Merry Christmas from Math Man (1 Viewer)

cutemouse

Account Closed
Joined
Apr 23, 2007
Messages
2,250
Gender
Undisclosed
HSC
N/A
Prove that:
You should really say because of periodicity.

By definition,



If you didn't use pv then you'd get something like a + exp(2pi k) term as well... (The complex exponential function is periodic)

EDIT: Just saw the posted solutions... You should really use pv (principal value) for the fore mentioned reasons. But this is all complex analysis (usually a second year university course).
 
Last edited:

math man

Member
Joined
Sep 19, 2009
Messages
503
Location
Sydney
Gender
Male
HSC
N/A
You should really say because of periodicity.

By definition,



If you didn't use pv then you'd get something like a + exp(2pi k) term as well... (The complex exponential function is periodic)

EDIT: Just saw the posted solutions... You should really use pv (principal value) for the fore mentioned reasons. But this is all complex analysis (usually a second year university course).
yeh this is the method i posted before...but i did forget to say pv...however these young minded 4u students need not worry about logarithms of complex numbers and the periodicity of the exponential function
 

cutemouse

Account Closed
Joined
Apr 23, 2007
Messages
2,250
Gender
Undisclosed
HSC
N/A
Complex analysis is a very beautiful course. Most of my friends hated it though.

You can do things like finding a solution to sin z = 2 (can't remember if this particular case has a solution, but it could have a complex solution) using sin z = (e^z - e^(-z) )/(2i) or by letting z=x+iy and using the double angle formulae and hyperbolic formulae.
 

Carrotsticks

Retired
Joined
Jun 29, 2009
Messages
9,494
Gender
Undisclosed
HSC
N/A
Complex analysis is a very beautiful course. Most of my friends hated it though.

You can do things like finding a solution to sin z = 2 (can't remember if this particular case has a solution, but it could have a complex solution) using sin z = (e^z - e^(-z) )/(2i) or by letting z=x+iy and using the double angle formulae and hyperbolic formulae.
Oh? I remember practising questions like that during Differential Calculus.

Or maybe because I was just practising random questions out of a general Calculus textbook, not a specific USYD syllabus one.

Indeed it has complex solutions lol.
 

math man

Member
Joined
Sep 19, 2009
Messages
503
Location
Sydney
Gender
Male
HSC
N/A
Complex analysis is a very beautiful course. Most of my friends hated it though.

You can do things like finding a solution to sin z = 2 (can't remember if this particular case has a solution, but it could have a complex solution) using sin z = (e^z - e^(-z) )/(2i) or by letting z=x+iy and using the double angle formulae and hyperbolic formulae.
yeh im doing adv complex analysis next yr..can't wait lol
 

seanieg89

Well-Known Member
Joined
Aug 8, 2006
Messages
2,662
Gender
Male
HSC
2007
Well you may be right for part ii)... root2 z1 may not lie on the original circle, however it does lie on another cirlce....but this is only a simple 4u question so it does not matter if it does or does not lie on the original circle....you just need to prove it lies on a circle, which is easily done.
I did not mean any part in particular. I simply meant that the premise of the question is always false. The four points corresponding to 0, z_2, z_3, z_1+z_2 CANNOT ever be concyclic. (Try to find z_1 for which they are...I can provide geometric justification for my claim if you wish). So the question is logically equivalent to something like:

If 1=0, prove that 69=42. (Silly example to make a point.)
 

seanieg89

Well-Known Member
Joined
Aug 8, 2006
Messages
2,662
Gender
Male
HSC
2007
Complex analysis is a very beautiful course. Most of my friends hated it though.

You can do things like finding a solution to sin z = 2 (can't remember if this particular case has a solution, but it could have a complex solution) using sin z = (e^z - e^(-z) )/(2i) or by letting z=x+iy and using the double angle formulae and hyperbolic formulae.
Complex analysis is indeed magical. If any of you are interested in number theory you should check out how complex analysis is used to prove the prime number theorem (eg the second last chapter of Tom Apostol's Introduction to Analytic Number Theory)...it is pretty mindblowing the first time you see connections between complex numbers and the primes.
 
Last edited:

seanieg89

Well-Known Member
Joined
Aug 8, 2006
Messages
2,662
Gender
Male
HSC
2007
Does this proof have to use methods only learnt in MX2? I have a general idea for a proof, but I think it requires the use of the max/min modulus principle (http://en.wikipedia.org/wiki/Maximum_modulus_principle), but that's meant to be something learnt in 2nd Year. I only know of it because I've been watching lectures and reading up on a bit of Complex Analysis for next semester.

EDIT: Woops, this proof does the whole FTA, not just the step you said.
Yeah generally a complete proof will utilise something relatively deep...such as the Jordan curve theorem from topology or Liouville's theorem from complex analysis. This is the first step in several proofs though I believe.

PS I do recall seeing a relatively elementary proof of FTA somewhere that required little more than the mean value theorem and the definition of partial derivatives :).
 

SpiralFlex

Well-Known Member
Joined
Dec 18, 2010
Messages
6,960
Gender
Female
HSC
N/A
I am sure I was right on the first page when I said it does not geometrically sustain itself. My interpretation is,

Makes a right angle. If the points are concylic, then the angle at must also be 90. However, the angle at is clearly less than 90.
 

math man

Member
Joined
Sep 19, 2009
Messages
503
Location
Sydney
Gender
Male
HSC
N/A
I did not mean any part in particular. I simply meant that the premise of the question is always false. The four points corresponding to 0, z_2, z_3, z_1+z_2 CANNOT ever be concyclic. (Try to find z_1 for which they are...I can provide geometric justification for my claim if you wish). So the question is logically equivalent to something like:

If 1=0, prove that 69=42. (Silly example to make a point.)
i see what you mean..but this question is only designed to test the critical thinking skills of my students
 

math man

Member
Joined
Sep 19, 2009
Messages
503
Location
Sydney
Gender
Male
HSC
N/A
I am sure I was right on the first page when I said it does not geometrically sustain itself. My interpretation is,

Makes a right angle. If the points are concylic, then the angle at must also be 90. However, the angle at is clearly less than 90.
Did you join the line between z3 and z1+z2 ?
 

IamBread

Member
Joined
Oct 24, 2011
Messages
757
Location
UNSW
Gender
Male
HSC
2011
I don't know if I'm making a mistake, but im getting angle z3 z2 (z1 +z2) = 90. Which means angle z3 (z1+z2) z2 can't be 90 and it therefore cannot be cyclic.
 

Carrotsticks

Retired
Joined
Jun 29, 2009
Messages
9,494
Gender
Undisclosed
HSC
N/A
I don't know if I'm making a mistake, but im getting angle z3 z2 (z1 +z2) = 90. Which means angle z3 (z1+z2) z2 can't be 90 and it therefore cannot be cyclic.
I think we established about 20 posts ago that the question is invalid lol.
 

deterministic

Member
Joined
Jul 23, 2010
Messages
423
Gender
Male
HSC
2009
Math man didn't seem convinced yet, no one had really stated a why though.
Or else you have 2 different perpendiculars from z3 to the vector z1 (one at z2+z1 and the other at z2) which is impossible

edit: nevermind, you already said it
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top