Solving Argand Diagram for [0,pi/2] & [-pi/2,pi] Intervals

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SUMMARY

The discussion focuses on solving the Argand diagram for the intervals [0, π/2] and [-π/2, π], specifically analyzing the angle given by arg[4/-ω² + 5ωi + 6]. Participants clarify that to determine the quadrant, one must express the complex number in the form x + iy and consider the signs of x and y. The conversation also touches on maximizing the expression x² + y², leading to the conclusion that the maximum occurs at ω = 0, while the limits as ω approaches ±∞ yield infinity.

PREREQUISITES
  • Understanding of complex numbers and their representation in the Argand diagram.
  • Familiarity with the arctangent function and its properties.
  • Basic knowledge of calculus, particularly in maximizing functions.
  • Experience with Laplace transforms and differential equations.
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  • Study the properties of the arctangent function in relation to complex numbers.
  • Learn how to derive and maximize functions using calculus techniques.
  • Explore the application of Laplace transforms in solving differential equations.
  • Investigate the geometric interpretation of complex numbers in the Argand plane.
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Students and professionals in mathematics, particularly those studying complex analysis, differential equations, and optimization techniques. This discussion is beneficial for anyone looking to deepen their understanding of Argand diagrams and related mathematical concepts.

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For this I started with the 1st interval [0,pi/2] and said

arctan [4/ ... ] = [0 , pi/2]

but studying the tan graph this is all positive values of the y-axis

so surely the range of values for this is (0 , inf)?

If I've done that right, how do I do the other interval [-pi/2 , pi]?

Because using that method gives the result (-inf , 0), but then that would also be true for the quadrant [0 , -pi/2] surely, which brings me back to my first answer also being true for [pi/2 , pi]..

How do I attempt this because this is obviously wrong..

This was under the chapter about Laplace / Fourier Transforms

And if you could shed any light on how to maximise too I'd be greatful, because my notes havn't touched on this..

The y(t) = ... was formed from solving a differential equation using laplace transforms, I didn't include that because I didn't think it was relevant. =)
 
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Hi Firepanda! :smile:

(have an omega: ω and a phi: φ and a pi: π :wink:)
Firepanda said:
For this I started with the 1st interval [0,pi/2] and said

arctan [4/ ... ] = [0 , pi/2] …

I don't really follow what you're doing :confused: …

to get arg, you separate 4/… into the form x + iy (not 1/(x + iy) !), and then use tan-1 :wink:
 
tiny-tim said:
Hi Firepanda! :smile:

(have an omega: ω and a phi: φ and a pi: π :wink:)I don't really follow what you're doing :confused: …

to get arg, you separate 4/… into the form x + iy (not 1/(x + iy) !), and then use tan-1 :wink:


Oh shoot I read the question wrong =)

I mean for [-π/2 , 0]

So I guess that's the values for (-inf , 0)..

AH yeah I see now I was doing something wrong in my method too

So how I'm interpreting the question is this:

Where does the angle given by arg[4/-ω2+5ωi+6] lie in the bottom right quadrant [-π/2 , 0] of an argand diagram?

So do I need to change [4/-ω2+5ωi+6] into the form x + iy?
 
Firepanda said:
So how I'm interpreting the question is this:

Where does the angle given by arg[4/-ω2+5ωi+6] lie in the bottom right quadrant [-π/2 , 0] of an argand diagram?

No, it's asking what are the values of ω for which 4/3 lies in the bottom right (4th) quadrant? (and then same question for the 3rd quadrant).
So do I need to change [4/-ω2+5ωi+6] into the form x + iy?

Yes! :smile: (or you'll end up in the wrong quadrant :rolleyes:)
 
tiny-tim said:
No, it's asking what are the values of ω for which 4/3 lies in the bottom right (4th) quadrant? (and then same question for the 3rd quadrant).


Yes! :smile: (or you'll end up in the wrong quadrant :rolleyes:)

Great ok,

I canceled it all down and got my

arctan 5ω/(ω2-6) (lets assume this is right, forget my working too much to type out :P)

then 5ω/(ω2-6) has to lie in (-inf,0) for my arctan 5ω/(ω2-6) to equal the interval (-pi/2 , 0)

so how do I find ω from this?
 
This is getting unnecessarily complicated …

you don't actually need to consider tan-1 since all you are interested in is the quadrant, which means all you need to consider is the signs of x and y (in x + iy) …

just draw the four quadrants, and mark them ++ +- -+ and -- :wink:
 
tiny-tim said:
This is getting unnecessarily complicated …

you don't actually need to consider tan-1 since all you are interested in is the quadrant, which means all you need to consider is the signs of x and y (in x + iy) …

just draw the four quadrants, and mark them ++ +- -+ and -- :wink:

now I got it! ty. Much simpler, and I have myself an answer

Any idea of how to maximise? because that part has me stumped
 
Firepanda said:
Any idea of how to maximise? because that part has me stumped

just maximise x2 + y2 :smile:
 
tiny-tim said:
just maximise x2 + y2 :smile:

So I use my x and y, and add up their squares..

Could you give me the basic method on how to do this, just so it clicks with me and I'm able to =)
 
  • #10
Firepanda said:
So I use my x and y, and add up their squares..

Could you give me the basic method on how to do this, just so it clicks with me and I'm able to =)

erm … I don't see what you think the difficulty is :confused:

add up the squares … what do you get?
 
  • #11
Firepanda said:
So I just add them up, then apply a value of ω from my previous intervals of ω that makes this the largest?

yes :smile:

(except I think the question is asking for any ω)
 
  • #12
tiny-tim said:
yes :smile:

(except I think the question is asking for any ω)

wouldnt that just be infinity then?

or do you mean I have to find another range of values?
 
  • #13
Firepanda said:
wouldnt that just be infinity then?

Nooo … what formula are you using for Aω? :confused:
 
  • #14
tiny-tim said:
Nooo … what formula are you using for Aω? :confused:

Well I was using my x+iy values

I then found x^2 + y^2

and I got ω^4 -12ω^2 +61

So I'm trying to find the value of ω where this is largest, and by the graph it zooms off the screen so I was assuming inf and -inf for my ω
 
  • #15
Firepanda said:
and I got ω^4 -12ω^2 +61

No, it's 1/(that) …

(except you've used 25 instead of 25ω2)
 
  • #16
Firepanda said:
About this thread

https://www.physicsforums.com/showthread.php?t=314510

I keep getting a page not found message whenever I try and reply in it..

So I just wanted to ask one more question

My x2+y2

= 16(ω4-12ω2+61)/(ω4+13ω2+36)

Do I just draw this graph?

It shows a maximum at ω=0

But as ω tends to -inf and +inf then the graph tends to +inf

So is my answer ω = +- inf

Or ω = 0? :)

Hi Firepanda! :smile:

I'm getting confused :confused: …

|1/A| = 1/|A|, so isn't it just 1/(ω4+13ω2+36)?

(and you can factor that :wink:)
 
  • #17
tiny-tim said:
Hi Firepanda! :smile:

I'm getting confused :confused: …

|1/A| = 1/|A|, so isn't it just 1/(ω4+13ω2+36)?

(and you can factor that :wink:)

Well you said this earlier

tiny-tim said:
just maximise x2 + y2 :smile:

So that's why I was doing that

So I thought I was to compute x2 + y2, then find the maximum on the graph
 
  • #18
Firepanda said:
Well you said this earlier

I meant I don't see where your fraction (ω4-12ω2+61)/(ω4+13ω2+36) comes from :confused:
 
  • #19
tiny-tim said:
I meant I don't see where your fraction (ω4-12ω2+61)/(ω4+13ω2+36) comes from :confused:

EDIT: my working out was riddled with mistakes so ill just put the answer!

I seeeeee!

Woohoo

16/(ω4+36+13ω2)

I have the roots of the denominator, what do i do with those?
 
Last edited:
  • #20
Firepanda said:
= 4(-ω2+6-5ωi)/(-ω2+6-5ωi)(-ω2+5ωi+6)
…
Ah I see now I didn't square the denominator :P
…
Woohoo

16/(ω4+36+13ω2)

he he :biggrin:

now you see you were going round in circles … |1/A| = 1/|A| ? :smile:
I have the roots of the denominator, what do i do with those?

hmm :rolleyes: … on second thoughts, factoring doesn't help … the roots are all imaginary, which is a teeny bit irrelevant since ω is real :redface:

 
Last edited by a moderator:
  • #21
tiny-tim said:
he he :biggrin:

now you see you were going round in circles … |1/A| = 1/|A| ? :smile:

Ah *facepalms* lol

tiny-tim said:
hmm :rolleyes: … on second thoughts, factoring doesn't help … the roots are all imaginary, which is a teeny bit irrelevant since ω is real :redface:


Aww took me so long to find those roots too!

How do I complete the square with this? 13 is odd so when I do

(ω2+b)2 = b2 + c

What is b?
 
Last edited by a moderator:
  • #22
try (ω2+b)2 + c, with b = 6.5 :wink:

… and now I'm off to bed … :zzz:​
 

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