Square root of k

Question

This can be a strange question but what is the result of $\sqrt1$ ? I mean if we consider that a square root of a real number should always be greater or equal to zero the answer should be 1. But if we write $1$ using complex numbers notation and then we use the formula for the n-th root we come up with two solutions $1$ and $-1$.

Answer 1

By convention the square root of a positive number is positive.

This can be a little confusing at first, since both sentences below are correct:

The square root of $1$ is $1$.

$1$ and $-1$ are square roots of $1$.

The square root refers to the positive one.

Answer 2

We always get two roots, this is due to the Fundamental theorem of Algebra. The positive root is called the principal root.

To take powers of complex numbers, use De Moivre's formula:

$$\left(\cos(x)+i\sin(x)\right)^n\;=\;\cos(nx) + i\sin(nx)$$

Answer 3

It's just convention/definition, but the symbol $\sqrt{.}$ refers to the positive square root, for real numbers. In complex analysis, you can talk about branches and what's actually going on is that there are two branches of the square root function, and $\sqrt{.}$ is the principal branch.

Answer 4

Those are different pairs of shoes.

As square root shall be a function, it must be unique. As long as we work with real square roots, the convention is easy to maintain: We define $\sqrt a$ (for $x\ge 0$) as the unqiue non-negative solution of $x^2=a$.

When we work with complex numbers, one can still make a similar convention, but this will fail to be a continous choice. For example, we could declare that with $\sqrt a$ we mean the unique solution of $z^2=a$ with positive real part or with zero real and nonnegatie imaginary part. This convention (or extension of the real convention) makes $\sqrt{-1}=i$ even though $-i$ is also a solution of $z^2+1=0$. What makes this discomforting is that for $a\approx -1$ we may have $\sqrt a\approx -i$ instead of $\sqrt a\approx i$ - the square root becomes discontinuous at all points of the negative real axis. But this dicontinuity is not the square root functions "fault", it is because f our convention. With a different convention, the discontinuity might occur at the positive real axis, or at one half of the imaginary axis, or at some wierdly shaped curve ... So we have difficulties with keeping square root a function and continuous: If we insist on it being a function it becomes discontinuous somewhere. Interestingly, ít also works the other way round: You can keep the square root continuous if you stop insisting it is a function, so to speak. To learn more about this, research Riemann surfaces.