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

##x=2e^((2kpii)/5) k=01234##
##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=01234##
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…