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Help with complex loci on Argand diagram

  1. Sep 27, 2014 #1
    Hi all, I'm hoping for some help/guidance on this problem:
    1. The problem statement, all variables and given/known data

    The complex number z is represented by the point P on the Argand diagram. Sketch and describe in words the locus of P if:

    i) ##|2-iz|=2##
    ii) ##|\dfrac{1+i}{z}-2|=2##

    The solutions read:
    i) ##|2-iz|=2 \Leftrightarrow |z+2i|=2##
    ii) ##|\frac{1+i}{z} -2| = 2 \Leftrightarrow |z - \frac{1+i}{2}| = |z|##

    In (i), apparently what they've done is multiplied the expression ##2-iz## within the modulus by ##i##. But it doesn't explain why that is valid...
    With (ii), I have no idea how they did that...
    Some help here would be appreciated - I haven't seen any formal information given on this in the book, and I can't see how the result is implied :(

    2. Relevant equations

    ##|z-a|## represents the scalar/real distance between two complex numbers ##z## and ##a## on the Argand diagram.

    3. The attempt at a solution

    (i) - Multiplication of expression inside modulus by ##i## yields answer, but I don't have the proof for why this can be done.
    (ii) - Not really sure how to start here.

    Thanks!
    Stephen :)
     
  2. jcsd
  3. Sep 27, 2014 #2

    NascentOxygen

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    Staff: Mentor

    Multiplying a complex number by i amounts to a rotation of +90 degrees on the Argand plane, but doesn't change the vector's length (modulus). Plot 5+i2 and -2+i5 to illustrate this.

    So perhaps in the interest of simplifying the expression, multiplying by i inside a modulus is a useful technique?
     
  4. Sep 27, 2014 #3

    Ray Vickson

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    Science Advisor
    Homework Helper

    In (i) they used two facts:
    (1) ##|z_1 z_2| = |z_1| \; |z_2|## for any two complex numbers ##z_1, z_2## (applied to ##z_1 = i##, ##z_2 = 2-iz##).
    (2) ## i \times -i = +1##.
     
  5. Sep 29, 2014 #4
    Thanks for the help :) I have figured it out. Thanks again :D Just proved it myself that if ##|z_1| = |z_2|##, then ##|kz_1|=|kz_2|, k\in C \cup R##
     
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