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QUESTION

# How do you find all the real and complex roots of x^5 - 32 = 0?

x=2e^((2kpii)/5), k=0,1,2,3,4

x^5-32=0

x^5=32

=root(5)(32)

The real solution is obviously 2.

Let's extend the equation to polar complexes:

x^5=32=32e^(2kpii), with k any integer, but pratically, from 0 to 4.

x=root(5)(32e^(2kpii))

x=2e^((2kpii)/5), k=0,1,2,3,4

MORE EXPLANATION OR ANOTHER WAY OF LOOKING AT IT:

An easy way to see this to imagine that the solution divided up around the circle of radius 2 in complex plane. Now divide the circle by 5 so the solutions will be: x_1 = 2/_ theta

1) Real x_1 = 2/_ (theta = 0)  2) Complex x_2 = 2/_ (theta = (2pi)/5)  Find the real and imaginary part: 2costheta + 2isintheta R = 2cos((2pi)/5); I = sin((2pi)/5)  3) Complex x_3 = 2/_ (theta = (4pi)/5)  this is 2((2pi)/5) Find the real and imaginary part: 2costheta + 2isintheta 4) Complex x_4 = 2/_ (theta = (6pi)/5)  this is 3((2pi)/5) Find the real and imaginary part: 2costheta + 2isintheta 4) Complex x_5 = 2/_ (theta = (8pi)/5)  this is 4((2pi)/5) Find the real and imaginary part: 2costheta + 2isintheta

Hope this help...