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Now, use Euler's formula to find the 3 roots.
(-64 i) ^1/3 = - 2√(3) - 2i???
Originally Posted by mous99 (-64 i) ^1/3 = - 2√(3) - 2i??? That is one of he cube roots of -64i. There are two others - do you know how to find them? You can use de Moivre's tehorem:
From this you can derive the fact that for ,
Here you have , , and . Set k = 0, 1, and 2 to find the three cube roots.
(-64i)^1/3 = (-64)^1/3 *(i)^1/3
= -4 (i)^1/3
= -4(cos π/6 + i sin π/6)
why the 0= 3π/ 6, not π/6???
Originally Posted by mous99 (-64i)^1/3 = (-64)^1/3 *(i)^1/3 Not quite right: for respectively.
Originally Posted by mous99 Find (-64i)^1/3. I don't understand why all the replies assume that the question is about cube roots.
I would have thought that it is straightforward complex expatiation: .
In this case the principal value:
Just possibly all replies assume that the question is about cube roots because the question specifically [b]asked[/b about a cube root.
Originally Posted by HallsofIvy Just possibly all replies assume that the question is about cube roots because the question specifically asked about a cube root. Where in Originally Posted by mous99 Find (-64i)^1/3. is there anything about cube roots?
Is that not just complex exponentiation?
Last edited by Plato; December 19th 2012 at 01:13 PM.
how to find θ= 3π/2 +2kπ,
because my answer for θ = π/2.
its no same with u.
(-64i)^1/3 = 4(-i)^1/3
z = re^iθ
z^3 = (r^3)(e^3θ )
so, r^3 = 64
r = 4
how to find 3θ = 3π/2 +2kπ???
z = (-64i)^1/3
z^3 = (-64i)1/3
r = √[(0)^2 + (-64)^2 ] =64
cos 3 π/2 = 0
sin 3π/2 = -1
z^3 = (r^3) ( e^(3π/2 + 2kπ)
theta = π / 2 + (2kπ/3)
when k = 0, theta = π/2 z0 = 4i
when k =1, theta = ??? (how to find this theta)
Did you consider putting k= 1 in the formula you are given?
If k= 0, as you say.
If k= 1,
If k= 2,
If k= 3, which will give the same complex number as with k= 0 because sine and cosine (and so ) have period .
Last edited by HallsofIvy; December 22nd 2012 at 03:11 PM.
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