Complex Number solutions for a Circle

Question

I have a circle of radius 5 with its center at the origin represented as $X^2+Y^2=25$. I get that it has a solution for all values ranging from $-5$ to $+5$.

My question is what does it mean when the equation returns a complex number? Example for $x=6$, I get $y = \pm i\sqrt{11}$. In doing this for all real number greater than $+5$ and less than $-5$, what is being returned/plotted and what plane is this plot on? Is this another circle on the imaginary plane, all values on the plane beyond a circle?

Answer 1

When you graph the solutions in "the plane", you are restricting yourself to look at solutions to the equation where both $x$ and $y$ are real. You could, for example, restrict further to only allow $x$ and $y$ to be rational numbers, and think about how those points fit in with all the real solutions.

To think geometrically about non-real complex solutions, you will need more (real) dimensions! You could restrict yourself, as it sounds like you are doing in the question, to just solutions where $x$ is real and $y$ is allowed to be complex. Then you will need another dimension/direction for the imaginary part of $y$. You could graph this in a "$z$" direction, so that solutions where $x>5$ not lie in the plane but above/below it. You will find that for $|x|>5$ the solutions will be points where $x^2 - z^2 = 25$ so that if you just look at the $(x,z)$ plane the solution set will look like a hyperbola.

Probably the most interesting thing to look at is when you allow both $x$ and $y$ to be complex... but graphing this would require more dimensions.

Answer 2

All the points that lie inside or on the boundary of the circle $$ \left\{ \ (x, y) \in \mathbb{R}^2 \ \colon \ x^2+ y^2 = 5^2 \ \right\} $$ lie to the right of the line $$ \left\{ \ (x, y) \in \mathbb{R}^2 \ \colon \ x = -5 \ \right\} $$ and to the left of the line $$ \left\{ \ (x, y) \in \mathbb{R}^2 \ \colon \ x = 5 \ \right\}. $$ Moreover, the circle intersects the former line at the point $(-5, 0)$ and the latter one at the point $(5, 0)$.

Answer 3

No, you do not get a point on another circle in complex plane if you let $x=6$ and $y= \pm i\sqrt {11} $

The complex plane is the same as the usual $xy$ plane and you can not plot $(6, i\sqrt {11})$ on the complex plane because this is not a single point any more.

One way to find more circles is to consider $x^2+y^2 = R^2 $ where R is the radius and you can pick any $R$ and graph your circle on the $xy$ plane or as you like to call it on the complex plane.

Answer 4

Let me denote by $A$ the circle you consider: $$ A=\{(x,y)\in\mathbb R^2;x^2+y^2=25\}. $$ This is the usual real circle and this is what a circle in the plane means unless something different is specifically explained. This set consists of all pairs $(x,y)$ of numbers $x$ and $y$ which are real numbers and satisfy the equation $x^2+y^2=25$.

I stress that it is part of the very definition of $A$ that the two numbers are real. You can ask whether a point $(x,y)$ can be in $A$ if $x=6$. As you found out, if $x=6$ and the equation $x^2+y^2=25$ is satisfied, then $y=\pm i\sqrt{11}$. But since $y$ is not in $\mathbb R$, the point $(x,y)$ is not in $A$. What fails is not the equation $x^2+y^2=25$ but the condition that $x\in\mathbb R$ and $y\in\mathbb R$. You could say that you have found a solution of the equation which is not in the plane $\mathbb R^2$.

You can also consider a different set: $$ B=\{(x,y)\in\mathbb C^2;x^2+y^2=25\}. $$ This set looks otherwise similar to $A$, but the two numbers $x$ and $y$ can now be complex. You can call $B$ a complexification of $A$. Since $\mathbb R\subset\mathbb C$, we have $A\subset B$. But there are points in $B$ which are not in $A$. For example, the point $(6,-i\sqrt{11})$ is in $B$ — both numbers are complex and the equation is satisfied — but not in $A$ as discussed above. However, the set $B$ is not what is usually called a circle. It is a sort of an extension of the circle, but in a weird way.

If you identify $\mathbb C$ with $\mathbb R^2$, then $B$ is a two-dimensional surface in $\mathbb R^4$, whereas $A$ is a one-dimensional curve in $\mathbb R^2$. Intersecting the surface with a suitable two-dimensional plane in $\mathbb R^4$ gives a circle ($B\cap\mathbb R^2=A$).

Answer 5

In addition to Joona's great answer, there's a nice visualization I think is worth sharing.

If we define our complex number $x$ for $\{(x,y)\in\mathbb C^2;x^2+y^2=25\}$ as $x + iz$ we have the following:

\begin{equation} (x+iz)^2+y^2=25 \end{equation} \begin{equation} (x^2+2xiz+i^2z^2)+y^2 = 25 \end{equation} \begin{equation} x^2+2xiz-z^2+y^2 = 25 \end{equation} \begin{equation} x^2+y^2+2xiz-z^2 = 25 \end{equation}

As we want to visualize a surface in three dimensions, we drop the imaginary part, leaving this as our circle equation:

\begin{equation} x^2+y^2-z^2 = 25 \end{equation}

If we would to graph this, for example in Geogebra, we see the following:

Circle equation surface

Which effectively, if interesected with $z=0$, gives us our circle in $ \mathbb R^2 $:

Surface intersected by $xy$ plane

Then, the points where $y = 0$ intersects the surface are the ones you're looking for. Here is the Geogebra plot if you like to play with it.

Lastly, here's how your point looks like considering the above definition:

And as Callus mentioned in his answer, we can see that the curve is indeed a hyperbola.