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How can I find arg z (using the CAST system maybe)? Watch

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    I hated the CAST system (still do) and got by fine in A Level Maths without it.

    But now it seems I need it.

    When finding arg z, should the axes be like this:

    + y axis is pi/2, pos x axis is 0, neg y axis is -pi/2 and neg x axis is + or - pi

    ?

    I (think I) know in CAST that you start from the pos x axis, which is 0, and work anticlockwise so the pos y axis is pi/2 and the neg x axis is pi and the neg y axis is 3pi/2 and the pos x axis can be 2pi (or 0 again)...

    Hope that makes sense..

    Basically, how do you use CAST/4 quadrants to calculate the value of arg(z)?

    Thank you!
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    (Original post by PhysicsGal)
    I hated the CAST system (still do) and got by fine in A Level Maths without it.

    But now it seems I need it.

    When finding arg z, should the axes be like this:

    + y axis is pi/2, pos x axis is 0, neg y axis is -pi/2 and neg x axis is + or - pi

    ?

    I (think I) know in CAST that you start from the pos x axis, which is 0, and work anticlockwise so the pos y axis is pi/2 and the neg x axis is pi and the neg y axis is 3pi/2 and the pos x axis can be 2pi (or 0 again)...

    Hope that makes sense..

    Basically, how do you use CAST/4 quadrants to calculate the value of arg(z)?

    Thank you!
    the first quadrant is alpha (e.g. the base value of the argument)
    The second quadrant is pi-alpha
    The third quadrant is -pi+alpha
    The bottom fourth quadrant is -alpha
    where alpha=tan^{-1}(\frac{y}{x})
    At least from FP1 this is what I did

    so e.g. z=2+3i which is clearly in the first quadrant so arg(z)=tan^{-1}(\frac{3}{2})
    Remember when working with the argument, when you place your values in the inverse tangent make them positive, otherwise the calculation will not come out well
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    (Original post by Robbie242)
    the first quadrant is alpha (e.g. the base value of the argument)
    The second quadrant is -pi+alpha
    The third quadrant is pi-alpha
    The bottom fourth quadrant is -alpha
    where alpha=tan^{-1}(\frac{y}{x})
    At least from FP1 this is what I did

    so e.g. z=2+3i which is clearly in the first quadrant so arg(z)=tan^{-1}(\frac{3}{2})
    Remember when working with the argument, when you place your values in the inverse tangent make them positive, otherwise the calculation will not come out well
    First time I am studying this so my questions might be simple/silly.

    Anyway.........

    for z = -3 + 3i:
    This is in the 2nd quadrant so according to your reply I would use -pi+alpha, where alpha is tan^-1(3/-3) = tan^-1(1) (you said to make it positive)
    which gives.....-2.4 (1 d.p. although it should obviously be in pi or surd form)
    But the answer is 3pi/4 which is positive and equivalent to 2.4 (1 d.p.)..

    Where has it gone wrong and changed into a negative?

    Your system is interesting - how did you figure it out/ie: how do I go about remembering it and understanding it?
    Thanks

    Edit:
    I think the 2nd quad. should be pi+alpha (not -pi+alpha) cos that seems to give the correct positive angle

    Thank you for your response btw!
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    (Original post by PhysicsGal)
    First time I am studying this so my questions might be simple/silly.

    Anyway.........

    for z = -3 + 3i:
    This is in the 2nd quadrant so according to your reply I would use -pi+alpha, where alpha is tan^-1(3/-3) = tan^-1(1) (you said to make it positive)
    which gives.....-2.4 (1 d.p. although it should obviously be in pi or surd form)
    But the answer is 3pi/4 which is positive and equivalent to 2.4 (1 d.p.)..

    Where has it gone wrong and changed into a negative?

    Your system is interesting - how did you figure it out/ie: how do I go about remembering it and understanding it?
    Thanks
    Let me give this one a go, I remember it just by remembering it, its quite a simple system but I may have wrote one of them down wrong.
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    (Original post by PhysicsGal)
    First time I am studying this so my questions might be simple/silly.

    Anyway.........

    for z = -3 + 3i:
    This is in the 2nd quadrant so according to your reply I would use -pi+alpha, where alpha is tan^-1(3/-3) = tan^-1(1) (you said to make it positive)
    which gives.....-2.4 (1 d.p. although it should obviously be in pi or surd form)
    But the answer is 3pi/4 which is positive and equivalent to 2.4 (1 d.p.)..

    Where has it gone wrong and changed into a negative?

    Your system is interesting - how did you figure it out/ie: how do I go about remembering it and understanding it?
    Thanks

    Edit:
    I think the 2nd quad. should be pi+alpha (not -pi+alpha) cos that seems to give the correct positive angle

    Thank you for your response btw!
    In radians I did -(-pi+tan^-1(1)) and it came out 3/4pi, it must be as follows then -(-pi+tan^-1(1))=pi-alpha I guess
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    Yeah, just flip the rules for the second and third quadrant.
    If 0≤x≤pi/2, then x less than pi is in the second quadrant and x more than pi is in the third.
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    (Original post by aznkid66)
    Yeah, just flip the rules for the second and third quadrant.
    If 0≤x≤pi/2, then x less than pi is in the second quadrant and x more than pi is in the third.
    So the third quadrant would be -pi+alpha then?
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    Why is this so difficult?

    Start with Z = (a + jb); Mod Z = Sq root(a^2 + b^2); arg Z = tan^-1 (b/a)
    i.e. simply Pythagoras' theorem.

    1) Draw your axis

    2) Convention is a positive (anti-clock) rotation of pi/2 (90deg) corresponds to +j. (i.e. on the cartesian +ve y axis)
    positive rotation of pi corresponds to J*J = j^2 = (-1) (i.e. 2 x90 deg. which places it on the cartesian -ve x axis)
    positive rotation of 3pi/4 corresponds to j*j*j = j^3 = -j (i.e. 3 x 90 deg. placing it on the cartesian -ve y axis)
    positive rotation of 2pi corresponds to j*j*j*j = j^4 = +1 (i.e. 4 x 90 deg which brings us back to the +ve x axis)

    3) Draw a sketch of your vector on the Argand (z) plane

    4) Forget what you calculator says.

    5) Tan(theta) is the slope of the vector i.e. (opposite/adjacent) therefore always y/x (cartesian); always b/a (complex)

    6) Angle theta is then tan^-1 (opposite/adjacent) therefore always tan^-1(y/x); always tan^-1 (b/a)

    7) Now look at your calculator. You can see exactly where the calculator angle is represented on your sketch.


    The key is that j is a rotation which is why the complex axis is called the z-plane because it maps complex numbers onto a cartesian axis and vice-versa.
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    Mhmm.
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    I think drawing a diagram of the 4 quadrants, and then just working the angle out from there is easiest (for me) cos these rules are just confusing and hard to remember!

    I just need to practice more questions and the working out of angles etc. will hopefully become 2nd hand nature.

    Thanks for the tips guys
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    Why not a mixture of both? Draw the diagram, then get the rules from the diagram

    z=a+bi

    If principle Arctan(b/a) is positive (a and b are the same sign):

    Case 1: a and b are positive.
    Case 3: a and b are negative.

    Spoiler:
    Show


    From the diagram, you can see that Case 1 has an angle of theta and Case 3 has an angle of pi+theta.

    If principle Arctan(b/a) is negative (a and b are different signs):

    Case 4: a is positive, b is negative.
    Case 2: a is negative, b is positive.

    Spoiler:
    Show


    From the diagram, you can see that Case 4 is theta and Case 2 is pi+theta.




    But drawing the diagram and working out the angle will be good too. With enough, you'll definitely be able to just draw the reference triangle and find the quadrant/manipulation in your head, if not have them learned by heart! ^_^
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    I never heard of this CAST business (aside from on this forum) and I looked it up and it seems stupid and pointless.

    All you need to remember is that all angles should measured from the positive x axis going anticlockwise and then if you remember the standard periodicity rules of the trig functions - it won't even matter which particular inverses you use of the trig functions since you will know how to adjust to get the correct answer.
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    (Original post by Mark85)
    I never heard of this CAST business (aside from on this forum) and I looked it up and it seems stupid and pointless.
    Sanity at last.
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    (Original post by Mr M)
    Sanity at last.
    I am assuming that this CAST business is part of the school curriculum?

    I think that the main issue when setting these things is the lack of appreciation for the practical difference in difficulty between formalizing something so as to make it foolproof and algorithmic and the unformalized version upon which it is based.

    I mean, if you presented Gaussian elimination for the first time in its purely formal sense without first talking about solving 2x2 systems naively then it would be much harder to pick up and work out what was going on... I mean, I assume most people solve larger linear systems pretty much thinking in the same way (albeit doing the order of substitutions in a more directed manner) and the point is that by understanding what is going on - there is no need to remember the algorithm because you know how it works and could thus derive it if needed.

    In those terms, this CAST thing is much of a muchness but I think it is symptomatic of that wider problem with the thinking behind the curriculum - essentially method over principle.
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    (Original post by Mark85)
    I am assuming that this CAST business is part of the school curriculum?
    Nope. Many people (mistakenly in my opinion) believe it makes the task of finding solutions to trigonometric equations easier. I'd never heard of someone using it for complex numbers before!

    (Original post by TenOfThem)
    ...
    ToT is a fan and will presumably be able to explain the merits of this approach.
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    (Original post by Mr M)
    Nope. Many people (mistakenly in my opinion) believe it makes the task of finding solutions to trigonometric equations easier. I'd never heard of someone using it for complex numbers before!
    Me neither


    ToT is a fan and will presumably be able to explain the merits of this approach.
    As I have said before, I think that there is a tendency for people to use the method they are first taught ... for me the unit circle was the the method I used at school

    I use it because it is visual and so am I
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    (Original post by TenOfThem)
    I use it because it is visual and so am I
    Never forget.
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    I did not mention, nor was I referring to learning styles
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    (Original post by TenOfThem)
    As I have said before, I think that there is a tendency for people to use the method they are first taught ... for me the unit circle was the the method I used at school

    I use it because it is visual and so am I
    With all due respect 'because it is visual' is really just weasal words... it means nothing. It is no more or less 'visual' then talking about measuring angles from the positive x axis anticlockwise.

    There is a very basic principle undermining this 'issue' (clarifying the domains and codomains of the trignometric functions and thus their inverse) and doing anything other than addressing it directly is obfuscating as opposed to clarifying. As you can see from posts on forums like this, the direct result of such teaching style is that people vaguely understand that there is an issue present and have a new buzzword to label it with but ultimately understand nothing.
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    (Original post by Mark85)
    With all due respect 'because it is visual' is really just weasal words... it means nothing. It is no more or less 'visual' then talking about measuring angles from the positive x axis anticlockwise.
    What are you talking about?



    I will be honest I am not particularly happy about being quoted on a thread that I have not posted on

    I do not see any need for me to defend my use of a particular system
 
 
 
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