How do you find all the complex fourth-roots in rectangular form of #w=25(cos(4pi)/3 + isin(4pi)/3)#?

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Answer 1 · 2017-08-07T17:45:44

#root(4)(w) = color(blue)(1.699), color(blue)(1.699i), color(blue)(-1.699), color(blue)(-1.699i)#

Let's first simplify the expression.

#cos(4pi) = 1#, so we have

#w = 25(1/3 + i(sin(4pi))/3)#

#sin(4pi) = 0#, so we have

#w = 25(1/3 + 0)#

Or

#color(red)(w = 25/3#

The #4# fourth roots of #25/3# are

#root(4)(25/3) = ul((sqrt5)/(root(4)(3))) ~~ color(blue)(ulbar(|stackrel(" ")(" "1.699" ")|)#
#ul((isqrt5)/(root(4)(3))) ~~ color(blue)(ulbar(|stackrel(" ")(" "1.699i" ")|)#
#ul(-(sqrt5)/(root(4)(3))) ~~ color(blue)(ulbar(|stackrel(" ")(" "-1.699" ")|)#
#ul(-(isqrt5)/(root(4)(3))) ~~ color(blue)(ulbar(|stackrel(" ")(" "-1.699i" ")|)#