Complex (1 Viewer)

fullonoob · Nov 24, 2009

Find the complex 5th roots of -1
not really sure, had my go

help plox
so far i have cos 5 @ + isin5@ = -1
cos 5@ = -cos2pi

tommykins · Nov 24, 2009

convert it to rcis@ form and use de-moivre's

bachviete · Nov 24, 2009

fullonoob said:
Find the complex 5th roots of -1
not really sure, had my go
help plox
so far i have cos 5 @ + isin5@ = -1
cos 5@ = -cos2pi

$z^5 = -1$

$z^5 = r^5\left(\cos5\theta + i\sin\theta5\right)$

$-1 = \cos\pi + i\sin\pi$

$\therefore r^5\left(\cos5\theta + i\sin\theta5\right) = 1\left(\cos\pi + i\sin\pi\right)$

equating terms

$r^5 = 1 \implies r = 1$

$cis 5\theta = cis \pi$

Generalising,

$cis 5\theta = cis\left(2\pi k + \pi)$

$\therefore 5 \theta = 2\pi k + \pi$

$\implies \theta = \frac{2\pi}{5}k + \frac{\pi}{5}$

$k = 0, z_{1} = 1 \, cis \frac{\pi}{5}$

$k = 1, z_{2} = 1 \, cis \frac{3\pi}{5}$

$k = 2, z_{3} = 1 \, cis \pi = -1$

$k = 3, z_{4} = 1 \, cis \frac{7\pi}{5}$

$k = 4, z_{5} = 1 \, cis \frac{9\pi}{5}$

addikaye03 · Nov 30, 2009

Quicker method:

z^5=-1

z=(-1)^1/5

z=(cis(pi+2kpi))^1/5

z=cis((2k+1)/5)pi where k=0,-+1,+-2

Then sub in k values to get roots

Complex (1 Viewer)

fullonoob

fail engrish? unpossible!

tommykins

i am number -e^i*pi

bachviete

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addikaye03

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