# Square root of complex numbers

I’m studying complex analysis.

I was reading a section about roots of complex numbers, and I found that $\sqrt{i}$ has two values. ($i$ : imaginary unit) However, for a non-zero real number, $\sqrt{a}$ is always one value. Moreover, $\sqrt{a}$ is a real number.

However, $\sqrt{i}=\pm(\sqrt{0.5}+\sqrt{0.5}i)$. This is not a number !! I think a number should have one value. If $\sqrt{i}$ is the number having positive sign, then it shouldn’t have the negative sign unless it is a zero.

I understand those are the numbers satisfying $x^2=i$. But.. what does $\sqrt{i}$ represent for?

Is it different from the sqaure root of real numbers?

• $\sqrt{1}=\pm 1.$ Isn't this a number? Those are two complex numbers sharing the same property. – Bumblebee Nov 24 '17 at 5:28
• I think it’s not.. I learned that sqrt(1) = 1, -sqrt(1) = -1 in my high school. – ylh0501 Nov 24 '17 at 5:30
• I understand they share the same property. But if we define sqrt(1)=+-1, then can we say ‘sqrt(1)’ is a real number? – ylh0501 Nov 24 '17 at 5:33
• $\sqrt$ is not a well-defined (single valued) function. That is the point! – Bumblebee Nov 24 '17 at 5:39
• It's well define for the reals because we arbitrarily chose to say. $\sqrt{23}$ is the positive square root. It was completely arbitrary. But we don't do the same for complex numbers. – fleablood Nov 24 '17 at 6:11

Every complex and real number except $0$ have two square roots. If $r$ is one of the square roots $-r$ is the other.

So it is not true that "$\sqrt i$ has two values". It's that, "$i$ has two values as square roots. And it's not true that positive real numbers have one square root. They have two.

So what does the symbol $\sqrt {}$ and what does "the square root" mean? Well, nothing really. It's convenience to have a single value "square root function" so we arbitrarily chose that the positive value of square roots of positive real would be "the" square root. And the negative square root was the "other".

We could do the same for complex square roots. We could arbitrarily decide the one with a no negative real component was "the" square root and the one with a negative component would be the "other". But what would be the point?

The main reason we do this for the reals is because the real numbers is convenience really. And none of that convenience is useful in the complex numbers. The complex numbers don't have an greater/less than ordering. You are going to learn that exponentiation is cylic and there are multiple logarithms of each number (don't worry about that; you'll learn it later).

So basically we say $\sqrt z$ to mean the set of the two complex numbers, $w$ and $-w$, so that $w^2=(-w )^2=z$. Or we write $\sqrt z =\pm w$ to mean that the number to be consider a square root of $z$ could be either of $w$ or $-w$.

In the reals, for positive $x$ we define $\sqrt x$ to be the positive value. It is therefore a function because it produces a unique output for each input. However, the equation $x^2=4$ has two solutions. It is not correct to take the square root of each side and tacitly assume the plus sign to get $x=2$ because $x=-2$ is also a solution of the equation. In the complex numbers there is no order, so no set of positive numbers. The equation $z^2=y$ has two solutions for $z$ which are negatives of each other. This behavior is because $(-1)^2=1$ so you can negate any solution to get another one. I see people avoid the square root sign when working in complex variables, I think for this reason. They use the $1/2$ power. As in the reals, this does not select a particular square root. As in the reals, every number (now not just every positive number) except $0$ has two square roots.

For any real or complex numbers $a=b^2$ we have also that $a=(-b)^2$ - and that applies whether or not $a$ is real or complex. So if $a\neq 0$ the equation $x^2=a^2$ will have two solutions $x=\pm a$. We can see this by rewriting the equation as $$(x+a)(x-b)=0$$ which has two roots.

Obviously a function has a single value, and if we want to turn the square root into a function we have to choose a single "principal" value out of the two possible values.

When we work in the real numbers only non-negative integers have a square root and the convention is to choose the positive square root of a positive real number. In the complex numbers every number can have a square root. If $b=a^2e^{2i\theta}$ we have the two solutions $\pm ae^{i\theta}$ to the equation $x^2=b$.

We could, for example choose $- \pi \lt 2\theta \le \pi$ or $0\le 2\theta \lt 2\pi$, and either would give a square root function. In the first case the square root would be the choice with real part $\ge 0$, resolved to the positive imaginary axis for negative reals. In the second case we would choose the solution with non-negative imaginary part, resolved to the positive real solution in the case of positive real numbers. Other choices are also possible.

If you look carefully and think geometrically, you will come to see that this involves tearing the plane down the negative real axis in the first case or the positive real axis in the second case, and that nearby numbers in the plane can have very different square roots. That is not always convenient - so it is sometimes useful to choose one definition over another so that the function is continuous throughout a particular region of interest.

Another way of resolving the issue is to consider the two values of the square root as belonging to two sheets of a single Riemann Surface (with a single value at the origin), which can preserve continuity.

The two values signal the need to take care, but mathematicians have developed tools to do this. As with all such tools it is necessary to learn how to use them and how to recognise the need. For example, in the real numbers the (real) cube root is a function. But when we move to complex numbers there are three possible values for the cube root, and a change of perspective is necessary.

For interest these three values come from the three solutions of the equation $x^3=1$. Obviously $x=1$ is one of these, and writing $x^3-1=(x-1)(x^2+x+1)=0$ we see that the other possible cube roots of $1$ are the solutions of $x^2+x+1=0$.