finding all values of z from z^3 -i=0 (1 Viewer)

cloud edwards · Apr 2, 2018

how does one go about finding all values of z from z^3-i=0

i got so far (not sure if right):

(-i)
(sqrt3)/2 + i/2

pikachu975 · Apr 2, 2018

z^3 = i
r^3 cis3theta = cispi/2

r = 1

3theta = pi/2 + 2kpi
Set k = -1, 0 and 1

captainhelium · Apr 2, 2018

I think this is how to do the question:

$$z^3 = i$

$z^3 = cis (\frac{\pi}{2} + 2k\pi)$

$2k\pi$ $is added since rotating a complex number by a multiple of$ $2\pi$ $is going to give you the same value I think.$

$z^3 = cis(\pi(\frac{1}{2}+2k))$

$Using De Moivre's Theorem$

$z = cis(\frac{\pi}{3}(\frac{1}{2}+2k))$

$Then, sub k=0, k=1, k=-1 into the above equation.$

$$ When k = 0$

$z = cis(\frac{\pi}{6})$

$$ When k = 1$

$z = cis(\frac{5\pi}{6})$

$$ When k = -1$

$z = cis(\frac{-\pi}{2}) = -i$

Hopefully that's correct. I'm kinda new to LaTeX so my answer might look a bit funny. Hope that helped!

cloud edwards · Apr 3, 2018

Thanks

finding all values of z from z^3 -i=0 (1 Viewer)

cloud edwards

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pikachu975

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captainhelium

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cloud edwards

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