How do you find the 4th root of #2 (cos (pi/4) + isin(pi/4)) #?

1 Answer
Jul 22, 2016

The principal #4#th root is:

#root(4)(2(cos(pi/4)+i sin(pi/4)))=root(4)(2)(cos(pi/16) + i sin(pi/16))#

There are #3# other #4#th roots.

Explanation:

In general, if #n# is a positive integer and #theta in (-pi, pi]# then the principal #n#th root is given by:

#root(n)(r(cos theta + i sin theta)) = root(n)(r)(cos (theta/n) + i sin (theta/n))#

So in our example:

#root(4)(2(cos(pi/4)+i sin(pi/4)))=root(4)(2)(cos(pi/16) + i sin(pi/16))#

Note that the primitive Complex #4#th root of #1# is #i#, so the other #4#th roots are:

#i root(4)(2)(cos(pi/16) + i sin(pi/16)) = root(4)(2)(cos((5pi)/16)+i sin((5pi)/16))#

#i^2 root(4)(2)(cos(pi/16) + i sin(pi/16)) = root(4)(2)(cos(-(7pi)/16)+i sin(-(7pi)/16))#

#i^3 root(4)(2)(cos(pi/16) + i sin(pi/16)) = root(4)(2)(cos(-(3pi)/16)+i sin(-(3pi)/16))#