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Using an Argand diagram, find the solution satisfying these pairs of equations

  1. May 27, 2012 #1
    1. Question

    Note: arg ( ) refers to the argument of the complex number, i.e. the angle. | | refers to the modulus, the length.

    Since I don't know how to show an Argand diagram here, I will describe my solution algebraically.


    1. (a)

    arg(z - 3) = -3∏/4, arg(z + 3) = -∏/2

    1. (b)

    arg(z - 4i) = ∏, |z+6| = 5

    2. Attempt

    1. (a)

    arg(z+3) creates an angle of 90° clockwise, so z must be complex number of the form z = -a-bi (via the parallelogram rule).

    Since z + 3 lies on the y-axis, -a = -3

    so z = -3 +(-b)i

    Let p be a complex number where p = 0+ci, c<0. P is passed through by the vector (z-3).

    arg(p-3) = -(∏-3∏/4) = -∏/4

    tan (-∏/4) = p/3
    p = -3

    Now, since z is a complex number such that z = -3 - bi, p must be the midpoint of of (z-3).

    So z = -3+(-3*2)i

    ∴ z = -3 - 6i


    1. (b)

    z + 6 creates a straight line, so

    arg(z + 6) = 0

    z+6 = 5(cos 0 + i sin 0) = 5
    z = -1

    Correct answer: -3 + 4i, -9 + 4i


    3. My Issues

    Is the working correct? My answer for 1.(a) is the same as the book, but I'm not sure whether the solution is "formal" enough.

    For 1.(b), I got stuck and don't know how to solve the rest of it.

    I am having problems geometrically and algebraically interpreting pairs of equations like arg(z+3)=a and arg(z-6)=b, where I have to find the points satisfying the solutions.

    Does anyone know where I can find videos for this type of question? Most of the videos I've found merely expounds on the very basics, where z is given.

    Thanks!
     
    Last edited: May 27, 2012
  2. jcsd
  3. May 29, 2012 #2

    Curious3141

    User Avatar
    Homework Helper

    Hi, Alshia, looks like your post has been left unanswered too long. Sorry about that! :smile:

    Not quite following you here. No sense in stating z = -a-bi, unless you *define* a and b to have certain properties, e.g. that both a and b are positive real numbers.

    What you can state with certainty is that (z+3) is purely imaginary, and has the form [itex]bi[/itex], where b is a negative real number.

    Hence z = -3 + bi.

    Essentially, you're saying the same thing, just that you assigned b to be positive (but did not state it, which is important).

    Honestly, I'm finding it difficult to follow the logic here. I would just have done this:

    z-3 = -6 + bi (using my definition of b as a negative number).

    If arg(z-3) = -3π/4, then it lies in the 3rd quadrant. Also, the vector of (z-3) makes an angle of ∏/4 (45 degrees) with the negative real axis. Therefore, (z-3) has real and imaginary parts that are equal in magnitude. Hence, the imaginary part of (z-3) = -6.

    So z - 3 = -6 -6i

    and z = -3 -6i.

    Do you find that easy to follow?

    No. |z| = r defines a locus of points representing complex numbers with a a modulus of r.

    What shape is formed by points that lie at some fixed distance from another point?

    Once you answer this, we'll take it from there.
     
  4. Jun 2, 2012 #3
    Regarding question 1(a), thanks. I understand it now. The logic is good. I'm wondering though, can you show me the proof that within an isoceles triangle, there are two equal angles? Doesn't matter if it is a right-hand or not, the results generalize anyway.

    Also, is it possible to draw z-3 without changing its form into z+(-3)? Since z = -3+bi, the vector (-3+0i) will be right above it. Hence drawing z-3 will create a right-angled triangle, where arg(z-3)=∏/2.

    Am I misinterpreting this?


    As for 1(b), it forms a circle with radius 5 and centre (0,0). Am I correct?
     
  5. Jun 2, 2012 #4
    For 1b, the first equation can tell you the direction of z-4i. Then just draw out the Argand diagram and use some trigonometry to find the direction of z. Then with the second equation , you can easily figure out z.
     
  6. Jun 2, 2012 #5
    Isoceles triangles are defined to have two equal sides, and hence, two equal angles.

    arg(z-3) = -3∏/4 is already stated in the problem. Changing that would make it wrong.

    It does form a circle with radius 5. But the center is not at (0,0), but instead at (-6,0). Now assume z = a + ib, and compare it to the first equation arg(z - 4i) = ∏. What does this tell you about b?
     
  7. Jun 3, 2012 #6
    @dimension10:

    Thanks for the tip! That reminds me about how to solve this using polar coordinates:

    z + 4i = r(cos ∏ + i sin ∏) = -r

    z = -r + 4i

    Since r > 0, a < 0.

    |z + 6| = |-r + 4i + 6|
    5 = √[(6-r)^2 + 4^2]
    25 = r^2 - 12r + 52
    0 = r^2 - 12r + 27

    Using quadratic formula or butterfly method,

    r = 3 or r = 9

    Since a = -r,

    a = -3 or a = -9


    ∴ z = -3 + 4i OR z = -9 + 4i


    @Infinitum:

    It does form a circle with radius 5. But the center is not at (0,0), but instead at (-6,0).

    Why is the centre (-6,0)? If we define z as vector OA, 6 as vector OB and the destination of z + 6 as the point C, then

    OA + OB = OC

    ...So z+6 is the position vector of point C (Note: O is the origin). Hence, we draw a circle with the origin being the centre.


    Now assume z = a + ib, and compare it to the first equation arg(z - 4i) = ∏. What does this tell you about b?

    Any vector which creates an angle of ∏ radians lies on the x-axis, therefore y = Im = 0.

    Equating imaginary parts,

    b-4 = 0, so b = 4.


    Isoceles triangles are defined to have two equal sides, and hence, two equal angles.

    So how does having two equal sides ensure that there will be two equal angles as well?


    arg(z-3) = -3∏/4 is already stated in the problem. Changing that would make it wrong.

    In hindsight, turns out drawing the vector z-3 as z+(-3) is correct. Anyway, not a big deal here.
     
  8. Jun 4, 2012 #7
    Ohh, you are talking about the vector z+6, I was talking about only the vector z, as this is what you are trying to find :wink:

    Yep! :approve:

    So you now know that z = a+4i, and |(a+6) + 4i| = 5. Can you solve for a?

    This is quite a basic result, which you should be able to prove :smile: Try taking a triangle with two equal sides, draw out a median, and prove that the two smaller triangles formed are congruent. Hence, their respective angles will be equal too.
     
  9. Jun 7, 2012 #8
    "Ohh, you are talking about the vector z+6, I was talking about only the vector z, as this is what you are trying to find ;)"

    Yeah I admit I made an ambiguous statement there, though it was implied I was talking about z+6 because the radius I stated was 5. Anyway, not a big deal here.

    Yes I can find the solutions now, problem solved.


    This is quite a basic result, which you should be able to prove :). Try taking a triangle with two equal sides, draw out a median, and prove that the two smaller triangles formed are congruent. Hence, their respective angles will be equal too.

    Apparently, from where I'm at, nobody in high school taught me the formal methods of proving mathematical equivalences. Heck, nobody explained Phytagorean Theorem's proof to me. I wasn't even familiar with terms like 'median' and 'congruent' until I started reading some geometry online.

    But anyway, back to the topic. Yes, I can see how to prove it, although it may not be a 'formal' proof.

    Let an isoceles triangle PQR have two sides with length a and one side with length b. Drawing the median (which is the perpendicular bisector of b) which connects P to S, where S is the midpoint of QR, we find that:

    Area PQS = 0.5 x h x 0.5b, and
    Area PRS = 0.5 x h x 0.5b

    Hence Area PQS and Area PRS are equal. If two triangles have equal area, the lengths of their sides and the angles within the triangles must be equal. Hence an isoceles triangle has two equal angles.


    ...My statements are ambiguous >.<.

    EDIT:

    Made a mistake. If two triangles have equal area AND lengths, then their angles must correspond to each other.

    My previous statement is false because there are many length combinations that form the same area. For example:

    0.5 x 3 x 4 = 6
    0.5 x 12 x 1 = 6
    0.5 x 6 x 2 = 6
     
    Last edited: Jun 7, 2012
  10. Jun 7, 2012 #9
    I faced the same situation. Schools here don't teach much, so you're left with books and internet to learn stuff, and somehow, I like it because it gives me freedom to do what I want to do :biggrin:

    Reminds me of Mark Twain's quote : 'Don't let schooling interfere with your education.'

    Yep! Area isn't a sufficient condition for congruence, as you note after the edit. :approve:

    There are actually postulates which show in which cases the triangle would be congruent. The one you used was the SSS postulate(each side matches each other in length)

    Here : http://www.mathopenref.com/congruenttriangles.html
     
  11. Jun 9, 2012 #10
    Only if you were studying from the start. If you start studying when exam is near...that would be stressful. Still, I do think we are in charge of our own education. Wish I realized that earlier.


    Interesting! By the way, is a postulate equivalent to an axiom?
     
  12. Jun 9, 2012 #11
    Ahh, but studying at the last minute isn't the mark of a dedicated student. If you really love to learn stuff, you probably wouldn't do it that way :wink:


    Yes, axioms and postulates are often used interchangeably, especially in Euclidean geometry.
     
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