How do you use demoivre's theorem to simplify ${(1 - i)}^{12}$ $(1-i)^12$ ?

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Answer 1 · 2016-08-13T23:55:06

$- 64$ $-64$

$z = 1 - i$ $z = 1 - i$ will be in 4th quadrant of argand diagram. Important to note for when we find the argument.

$r = \sqrt{1^{2} + {(- 1)}^{2}} = \sqrt{2}$ $r = sqrt(1^2 + (-1)^2) = sqrt(2)$

$θ = 2 π - {tan}^{- 1} (1) = \frac{7 π}{4} = - \frac{π}{4}$ $theta = 2pi - tan^(-1)(1) = (7pi)/4 = -pi/4$

$z = r (cos θ + i sin θ)$ $z = r(costheta + isintheta)$

$z^{n} = r^{n} (cos n θ + i sin n θ)$ $z^n = r^n(cosntheta + isinntheta)$

$z^{12} = {(\sqrt{2})}^{12} (cos (- 12 \frac{π}{4}) + i sin (- 12 \frac{π}{4}))$ $z^12 = (sqrt(2))^12(cos(-12pi/4) + isin(-12pi/4))$

$z^{12} = 2^{\frac{1}{2} \cdot 12} (cos (- 3 π) + i sin (- 3 π))$ $z^12 = 2^(1/2*12)(cos(-3pi) + isin(-3pi))$

$z^{12} = 2^{6} (cos (3 π) - i sin (3 π))$ $z^12 = 2^6(cos(3pi) - isin(3pi))$

$cos (3 π) = cos (π) = - 1$ $cos(3pi) = cos(pi) = -1$

$sin (3 π) = sin (π) = 0$ $sin(3pi) = sin(pi) = 0$

$z^{12} = - 2^{6} = - 64$ $z^12 = -2^6 = -64$

How do you use demoivre's theorem to simplify (1-i)^12(1−i)12(1-i)^12?