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# UKC

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The solution(s) to x^2 -4 = 0 shows you where the curve f(x) = x^2 -4 crosses the x-axis (+2 & -2).

The curve f(x) = x^2 +4 does not cross the x-axis and the solutions to x^2 + 4 = 0 are complex (+2i & -2i).

What does this represent graphically?

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## / Maths Question : Complex Numbers |

lowersharpnose - * * on 02 Oct 2012

The solution(s) to x^2 -4 = 0 shows you where the curve f(x) = x^2 -4 crosses the x-axis (+2 & -2).

The curve f(x) = x^2 +4 does not cross the x-axis and the solutions to x^2 + 4 = 0 are complex (+2i & -2i).

What does this represent graphically?

MG - * * on 02 Oct 2012

In reply to lowersharpnose: A curve that doesn't cross the x-axis for real numbers?? Are looking for a physical interpretation?

loundsy - * * on 02 Oct 2012

In reply to lowersharpnose:

you can draw them on an argand diagram but not on a normal xy axis

you can draw them on an argand diagram but not on a normal xy axis

Ramblin dave - * * on 02 Oct 2012

In reply to lowersharpnose:

Basically the same thing, except than rather than a function that takes the real number line (the x axis) to another real number line (the y axis), f(x) is being thought of as a function that takes a complex plane to another complex plane, which is a bit harder to visualize because it's four dimensional.

Basically the same thing, except than rather than a function that takes the real number line (the x axis) to another real number line (the y axis), f(x) is being thought of as a function that takes a complex plane to another complex plane, which is a bit harder to visualize because it's four dimensional.

Jon Stewart - * * on 02 Oct 2012

In reply to Ramblin dave:

I wondered why I was having trouble.

> (In reply to lowersharpnose)

> ...which is*a bit* harder to visualize because it's four dimensional.

> ...which is

I wondered why I was having trouble.

Ramblin dave - * * on 02 Oct 2012

In reply to Jon Stewart:

One way of handling it is:

draw a (merely 3d) graph where the x and y axes are the complex plane and the z axis is the real part of f(x+iy)

do the same for the imaginary part of f(x+iy)

each of these has a line in the plane z = 0 where the graph crosses it. This represents the points in the complex plane for which the real and imaginary parts respectively of f(x+iy) are zero.

Draw these two lines on the same plane. f(z) = 0 at the points where they meet.

Often people seem to visualize complex valued functions by looking at a 3d graph of the absolute value, which can help a bit...

One way of handling it is:

draw a (merely 3d) graph where the x and y axes are the complex plane and the z axis is the real part of f(x+iy)

do the same for the imaginary part of f(x+iy)

each of these has a line in the plane z = 0 where the graph crosses it. This represents the points in the complex plane for which the real and imaginary parts respectively of f(x+iy) are zero.

Draw these two lines on the same plane. f(z) = 0 at the points where they meet.

Often people seem to visualize complex valued functions by looking at a 3d graph of the absolute value, which can help a bit...

lowersharpnose - * * on 02 Oct 2012

In reply to Ramblin dave:

Thanks.

I think that s the sort of image I was after, though I am going to have to work at it to visualise what is going on.

Thanks.

I think that s the sort of image I was after, though I am going to have to work at it to visualise what is going on.

Pero - * * on 02 Oct 2012

In reply to lowersharpnose: It's a great idea to look at the absolute value of the function. With z = x + yi, and f(z) = z^2+4 it gives a sort of bowl with a wave around the x-axis. On each line of constant imaginary part of z, the curve is like x^2 + c, where c is a constant, with a minimum on the y-axis (x = 0).

As y reduces towards 2, the bottom of the bowl reduces in minimum height, until it reaches 0 (at y=2), then bounces up again to peak at 4 (y=0), before reducing again to reach 0 at y=-2, then bounces up again and gets further away from the ground again.

Alternatively, think of a conventional bowl, touching the ground at the point (0,0). Do a sort of upward karate chop at this point, leaving the bowl touching the ground at (0,2) and (0,-2) with an upward wave-like curve where it used to touch the ground!

As y reduces towards 2, the bottom of the bowl reduces in minimum height, until it reaches 0 (at y=2), then bounces up again to peak at 4 (y=0), before reducing again to reach 0 at y=-2, then bounces up again and gets further away from the ground again.

Alternatively, think of a conventional bowl, touching the ground at the point (0,0). Do a sort of upward karate chop at this point, leaving the bowl touching the ground at (0,2) and (0,-2) with an upward wave-like curve where it used to touch the ground!

DancingOnRock - * * on 02 Oct 2012

In reply to lowersharpnose: Graphically it's a standard x^2 curve that crosses the Y axis at 0,4 and goes upwards on both sides. Hence why it doesn't cross the X-axis.

whispering nic - * * on 02 Oct 2012

In reply to lowersharpnose: This is all very well but what is the surface area of a Pringle?

Talius Brute on 02 Oct 2012 - *client-80-4-216-14.cht-bng-011.adsl.virginmedia.net*

Pero - * * on 02 Oct 2012

... a karate chop along the x-axis, of course!

Tom Last - * * on 02 Oct 2012

In reply to Talius Brute:

Don't worry mate, they're just making it up.

> (In reply to lowersharpnose)

>

> How do you all know this shit? Very impressive!

>

> How do you all know this shit? Very impressive!

Don't worry mate, they're just making it up.

lowersharpnose - * * on 02 Oct 2012

In reply to Pero:

Thanks.

I really would appreciate a picture of this. You are describing something that I can't easily see. More work required on my part, I guess.

Thanks.

I really would appreciate a picture of this. You are describing something that I can't easily see. More work required on my part, I guess.

lowersharpnose - * * on 02 Oct 2012

birdie num num - * * on 03 Oct 2012

In reply to lowersharpnose:

THIS THREAD was started in off belay and should have been started in Gobbeldygook, Please try to post in the correct forums etc.

THIS THREAD was started in off belay and should have been started in Gobbeldygook, Please try to post in the correct forums etc.

Robert Durran - * * on 03 Oct 2012

In reply to lowersharpnose:

One nice thing is that for the quadratic funtion f(z)= (z-a)^2 +b (a,b real), which, for real z, has a graph which is a parabola with minimum f(z)=b at x=a, takes real values for values of z on the real axis and on a line parallel to the imaginary axis through z=a. A graph of these real values plotted perpendicular to the complex plane consists of the original parabola and another one in a plane perpendicular to the original parabola with a max f(z)=b at z=a. ie "upside down", rotated 90 degrees around its axis of symmetry, and touching the original parabola at its minimum point. For your funtion with a=0 and b=4, this second parabola cuts the complex plane at 2i and -2i.

I don't think I've made that very clear, but it is late...

One nice thing is that for the quadratic funtion f(z)= (z-a)^2 +b (a,b real), which, for real z, has a graph which is a parabola with minimum f(z)=b at x=a, takes real values for values of z on the real axis and on a line parallel to the imaginary axis through z=a. A graph of these real values plotted perpendicular to the complex plane consists of the original parabola and another one in a plane perpendicular to the original parabola with a max f(z)=b at z=a. ie "upside down", rotated 90 degrees around its axis of symmetry, and touching the original parabola at its minimum point. For your funtion with a=0 and b=4, this second parabola cuts the complex plane at 2i and -2i.

I don't think I've made that very clear, but it is late...

Robert Durran - * * on 03 Oct 2012

In reply to lowersharpnose:

I think it is like an arse sitting on the complex plane with the arse crack running above and parallel to the real axis with the anus directly over the origin. The arse makes contact with the complex plane at +2i and -2i. Assume a very firm, pert arse that doesn't squash at all on contact. Think Kylie. That's how I'm picturing it anyway. I hope this helps.

> (In reply to Pero)

> I really would appreciate a picture of this. You are describing something that I can't easily see. More work required on my part, I guess.

> I really would appreciate a picture of this. You are describing something that I can't easily see. More work required on my part, I guess.

I think it is like an arse sitting on the complex plane with the arse crack running above and parallel to the real axis with the anus directly over the origin. The arse makes contact with the complex plane at +2i and -2i. Assume a very firm, pert arse that doesn't squash at all on contact. Think Kylie. That's how I'm picturing it anyway. I hope this helps.

Pero - * * on 03 Oct 2012

In reply to lowersharpnose: Using Excel, you can draw a graph. First, calculate the absolute value for the function around (0,0). I did it from -4 to +4 on the x and y axis. The formula is:

SQRT(x^4 + y^4 + 2x^2y^2 + 8(x^2 - y^2) + 16)

Then choose a 3D surface graph of this. However, as the bit you want to see is at the bottom, it's hidden, so I changed the formula to -SQRT, so that Kylie is on her front with her bottom nicely in the air!

Will email you the excel file.

SQRT(x^4 + y^4 + 2x^2y^2 + 8(x^2 - y^2) + 16)

Then choose a 3D surface graph of this. However, as the bit you want to see is at the bottom, it's hidden, so I changed the formula to -SQRT, so that Kylie is on her front with her bottom nicely in the air!

Will email you the excel file.

Pero - * * on 03 Oct 2012

... looks a bit more like An Teallach than Kylie, though!

lowersharpnose - * * on 03 Oct 2012

Robert Durran - * * on 03 Oct 2012

In reply to Pero:

Each aesthetic in their own way....

> ... looks a bit more like An Teallach than Kylie, though!

Each aesthetic in their own way....

Jim C - * * on 03 Oct 2012

In reply to birdie num num:

Currently old folks (like me) say that the 'young uns' these days just type a question into a smartphone (used to be a scientific calculator) and the answer comes up (probably on a video in U-Tube.)

Is this just lies that I have been puting around?

> (In reply to lowersharpnose)

> THIS THREAD was started in off belay and should have been started in Gobbeldygook, Please try to post in the correct forums etc.

> THIS THREAD was started in off belay and should have been started in Gobbeldygook, Please try to post in the correct forums etc.

Currently old folks (like me) say that the 'young uns' these days just type a question into a smartphone (used to be a scientific calculator) and the answer comes up (probably on a video in U-Tube.)

Is this just lies that I have been puting around?

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