ \sqrt{2} }{2}+\frac{ \sqrt{2} }{2}i) = \\
2^{13}\sqrt{2}(- \frac{ \sqrt{2} }{2} + \frac{ \sqrt{2} }{2}i) = \\
2^{13}(-1+1i) = -8192+8192i \\}\)
\(\displaystyle{ Poprawka!}\)
W 4. linijce jest błąd.
Bo przecież \(\displaystyle{ cos(\pi - \frac{\pi}{4}) = -cos(\frac{\pi}{4})}\), a nie tak jak tam napisałem.
