Can imaginary number coordinate system co exist with our Cartesian 2D or 3D coordinate system?
I’m not sure what is meant by “co-exist”, so let me give an absurd example.When in highschool (back in the seventies :-), in algebra class, we were studying 2D functions y = y(x). At one moment it was proposed to extend them to complex coordinates x and y, by considering they belonged ‘’somehow’ to the (x,y) plane (!).Now then, let’s take these ‘isotrope straight lines’ y = ix, and apply to them the same usual formulas as for real coordinates. We would then ‘find out’ that- isotrope lines are perpendicular to themselves:the perpendicular to y = Ax is y = (- 1/A) x, so the perpendicular to y = ix is y = (- 1/i) x … = ix !- the distance between any two points of an isotrope line is zero:For y = Ax, dy = A dx, the distance is ds = SQR(dx^2 + dy^2) = SQR(1 + A^2) dx. So for y = ix, dy = i dx, and ds = SQR(1 + i^2) dx … = 0 !I was the only one who didn’t take these teachings for granted, but it took me quite a while to conclude that complex functions *do* have a ‘real’ geometric existence, only not in ‘real’ space but in an extended ‘complex’ space. For each additional imaginary variable, an additional geometric axis has to be ‘imagined’. Since complex numbers are two-dimensional, the ‘real’ n-space of n real variables corresponds to a ‘complex’ (n+n)-space of n complex variables.This of course posed the problem of graphic rendering of complex functions. The ‘minimal’ 2D case y = y(x), x and y complex, would already take 4 dimensions. Back then (and even still now…) many would refrain from even considering any graphics on these. But realising that a complex function is a *surface* in 4D (it is indeed a ‘map’ of either the complex x- or y-plane) , and that 4D surfaces oughtn’t be much harder than 3D surfaces to project into a picture, I struggled and little by little succeeded in putting my first functions on mm-paper, with the help of a pocket calculator.So the final answer to the question is: n imaginary and n real coordinates co-exist in a complex-structured (n+n)-space, with 2n axes, n real and n imaginary.Representing complex functions y = y(x) is not only possible but it does yield beautiful results. It’s the main topic of my webpages and YT-videos found here:brolproefQB-ComplexMy typical example: the Circle-Hyperbola. In complex coordinates they are the same function (differently oriented in the real plane):
Geometry Parallelogram COORDINATES?
Please help me when this problem. Its a parallelogram and I'm confused. Its called parallelogram "CDEF". I need to find the coordinates for "D" and the coordinates they already gave me are as follows: C: 0,s (yes s the letter) E: u,0 (yes u the letter) F: -t,0 (yes a NEGATIVE t...I really don't get it) please help and explain. Thank you so much.
Find the distance d (in the complex plane) between z = 5 + 10i and w = 11 - 6 i ?
I am new to complex number, not sure if there is more to this, still wrapping my head around the general idea. I know this problem probably involves the distance formula: D = √( (y2 - y1)^2 + (x2 - x1)^2 ) Just not sure how to go about it. Please help Thank you
Show that the points (a, b) and (b, a) are the same distance from the origin.?
Treat the letters (variables) like you would treat numbers. The distance from (0, 0) to (a, b) is d = √((a - 0)² + (b - 0)²) = √(a² + b²). Now try it with (0, 0), and (b, a). See what you get. ***P.S. That distance √(a² + b²) can not be simplified further. Leave it as it is.
1. Reasoning How does the Distance Formula ensure that the distance between two different points is positive?
1. You square the difference of the x-coordinates and difference of the y-coordinates. Whenever you square any number, positive or negative, the result is positive. Furthermore, when you take the square root of the sum of the differences, you are only looking for the positive square root. distance = sqrt( (x2-x1)^2 + (y2-y1)^2 ) 2. The midpoint formula is averaging the x-coordinates and y-coordinates of the two given points. In the case of a number line, you don't have y-coordinates, so you can just apply the midpoint formula to the coordinates on the number, which again is just the average of the two given coordinates. midpoint = (x1+x2)/2
Find the distance between points P(8,2) and O(3,8) to the nearest tenth?
Distance between points in 2 dimensions can easily be determined by using the Pythagorean Theorem. This can be simplified to: C^2 = A^2 + B^2 C^2 = (8 - 3)^2 + (2 - 8)^2 C^2 = 5^2 + (-6)^2 C^2 = 25 + 36 C^2 = 61 C = square root of 61 C = 7.8 (to the nearest tenth) =================================== I hope this is helpful.
How to find the distance between the points (a,b) & (0,0)? I don't get it?
You do the same thing, except you will not get a number as an answer, you will get an answer with a and b in the answer. You should have square root (a-0)^2 + (b-0)^2 Since a - 0 = a, and b - 0 = b, it can be simplified to square root a^2 + b^2