Sprawdzenie zadania

[ostrowwlkp]

\(\displaystyle{ z= \left( \sqrt{2}+ \sqrt{2}i \right) ^{12} \\ \cos\phi= \frac{ \sqrt{2} }{2} \\ \sin\phi= \frac{ \sqrt{2} }{2}}\)

1 cwiartka

\(\displaystyle{ \alpha _{0}= \frac{ \pi }{4} \\ z=2 ^{12} \left( \cos 1 2 \cdot \frac{ \pi }{4} + i\sin 12 \cdot \frac{ \pi }{4} \right) \\ z=4096 \left( \cos 3 \pi + i\sin 3 \pi \right) z=4096 \left( \cos \pi + i\sin \pi \right)\\ z=4096 \left( - \frac{ \sqrt{2} }{2} +i0 \right)\\ z=-2048 \sqrt{2}}\)

[Chromosom]

\(\displaystyle{ \cos\pi\neq\tfrac{\sqrt2}2}\) reszta dobrze