December 3rd, 2016, 07:13 AM  #1 
Newbie Joined: Dec 2016 From: HK Posts: 8 Thanks: 0  simple expressions
(âˆš3  j)^8 = 256e^j*(2pi/3) i am confused how does it work 
December 3rd, 2016, 08:48 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,109 Thanks: 983 
$z^n = r^n\bigg[\cos(n \theta) + i\sin(n\theta)\bigg]$ 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 $\dfrac{4\pi}{3} + 2\pi = \dfrac{2\pi}{3}$ $(\sqrt{3}i)^8 = 256\bigg[\cos\left(\dfrac{2\pi}{3}\right) + i\sin\left(\dfrac{2\pi}{3}\right)\bigg]$ $(\sqrt{3}i)^8 = 256 \cdot e^{i \cdot \frac{2\pi}{3}}$ 
December 3rd, 2016, 10:09 AM  #3 
Newbie Joined: Dec 2016 From: HK Posts: 8 Thanks: 0 
Thank you so much !! just started learning calculus

December 3rd, 2016, 10:15 AM  #4  
Newbie Joined: Dec 2016 From: HK Posts: 8 Thanks: 0  Quote:
 
December 3rd, 2016, 01:08 PM  #5 
Math Team Joined: Jul 2011 From: Texas Posts: 2,109 Thanks: 983  

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