Question

(-1+і)^10=???
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Answer 1

(i - 1)¹⁰ = ((i-1)²)⁵ = (i² - 2i + 1)⁵ = (-1-2i+1)⁵ = (-2i)⁵ = -32i⁵ = -32*(-1)*(-1)*i =

= -32i

Answer 2

$\displaystyle(-1+i)^{10}=\left(\sqrt{(-1)^2+1^2}\cdot\left(-\frac1{\sqrt{(-1)^2+1^2}}+\frac{i}{\sqrt{(-1)^2+1^2}}\right)\right)^{10}=\\=(\sqrt2)^{10}\left(\cos\frac{3\pi}4+i\sin\frac{3\pi}4\right)^{10}=32\left(\cos\frac{30\pi}4+i\sin\frac{30\pi}4\right)\\\cos\frac{30\pi}4=\cos\frac{15\pi}2=\cos\left(8\pi-\frac\pi2\right)=\cos\left(-\frac\pi2\right)=0\\ \sin\frac{30\pi}4=\sin\left(-\frac{\pi}2\right)=-1\\ \boxed{(-1+i)^{10}=-32i}$

Использована формула Муавра:
$(\cos a+i\sin a)^n=\cos na+i\sin na$