SOLUTION: Use de Moivre's Theorem to find the square root of -5+12i. r=13 √ 25+144 = √ 169 arctan= (12 ÷ 5) = 1.1760... 13(cos 1.7600... +2πn + isin 1.1760...) 13

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Question 238741: Use de Moivre's Theorem to find the square root of -5+12i.
r=13 √ 25+144 = √ 169
arctan= (12 ÷ 5) = 1.1760...
13(cos 1.7600... +2πn + isin 1.1760...)
13^(1/2)(((cos 1.1760... + 2πn)/2)+((isin1.76)/2)
This is what I have done so far, please tell me if it is correct, and I am unsure as to what steps to make next.
Answer by Alan3354(21601) About Me  (Show Source):
You can put this solution on YOUR website!
Use de Moivre's Theorem to find the square root of -5+12i.
r=13 √ 25+144 = √ 169
------------------------
angle A = arccos(-5/13)
z = 13(cos(A) + isin(A)) A =~ 112.62º
sqrt(z) = sqrt(13)*(cos(A/2) + isin(A/2))
sqrt(z) = sqrt(13)*(0.5547 + i*0.8321)
= 2 + 3i
----------
arctan= (12 ÷ 5) = 1.1760...
13(cos 1.7600... +2πn + isin 1.1760...)
13^(1/2)(((cos 1.1760... + 2πn)/2)+((isin1.76)/2)