Several equal but different ones (1)?

[Izy Amisheeva]

TL;DR

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

e^{i α/2} = e^{i π}

i α / 2 = i pi

so: α = 2 pi

so the solution will be e^{2 i π}

This solution actually works when replaced in the equation.

The question is:

e^{2 i π} works, but 1 which is the same does not work, just as e^{4 i π}does 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

 

[mathman]

There are many solutions: $e^{(4n+2)\pi i}$, for all integer $n$.

 

[DaveE]

√1 = ±1

 

[fresh_42]

The equation to be solved is $x^2+1=0$ which has two solutions in the algebraic closure of $\mathbb{R}$.

If you deal with complex numbers, then you cannot use all formulas you're accustomed to in the reals, especially powers. Have a read:
https://www.physicsforums.com/insights/things-can-go-wrong-complex-numbers/

 

[Mark44]

√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.

 

[Stephen Tashi]

(√x )+ 1 = 0

e^{i α/2} = e^{i 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 e^{2 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.

 

[HallsofIvy]

√1 = ±1

No, √1= 1. The square root of a positive real number, √a, is defined as the positive number, x, such that $x^2= a$.