what are the solutions to: x^n - a = 0?

Question

how meny complex solution are?

Answer 1

In the complex field, there are always n solutions to such a problem.

Step one: find all n roots of 1, then multiply each one by the n'th root of the absolute value of a.

e.g., x^8 - 16 = 0

the 8 roots of 1:

1
-1
i
-i
(√2/2)(1+i)
(√2/2)(1-i)
(√2/2)(-1+i)
(√2/2)(-1-i)

the 8th root of 16 = √2

Multiply each root of 1 by the value √2

you get:
√2
-√2
i√2
-i√2
(1+i)
(1-i)
(-1+i)
(-1-i)

check:
(-1-i)^8 = (((-1-i)^2)^2)^2)
= ((2i)^2)^2
= (-4)^2
= 16

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√ is the square root sign on my screen.
i is the square root of -1
(or, the better definition: i is such that i^2 = -1)

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some 'curves'

if n = 0, then a must be 1; in that case, any x (other than 0) is OK.

if n is negative, then use the inverse, and you get a positive n problem.

x^(-4) -a = 0
x^(-4) = a
x^4 = 1/a (for a not equal to zero)

Answer 2

x ^n = a
so x = nth root of a if n is even there are two solutions. in n is odd number > 1 then there is a complex solution