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Show that (-1+i)^7=-8(1+i)
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did you get it?
Yeah I just expanded it.
you love binomial thm so much !
\[\text{ \let } x=-1+i=i-1 \\ x^2=(i-1)^2=i^2-2i+1=-1-2i+1=-2i \\ x^7=x^{2 \cdot 3}x\]
\[x^7=(x^2)^3x =(-2i)^3(-1+i)\]
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Oh okay. thanks!
de moivre's thm is also fun, try it :)
\[-1+i = \sqrt{2}\angle \frac{3\pi}{4} \implies (-1+i)^7 = \sqrt{2}^7 \angle\dfrac{7\cdot 3\pi}{4} = 8\sqrt{2} \angle \dfrac{5\pi}{4} = \cdots \]
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