Find all the values of (-2+2i)^(1/3)

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SUMMARY

The discussion focuses on finding all the values of the complex number (-2+2i) raised to the power of 1/3. The solution involves expressing the number in polar form as 2(-1+i) and applying the formula for roots of complex numbers, specifically u^{1/n}=|u|^{1/n} e^{i/n(\theta+2n\pi i)}. The argument θ is determined to be 3π/4, leading to the calculation of the cube roots using m=0, 1, and 2 to obtain the complete set of values.

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Homework Statement



Find all the values of (-2+2i)1/3

Homework Equations



eib = cos(b) + i * sin(b)

The Attempt at a Solution





(-2+2i)1/3 = (2(-1+i))1/3 = 21/3 (√2)1/3 (eiπ(3/4 + 2m))1/3 = 21/2 (eπi)1/4 + 2m/3 = ... Where am I going with this? I don't even know.
 
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You need to use the definition:

[tex]u^{1/n}=|u|^{1/n} e^{i/n(\theta+2n\pi i)}[/tex]

where [itex]\theta[/itex] is the argument of u
 
Jamin2112 said:

Homework Statement



Find all the values of (-2+2i)1/3

Homework Equations



eib = cos(b) + i * sin(b)

The Attempt at a Solution





(-2+2i)1/3 = (2(-1+i))1/3 = 21/3 (√2)1/3 (eiπ(3/4 + 2m))1/3 = 21/2 (eπi)1/4 + 2m/3 = ... Where am I going with this? I don't even know.
Change it back to a+bi form using m=0,1,2 .
 

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