Question #67a77

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Question #67a77

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Answer 1 · 2016-08-01T19:20:43

$z^{11} = 32 + 32 i$ $z^11 = 32 + 32i$

Explanation:

De Moivre's Theorem states that for complex number

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

$z^{n} = r^{n} (cos (n θ) + i sin (n θ))$ $z^n = r^n(cos(ntheta) + isin(ntheta))$

So we need to get our complex number into modulus-argument form.

For $z = x + y i$ $z = x+yi$

$r = \sqrt{x^{2} + y^{2}} and θ = {tan}^{- 1} (\frac{y}{x}) (usually!)$ $r= sqrt(x^2+y^2) and theta = tan^(-1)(y/x) " (usually!)"$

I say usually because the number may be in a different quadrant and require some action.

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

$θ = {tan}^{- 1} (\frac{1}{- 1}) = π - {tan}^{- 1} (1) = \frac{3 π}{4}$ $theta = tan^(-1)((1)/(-1)) = pi - tan^(-1)(1) = (3pi)/4$

So $z = \sqrt{2} (cos (\frac{3 π}{4}) + i sin (\frac{3 π}{4}))$ $z = sqrt(2)(cos((3pi)/4) + isin((3pi)/4))$

$z^{11} = {(\sqrt{2})}^{11} (cos (\frac{33 π}{4}) + i sin (\frac{33 π}{4}))$ $z^(11) = (sqrt(2))^11(cos((33pi)/4) + isin((33pi)/4))$

$z^{11} = 2^{\frac{11}{2}} (cos (\frac{π}{4}) + i sin (\frac{π}{4}))$ $z^11 = 2^(11/2)(cos((pi)/4) + isin((pi)/4))$

$z^{11} = 2^{\frac{11}{2}} (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i) = 2^{\frac{11}{2}} (2^{- \frac{1}{2}} + 2^{- \frac{1}{2}} i)$ $z^11 = 2^(11/2)(1/(sqrt(2)) + 1/(sqrt(2))i) = 2^(11/2)(2^(-1/2) + 2^(-1/2)i)$

$z^{11} = 2^{\frac{11}{2} - \frac{1}{2}} + 2^{\frac{11}{2} - \frac{1}{2}} i = 2^{5} + 2^{5} i$ $z^11 = 2^(11/2-1/2) + 2^(11/2-1/2)i = 2^5 + 2^5i$

$z^{11} = 32 + 32 i$ $z^11 = 32 + 32i$

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Question #67a77

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