nth root solution?

Question

How do you calculate the nth root of a negative numer. i.e. 8th root of -16

Answer 1

-16 can be written as 16*[cos(pi)+isin(pi)]
De Moivre's theorm...>
-16(power of 1/8)=16^(1/8)[cos2*k*pi/8+ isin2*k*pi/8] ..k=0 to 7 giving a set of 8 roots symnetrically disposed about the axis of the argand diagram, separated each by 45deg=pi/4, with IZI= sqrt(2)..I think!

Answer 2

If n is odd, then the root has the same sign as the number under the radical, i.e. , the 3rd 5th 7th 9th root, etc.

If n is even, i.e. square root, 4th root, 6th root, 8th root, 10th root, etc., then only positive numbers have real roots, the negative numbers will involve imaginary numbers, and need the symbol for a negative root (√-1) which is 'i'.

Your example is an even 'n', so your answer will use the 'i'. The 8th root of -16 = the 4th root of -4, or the 2nd (square) root of -2, which is i√2

Answer 3

16 = 2^4 = [ 2^(1/2)]^8 = (√2)^8, so
-16 = (i√2)^8
since i^8 = (i^4)² = 1² = 1

so 8th root is i√2

Of course, DeMoivre's Theorem will give you the other 7 8th roots. See the link, down near the end of the article.

Answer 4

The polynomial additivity equation would not enable for a subgroup of the slope consequently removing 3 of the strategies, the merely proper would effectively make the alogorithm invalid commencing to be a hollow indoors the consistency of the effect. My answer would consequently be e) Snickers Bar.

Answer 5

i = SQ RT OF -1