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December 3rd, 2016, 06:13 AM   #1
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simple expressions

(√3 - j)^8 = 256e^j*(2pi/3)

i am confused how does it work
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December 3rd, 2016, 07:48 AM   #2
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$z^n = r^n\bigg[\cos(n \theta) + i\sin(n\theta)\bigg]$

where $z = x + iy \implies r = \sqrt{x^2+y^2}$

$z = \sqrt{3} - i \implies r = \sqrt{3 + 1} = 2$

$\theta = \arctan\left(\dfrac{y}{x}\right) = \arctan\left(\dfrac{-1}{\sqrt{3}}\right) = -\dfrac{\pi}{6}$

$z^8 = 2^8\bigg[\cos\left(8 \cdot \dfrac{-\pi}{6}\right) + i\sin\left(8 \cdot \dfrac{-\pi}{6}\right)\bigg]$

$(\sqrt{3}-i)^8 = 256\bigg[\cos\left(\dfrac{-4\pi}{3}\right) + i\sin\left(\dfrac{-4\pi}{3}\right)\bigg]$

note $-\dfrac{4\pi}{3} + 2\pi = \dfrac{2\pi}{3}$

$(\sqrt{3}-i)^8 = 256\bigg[\cos\left(\dfrac{2\pi}{3}\right) + i\sin\left(\dfrac{2\pi}{3}\right)\bigg]$

$(\sqrt{3}-i)^8 = 256 \cdot e^{i \cdot \frac{2\pi}{3}}$
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December 3rd, 2016, 09:09 AM   #3
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Thank you so much !! just started learning calculus
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December 3rd, 2016, 09:15 AM   #4
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Quote:
Originally Posted by skeeter View Post
$z^n = r^n\bigg[\cos(n \theta) + i\sin(n\theta)\bigg]$

where $z = x + iy \implies r = \sqrt{x^2+y^2}$

$z = \sqrt{3} - i \implies r = \sqrt{3 + 1} = 2$

$\theta = \arctan\left(\dfrac{y}{x}\right) = \arctan\left(\dfrac{-1}{\sqrt{3}}\right) = -\dfrac{\pi}{6}$

$z^8 = 2^8\bigg[\cos\left(8 \cdot \dfrac{-\pi}{6}\right) + i\sin\left(8 \cdot \dfrac{-\pi}{6}\right)\bigg]$

$(\sqrt{3}-i)^8 = 256\bigg[\cos\left(\dfrac{-4\pi}{3}\right) + i\sin\left(\dfrac{-4\pi}{3}\right)\bigg]$

note $-\dfrac{4\pi}{3} + 2\pi = \dfrac{2\pi}{3}$

$(\sqrt{3}-i)^8 = 256\bigg[\cos\left(\dfrac{2\pi}{3}\right) + i\sin\left(\dfrac{2\pi}{3}\right)\bigg]$

$(\sqrt{3}-i)^8 = 256 \cdot e^{i \cdot \frac{2\pi}{3}}$
but why +2pi to -4pi/3 and where did that 2pi come from
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December 3rd, 2016, 12:08 PM   #5
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Quote:
Originally Posted by hkg277340 View Post
but why +2pi to -4pi/3 and where did that 2pi come from
review co-terminal angles from your trig course ...
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