FP2 Complex numbers: The locus of a point moving along a half-line

Watch
Maths&physics
Badges: 15
Rep:
?
#1
Report Thread starter 3 years ago
#1
the step between:

arg(...) = -3(pi)/4

and

tan (..) = .../...

im using ... to save time.

I know arg represents or means angle but where did tan come from? and why is the imaginary part divided by the real part .. does it represent opposite over adjacent? if so, how and why? thanks
Attached files
1
reply
Zacken
Badges: 22
Rep:
?
#2
Report 3 years ago
#2
(Original post by Maths&physics)
the step between:

arg(...) = -3(pi)/4

and

tan (..) = .../...

im using ... to save time.

I know arg represents or means angle but where did tan come from? and why is the imaginary part divided by the real part .. does it represent opposite over adjacent? if so, how and why? thanks
Think about a point x+iy in the complex plane. By connecting x+iy to the origin and the real axis you have a right angled triangle with base x and height y. So if the angle the line from the origin to x+iy makes with the real axis is theta than tan theta = opposiite/adjacent = y/x. But theta is precisely the argument.

This is what's happening here except with x,y replaced with other numbers x+2, y-2 or whatever it is.

[Note this is not wholly correct, you usually need a +- pi to get the argument right due to the signs in various quadrants, but the locus stuff takes care of that]
1
reply
Maths&physics
Badges: 15
Rep:
?
#3
Report Thread starter 3 years ago
#3
(Original post by Zacken)
Think about a point x+iy in the complex plane. By connecting x+iy to the origin and the real axis you have a right angled triangle with base x and height y. So if the angle the line from the origin to x+iy makes with the real axis is theta than tan theta = opposiite/adjacent = y/x. But theta is precisely the argument.

This is what's happening here except with x,y replaced with other numbers x+2, y-2 or whatever it is.

[Note this is not wholly correct, you usually need a +- pi to get the argument right due to the signs in various quadrants, but the locus stuff takes care of that]
is it too much trouble to create an illustration to help me understand please? thanks
0
reply
Zacken
Badges: 22
Rep:
?
#4
Report 3 years ago
#4
(Original post by Maths&physics)
is it too much trouble to create an illustration to help me understand please? thanks
Image
2
reply
Maths&physics
Badges: 15
Rep:
?
#5
Report Thread starter 3 years ago
#5
(Original post by Zacken)
Image
thanks a lot but in this case, the theta = -3pi/4, which is too large to create a right angle triangle.
0
reply
DFranklin
Badges: 18
Rep:
?
#6
Report 3 years ago
#6
(Original post by Maths&physics)
thanks a lot but in this case, the theta = -3pi/4, which is too large to create a right angle triangle.
But the same math holds; sin, cos and tan are defined to make this work for arbitrary angles.
1
reply
Maths&physics
Badges: 15
Rep:
?
#7
Report Thread starter 3 years ago
#7
(Original post by DFranklin)
But the same math holds; sin, cos and tan are defined to make this work for arbitrary angles.
I see thats its y + 2 and x-3, which is the opposite of what you'd expect from the initial example you gave me - which I understand perfectly, thanks to zacken. Could you please draw an example showing how that same logic (of the right angle triangle) applies in this instance please? thanks
0
reply
Maths&physics
Badges: 15
Rep:
?
#8
Report Thread starter 3 years ago
#8
Attachment 704912
(Original post by Zacken)
Image
(Original post by DFranklin)
But the same math holds; sin, cos and tan are defined to make this work for arbitrary angles.
so, tan(150) = - (root 3 / 3)


I drew that on the on the real and imaginary planes (see attachment), with only one value being a negative: the adjacent in this case.

when I done the inverse tan of (root3/-3) it gave me - 30, which when added to 180 gives use 150.

the same logic is applicable to the this particular issue with the complex numbers or have I got it completely wrong?
0
reply
DFranklin
Badges: 18
Rep:
?
#9
Report 3 years ago
#9
(Original post by Maths&physics)
Attachment 704912



so, tan(150) = - (root 3 / 3)


I drew that on the on the real and imaginary planes, and I drew this out, with only one value being a negative (the adjacent in this case)

when I done the inverse tan of (root3/-3) it gave me - 30, which when added to 180 gives use 150.

the same logic is applicable to the this particular issue or have I got it completely wrong?
To be honest, I have no idea what you are trying to say (or do) here.

There are issues(*) in deciding the correct angle if all you know is the tangent (or the ratio (y-a)/(x-b)). But in your problem you already know the angle -3\pi/4 and you're trying to get to the ratio (x-a)/(x-b). In which case you can just take the tangent, and that's fine. You should never be worrying about "adding 180 etc."

(*) The issue comes from the fact that if you want to find arg(x+iy) by simply looking at (x/y), then you obviously can't tell the difference between (x+iy) and -(x+iy) (since the second gives you the ratio (-x/-y) which is the same as (x/y)). So if you want the argument, you need to be careful.

To be honest, I think it would be helpful if you posted your original question in full. There are subtleties depending on exactly how you are supposed to be dealing with the fact that \arg(z-w) = \theta only defines a half-line, whereas something of the form \frac{y-b}{x-a} = \tan \theta defines a full line (possibly excluding the point (a, b)).
0
reply
Maths&physics
Badges: 15
Rep:
?
#10
Report Thread starter 3 years ago
#10
(Original post by DFranklin)
To be honest, I have no idea what you are trying to say (or do) here.

There are issues(*) in deciding the correct angle if all you know is the tangent (or the ratio (y-a)/(x-b)). But in your problem you already know the angle -3\pi/4 and you're trying to get to the ratio (x-a)/(x-b). In which case you can just take the tangent, and that's fine. You should never be worrying about "adding 180 etc."

(*) The issue comes from the fact that if you want to find arg(x+iy) by simply looking at (x/y), then you obviously can't tell the difference between (x+iy) and -(x+iy) (since the second gives you the ratio (-x/-y) which is the same as (x/y)). So if you want the argument, you need to be careful.

To be honest, I think it would be helpful if you posted your original question in full. There are subtleties depending on exactly how you are supposed to be dealing with the fact that \arg(z-w) = \theta only defines a half-line, whereas something of the form \frac{y-b}{x-a} = \tan \theta defines a full line (possibly excluding the point (a, b)).
the full question is attached. im just trying to understand how (tan theta = opposite / adjacent) in this particular case. I understand it when it comes to the initial example where all the values where positive, but not here. Name:  Screen Shot 2017-11-20 at 17.42.19.jpg
Views: 235
Size:  117.3 KB
0
reply
DFranklin
Badges: 18
Rep:
?
#11
Report 3 years ago
#11
(Original post by Maths&physics)
the full question is attached. im just trying to understand how (tan theta = opposite / adjacent) in this particular case
Well, I would say that talking about opposite / adjacent is thinking about this in the wrong way.

If \arg(x+iy) = \theta and R=\sqrt{x^2+y^2}, then

x+iy = R(\cos \theta + i\sin \theta).

This is the absolutely fundamental result that you should know off by heart.

So then obviously \dfrac{y}{x} = \dfrac{\sin \theta}{\cos \theta} =\tan \theta.

I will point out that the cartesian line you end up with is NOT identical to the half-line defined originally (since it isn't a half-line). The question seems to be ignoring this.

Are you happy with the direction of the red line in the diagram? [If you are hung up on opposite/adjacent I would have thought you'd have questions about this too].
1
reply
Maths&physics
Badges: 15
Rep:
?
#12
Report Thread starter 3 years ago
#12
(Original post by DFranklin)
Well, I would say that talking about opposite / adjacent is thinking about this in the wrong way.

If \arg(x+iy) = \theta and R=\sqrt{x^2+y^2}, then

x+iy = R(\cos \theta + i\sin \theta).

This is the absolutely fundamental result that you should know off by heart.

So then obviously \dfrac{y}{x} = \dfrac{\sin \theta}{\cos \theta} =\tan \theta.

I will point out that the cartesian line you end up with is NOT identical to the half-line defined originally (since it isn't a half-line). The question seems to be ignoring this.

Are you happy with the direction of the red line in the diagram? [If you are hung up on opposite/adjacent I would have thought you'd have questions about this too].
im familiar withx+iy = R(\cos \theta + i\sin \theta).

why are you dividing y/x?.... So then obviously \dfrac{y}{x} = \dfrac{\sin \theta}{\cos \theta} =\tan \theta. ???

no, I don't understand the direction of the line.
0
reply
DFranklin
Badges: 18
Rep:
?
#13
Report 3 years ago
#13
(Original post by Maths&physics)
im familiar withx+iy = R(\cos \theta + i\sin \theta).

why are you dividing y/x?.... So then obviously \dfrac{y}{x} = \dfrac{\sin \theta}{\cos \theta} =\tan \theta.
Because you want the gradient of the line.

no, I don't understand the direction of the line.
As suspected. In which case your problem is your understanding of sin, cos, tan and circular measure, which you cover in C2.
0
reply
Maths&physics
Badges: 15
Rep:
?
#14
Report Thread starter 3 years ago
#14
(Original post by DFranklin)
Because you want the gradient of the line.

As suspected. In which case your problem is your understanding of sin, cos, tan and circular measure, which you cover in C2.
since when are we trying to find the gradient? that was never explained to me. or do you mean gradient/angle?

I done c2 a while ago now. how would I determine the direction of the line?
0
reply
DFranklin
Badges: 18
Rep:
?
#15
Report 3 years ago
#15
(Original post by Maths&physics)
since when are we trying to find the gradient? that was never explained to me. or do you mean gradient/angle?
If a line though (a, b) has gradient G, then you can write it in the form \dfrac{y-b}{x-a} = G. Therefore if you want to put it in this form, you want to know the gradient. This should have been covered in C1.

I done c2 a while ago now. how would I determine the direction of the line?
If you've done C2, you should know that the direction of a (half-line) making an angle theta with the x-axis (theta measured counter clockwise) is (\cos \theta, \sin \theta).
0
reply
Maths&physics
Badges: 15
Rep:
?
#16
Report Thread starter 3 years ago
#16
(Original post by DFranklin)
If a line though (a, b) has gradient G, then you can write it in the form \dfrac{y-b}{x-a} = G. Therefore if you want to put it in this form, you want to know the gradient. This should have been covered in C1.

If you've done C2, you should know that the direction of a (half-line) making an angle theta with the x-axis (theta measured counter clockwise) is (\cos \theta, \sin \theta).
So I now see how we are working out the gradient.

But what happens to the i from y or even sin when we do y/x or sin/cos???

And thanks. I think I know the direction of the line now.
0
reply
DFranklin
Badges: 18
Rep:
?
#17
Report 3 years ago
#17
When you think of x+iy as a point in the plane, you map it to the point (x, y). So when you want to work out geometry, the i has already gone.
0
reply
Maths&physics
Badges: 15
Rep:
?
#18
Report Thread starter 3 years ago
#18
(Original post by Zacken)
......
(Original post by DFranklin)
When you think of x+iy as a point in the plane, you map it to the point (x, y). So when you want to work out geometry, the i has already gone.
why is it written arg(Z-Z1) = theta?

isn't arg = theta?

in this case is it: arg/this angle of these varying values of Z = this absolute value of theta?

because im interpreting it as theta(Z-Z1) = theta
0
reply
DFranklin
Badges: 18
Rep:
?
#19
Report 3 years ago
#19
(Original post by Maths&physics)
why is it written arg(Z-Z1) = theta?
Why is *what* written as arg(Z-Z1) = theta? What is this Z1 that you haven't mentioned until this post?

isn't arg = theta?
No. arg is a function, while (in this case) theta is the result of applying that function to a particular complex number.

What you've written is like saying "\sqrt(4) = 2, so \sqrt{} = 2". (And it makes about as much sense).
0
reply
Maths&physics
Badges: 15
Rep:
?
#20
Report Thread starter 3 years ago
#20
(Original post by DFranklin)
Why is *what* written as arg(Z-Z1) = theta? What is this Z1 that you haven't mentioned until this post?

No. arg is a function, while (in this case) theta is the result of applying that function to a particular complex number.

What you've written is like saying "\sqrt(4) = 2, so \sqrt{} = 2". (And it makes about as much sense).
ok, but the arg of z is the angle the R of Z creates with the positive real axis in an anti clockwise direction?
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Should there be a new university admissions system that ditches predicted grades?

No, I think predicted grades should still be used to make offers (686)
33.89%
Yes, I like the idea of applying to uni after I received my grades (PQA) (860)
42.49%
Yes, I like the idea of receiving offers only after I receive my grades (PQO) (387)
19.12%
I think there is a better option than the ones suggested (let us know in the thread!) (91)
4.5%

Watched Threads

View All
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise