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# 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...