Graphical representation of complex roots to equations

AI Thread Summary
Complex roots of equations, such as those represented by g(x) = x² + 4, do not intersect the x-axis, making their graphical representation more complex. A geometric interpretation of these roots requires four dimensions, but can be projected into three dimensions for visualization. Techniques include showing a transformation of a grid in the complex plane before and after applying the function, illustrating how complex roots can be identified. The discussion highlights that the roots correspond to points in a transformed grid, where the spacing and angles change. Understanding these concepts can enhance comprehension of complex numbers and their graphical representations.
RK1992
Messages
87
Reaction score
0
I've never properly studied complex numbers but I will soon (in September). Basically:

We get taught from a young age that:
the real root of f(x)=x²-4 is where the graph of y=f(x) cuts the x axis

But is there a graphical representation of a complex root?

What's so special about the value x= +/- 2i if g(x)=x²+4 ? Is there a 3D graphical representation of this root?

Thanks in advance.
 
Mathematics news on Phys.org
Well, remember that "real root of f(x)" really means a number a such that f(a)=0. It doesn't need any geometric interpretation to make sense.

Some people like to think in terms of geometry rather than algebra; so the particular correspondence between them you invoked says that the roots of f(x) correspond to the intersection of the parabola defined by y=f(x) and the line y=0, just like you described.

The similar geometric interpretation for complex-valued functions of complex numbers requires 4 dimensions to draw. We can often get away with projecting onto a three-dimensional image, though. (Of course, we have to project that onto a two-dimensional image so that we can draw it, and things get messy)

Another common way is to instead draw a before and after picture of the complex plane. e.g. put a grid on the "before" picture, and in the "after" picture we see a picture of how f transformed the grid.


For this function, a good "before" picture is to make the grid out of rays emanating from the origin and circles whose center is the origin.

The after picture consists of lines passing through the point -4 + 0i, and circles centered on that point. The spacing between the circles is unchanged. However, the spacing between the lines has doubled, and the grid overlaps itself -- e.g. the two rays emanating from the origin at angles x° and (x+180)° both map to the same ray emanating from -4 + 0i at an angle of (2x)°.

In this picture, we can estimate the roots of f by looking at 0 + 0i in the "after" picture, identifying the grid points that lie there, and then finding where they came from in the "before" picture.
 
A simpler (or more simple minded) way of looking at it:

The equation x^2- 2ax+ a^2+ b^2= (x-a)^2+ b^2= 0 has roots [math]x= a\pm bi[/math]. It has complex roots, of course, because its graph does not cross the x-axis.

The vertex of the graph is where x= a so that itex]y= (x-a)^2+ b^2= b^2[/itex]. That is, the graph goes down to (a, b^2) and then back up again. In general, if the graph of y= x^2- ax+ b lies entirely above the x-axis, and its vertex is at (x_0, y_0), then its roots are x_0\pm i\sqrt{y_0}.
 
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...
Back
Top