I Several equal but different ones (1)?

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I ran into a little problem trying to solve the (√x )+ 1 = 0 equation.
Summary: I ran into a little problem trying to solve the (√x )+ 1 = 0 equation.

Obviously there will be no solution in R, so I tried the following

(√x )+ 1 = 0

ei α/2 = ei π

i α / 2 = i pi

so: α = 2 pi

so the solution will be e2 i π

This solution actually works when replaced in the equation.

The question is:

e2 i π works, but 1 which is the same does not work, just as e4 i πdoes not work either.

It seems that of the infinite possibilities of complex representation of 1 only one of them is solution!

Anyone could explain this?

Tanks
 
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There are many solutions: ##e^{(4n+2)\pi i}##, for all integer ##n##.
 
√1 = ±1
 
DaveE said:
√1 = ±1
This isn't true. ##\sqrt 1 = +1## only, if we're talking about the real square root function that maps nonnegative real numbers to the same set of numbers.

In any case, the original equation was ##\sqrt x + 1 = 0##, or ##\sqrt x = -1##. If we square both sides, we get ##x = 1,## but this is not a solution of the first equation. In short, the equations ##\sqrt x + 1 = 0## and ##x = 1## are not equivalent.
 
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Izy Amisheeva said:
(√x )+ 1 = 0

ei α/2 = ei π

i α / 2 = i pi

This is not equivalent to the previous equation since ##e^A = e^B## does not imply ##A = B##

so the solution will be e2 i π
This solution actually works when replaced in the equation.

The solution for ##\alpha## works when replaced in the equation ##i \alpha / 2 = i \pi##, but that equation is not equivalent to the original equation.
 
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DaveE said:
√1 = ±1
No, √1= 1. The square root of a positive real number, √a, is defined as the positive number, x, such that x^2= a.
 
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