Turn on thread page Beta
    • Community Assistant
    • Thread Starter
    Offline

    17
    ReputationRep:
    Community Assistant
    Suppose we had an equation arg(z-3/z)=pi/4.

    The textbook and examsolutions require sketching and stuff but I'm terrible sketching (which is why I take the long method in solving mod inequalities by playing around with the definition and using a very long method) but for this method, I encountered a new method that is easier (For me) but I'm not too sure as whether or not this would get the full marks as I'm not sure if it'd be on the alternative solutions.

    arg(z-3 / z) = pi/4 --> arg(z-3) - arg = pi/4 --> arg[(x-3)+iy] - arg(x+iy) = pi/4

    Taking tans of both sides and using tan(A-B) = tan(A)-tan(B) / 1+tan(A)(B) yields y/x-3 - y/x / 1 + y/x-3 . y/x = 1 --> x^2+y^2-3x-3y = 0 --> (x-3/2)^2 + (y-3/2)^2 = 9/2 e.g a circle radius 3/root 2 and centre (3/2,3/2) but z-3/z needs to have both real and imaginary parts positive to have an argument of pi/4 hence z-3/z = (x-3)+iy / x+iy --> x(x-3)+y^2 > 0 e.g. (x-3/2)^2 + y^2 > 9/4.

    The final solution to the question is the locus is the part of the circle (3/2,3/2) radius 3/root 2 which lies outside the other circle centre (3/2,0) and radius 3/2 (this can help sketching it)

    Does anybody have a link to learn a much better, quicker method than the textbook explanation an this?
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by thekidwhogames)
    Suppose we had an equation arg(z-3/z)=pi/4.

    The textbook and examsolutions require sketching and stuff but I'm terrible sketching (which is why I take the long method in solving mod inequalities by playing around with the definition and using a very long method) but for this method, I encountered a new method that is easier (For me) but I'm not too sure as whether or not this would get the full marks as I'm not sure if it'd be on the alternative solutions.

    arg(z-3 / z) = pi/4 --> arg(z-3) - arg = pi/4 --> arg[(x-3)+iy] - arg(x+iy) = pi/4

    Taking tans of both sides and using tan(A-B) = tan(A)-tan(B) / 1+tan(A)(B) yields y/x-3 - y/x / 1 + y/x-3 . y/x = 1 --> x^2+y^2-3x-3y = 0 --> (x-3/2)^2 + (y-3/2)^2 = 9/2 e.g a circle radius 3/root 2 and centre (3/2,3/2) but z-3/z needs to have both real and imaginary parts positive to have an argument of pi/4 hence z-3/z = (x-3)+iy / x+iy --> x(x-3)+y^2 > 0 e.g. (x-3/2)^2 + y^2 > 9/4.

    The final solution to the question is the locus is the part of the circle (3/2,3/2) radius 3/root 2 which lies outside the other circle centre (3/2,0) and radius 3/2 (this can help sketching it)
    Seems like a fine approach. What approach does the book suggest?

    To make your approach the best it can be:
    You can skip the tan compound angle by just some quick and accurate manipulation to get \dfrac{z-3}{z} into the form A+iB

    z=x+iy so \omega = \dfrac{z-3}{z} = 1 - \dfrac{3(x-iy)}{x^2+y^2} = \left( 1-\dfrac{3x}{x^2+y^2} \right) + i \left( \dfrac{3y}{x^2+y^2} \right)

    Argument of this is \frac{\pi}{4} which implies \dfrac{\frac{3y}{x^2+y^2}}{1-\frac{3x}{x^2+y^2}} = 1 hence 3y = x^2+y^2 - 3x \Rightarrow (x-\frac{3}{2})^2 + (y-\frac{3}{2})^2 = \frac{9}{2}.

    We know that our z values lie on this circle.
    Then just sketch that on the complex plane. And as you say, we require \Re (\omega) > 0 and \Im (\omega) > 0. So, 1-\dfrac{3x}{x^2+y^2} > 0 hence x^2+y^2 > 3x \Rightarrow (x-\frac{3}{2})^2 + y^2 > \frac{9}{4}, as well as \dfrac{3y}{x^2+y^2} > 0 so y>0

    These conditions yield the loci, which are marked orange here:
    Spoiler:
    Show








    • Community Assistant
    • Thread Starter
    Offline

    17
    ReputationRep:
    Community Assistant
    (Original post by RDKGames)
    Seems like a fine approach. What approach does the book suggest?

    To make your approach the best it can be:
    You can skip the tan compound angle by just some quick and accurate manipulation to get \dfrac{z-3}{z} into the form A+iB

    z=x+iy so \omega = \dfrac{z-3}{z} = 1 - \dfrac{3(x-iy)}{x^2+y^2} = \left( 1-\dfrac{3x}{x^2+y^2} \right) + i \left( \dfrac{3y}{x^2+y^2} \right)

    Argument of this is \frac{\pi}{4} which implies \dfrac{\frac{3y}{x^2+y^2}}{1-\frac{3x}{x^2+y^2}} = 1 hence 3y = x^2+y^2 - 3x \Rightarrow (x-\frac{3}{2})^2 + (y-\frac{3}{2})^2 = \frac{9}{2}.

    We know that our z values lie on this circle.
    Then just sketch that on the complex plane. And as you say, we require \Re (\omega) > 0 and \Im (\omega) > 0. So, 1-\dfrac{3x}{x^2+y^2} > 0 hence x^2+y^2 > 3x \Rightarrow (x-\frac{3}{2})^2 + y^2 > \frac{9}{4}, as well as \dfrac{3y}{x^2+y^2} > 0 so y>0

    These conditions yield the loci, which are marked orange here:
    Spoiler:
    Show












    I think the book does it by considering half lines and playing around with circle theorems to get the loci and equation. Thanks for the tips! Would you recommend using this method in FP2 problems for 4/5 marks because it's easy to do but longer?
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by thekidwhogames)
    I think the book does it by considering half lines and playing around with circle theorems to get the loci and equation. Thanks for the tips! Would you recommend using this method in FP2 problems for 4/5 marks because it's easy to do but longer?
    If you can confidently carry it out with no mistakes, sure.

    Though it's useful to get to know the half-line + circle theorem approaches.
    • Community Assistant
    • Thread Starter
    Offline

    17
    ReputationRep:
    Community Assistant
    (Original post by RDKGames)
    If you can confidently carry it out with no mistakes, sure.

    Though it's useful to get to know the half-line + circle theorem approaches.
    Alright thanks for the help!
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: March 3, 2018

University open days

  1. University of Bradford
    University-wide Postgraduate
    Wed, 25 Jul '18
  2. University of Buckingham
    Psychology Taster Tutorial Undergraduate
    Wed, 25 Jul '18
  3. Bournemouth University
    Clearing Campus Visit Undergraduate
    Wed, 1 Aug '18
Poll
How are you feeling in the run-up to Results Day 2018?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.