Question

вычислите (1-i√3) ^ 6 / (√2-i√2) ^ 20
(комп числа)

Answer 1

$1 - i\cdot\sqrt{3} = \sqrt{1^2 + (\sqrt{3})^2}\cdot (\frac{1}{2} + i\cdot\frac{-\sqrt{3}}{2}) =$

$= 2\cdot( \cos(-\frac{\pi}{3}) + i\cdot\sin(-\frac{\pi}{3}))$

$(1 - i\cdot\sqrt{3})^6 = 2^6\cdot (\cos(-\frac{\pi}{3}\cdot 6) + i\cdot\sin(-\frac{\pi}{3}\cdot 6)) =$

$= 2^6\cdot (\cos(-2\pi) + i\cdot\sin(-2\pi)) =$

$= 2^6\cdot (\cos(0) + i\cdot\sin(0))  = 2^6\cdot ( 1 + i\cdot 0) = 2^6$

$\sqrt{2} - i\cdot\sqrt{2} = 2\cdot( \frac{\sqrt{2}}{2} + i\cdot(-\frac{\sqrt{2}}{2})) =$

$= 2\cdot ( \cos(-\frac{\pi}{4}) + i\cdot\sin(-\frac{\pi}{4}) )$

$(\sqrt{2} - i\cdot\sqrt{2})^{20} = 2^{20}\cdot (\cos(-\frac{\pi}{4}\cdot 20) + i\cdot\sin(-\frac{\pi}{4}\cdot 20)) =$

$= 2^{20}\cdot (\cos(-5\pi) + i\cdot\sin(-5\pi)) =$

$= 2^{20}\cdot (\cos(\pi) + i\cdot\sin(\pi)) =$

$= 2^{20}\cdot (-1 + i\cdot 0) = -2^{20}$

$\frac{(1 - i\cdot\sqrt{3})^6}{(\sqrt{2} - i\cdot\sqrt{2})^{20}} =$

$= \frac{2^6}{-2^{20}} = -2^{6 - 20} = -2^{-14}$