SOLUTION: Please help me solve this (16i)^0.25

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Question 899380: Please help me solve this (16i)^0.25
Found 2 solutions by Fombitz, Edwin McCravy:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Using DeMoivre's formula,

%280%2B16i%29%5E%280.25%29=2%28cos%2845%2F2%29%2Bisin%2845%2F2%29%29
%280%2B16i%29%5E%280.25%29=2%280.9238%2B0.3827i%29
%280%2B16i%29%5E%280.25%29=1.848%2B0.7654i

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
There are 4 answers.  The other tutor only gave one.
He used degrees, I'll use radians:

%2816i%29%5E0.25

%280%2B16i%29%5E0.25

The modulus is 16 and the argument is pi%2F2%2B2pi%2An

16%28cos%28pi%2F2%2B2pi%2An%29%2Bi%2Asin%28pi%2F2%2B2%2Api%2An%29%29%5E%281%2F4%29

Now we use DeMoivre's theorem:



2%28cos%28pi%2F8%2B2pi%2An%2F4%29%2Bi%2Asin%28pi%2F8%2B2pi%2An%2F4%29%29

2%28cos%28pi%2F8%2B4pi%2An%2F8%29%2Bi%2Asin%28pi%2F8%2B4pi%2An%2F8%29%29

2%28cos%28%28pi%2B4pi%2An%29%2F8%29%2Bi%2Asin%28%28pi%2B4pi%2An%29%2F8%29%29

2%28cos%28%28%281%2B4n%29pi%29%2F8%29%2Bi%2Asin%28%28%281%2B4n%29pi%29%2F8%29%29

Now use n=0,1,2,3 to get 4 answers:

2%28cos%28pi%2F8%29%2Bi%2Asin%28pi%2F8%29%29

2%28cos%285pi%2F8%29%2Bi%2Asin%285pi%2F8%29%29

2%28cos%289pi%2F8%29%2Bi%2Asin%289pi%2F8%29%29

2%28cos%2813pi%2F8%29%2Bi%2Asin%2813pi%2F8%29%29

Edwin