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Homework Help: Complex Number, properties of moduli

  1. Jun 27, 2009 #1
    1. The problem statement, all variables and given/known data
    Hello! I'm lost on how to start this, I've got formulas given to me from the text, but I have no idea on how to piece everything together. So I need to use established properties of moduli to show that when [tex]\left.\left|z_{3}\right|\neq\left|z_{4}\right|[/tex],

    2. Relevant equations
    Re is the real component of z where z=(x,y) so x = Re z , y= Im z

    z can be written as:

    Vector properties
    [tex]\left.|z| = \sqrt{x^{2}+y^{2}}[/tex]

    [tex]\left.Re(z) \leq |Re(z)| \leq |z|[/tex]
    [tex]\left.|z_{1}\pm z_{2}|\leq|z_{1}|+|z_{2}|[/tex]
    [tex]\left.|z_{1}\pm z_{2}|\geq||z_{1}|-|z_{2}||[/tex]

    3. The attempt at a solution
    Well what I attempted to do is to say the [tex]\left.Re(z_{1}+z_{2})[/tex] portion can be turned into [tex]\left.x_{1}+x_{2}[/tex].
    I can also argue(given the properties) the left denominator is bigger or equal to the right hand side denominator since [tex]\left.|z_{1}\pm z_{2}|\geq||z_{1}|-|z_{2}||[/tex] but it doesn't really help since the left term is suppose to be smaller than the right hand term.

    I can also say that the term [tex]\left.|z_{1}|+|z_{2}|[/tex] is given by [tex]\left. \sqrt{x^{2}_{1}+y^{2}_{1}}+\sqrt{x^{2}_{2}+y^{2}_{2}}[/tex] which is bigger than just the [tex]\left.x_{1}+x_{2}[/tex] term?

    But these methods are just me trying to prove it going reverse. But I'm just really lost trying to prove it starting with the simple [tex]\left.\left|z_{3}\right|\neq\left|z_{4}\right|[/tex] because the question gives us a result that has 2 addition z terms(where does z1 and z2 come from?)? Please give me some advice on how I can approach this.

    Last edited: Jun 27, 2009
  2. jcsd
  3. Jun 27, 2009 #2
    But you do want the left denominator to bigger than or equal to the right denominator. Remember, when you take reciprocals of both sides, you have to flip the inequality. Thus the last inequality you listed under relevant equations takes care of the denominators. As for the numerators, apply Re(z) <= |z| then the basic triangle inequality (listed before the last inequality under relevant equations).
  4. Jun 27, 2009 #3
    Thanks for the reply! I understand what you mean, but would that be the extent of "showing". I guess sometimes I'm just not sure how to properly write these proof type problems.

    On a somewhat related note, I'm suppose to graph
    [tex]\left.|z+i|\leq3[/tex] and [tex]\left.|z+i|\geq3[/tex]
    I know that normally, if we have something like :
    [tex]\left.|z-1+i|=1[/tex], it means we have a radius of 1, with a center at (1,-i)
    I also know that we can say:
    [tex]\left.|z+i|\leq |z| + i = 3[/tex]
    [tex]\left.|z+i|\geq ||z| - i| = 3[/tex]

    So my question is for, [tex]\left.|z+i|\leq3[/tex]:
    Would it simply be a circle that is shifted down by i with a radius of 3?
  5. Jun 27, 2009 #4


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    |z+i|< 3

    If |z+i|=3 you have a circle of radius three centered at -i. So if |z+i|< 3 then z must be closer to -i than the points on the circle. Does that answer the question?

    For the OP, a good way to write those proofs is to start with

    [tex]Re(z_1 + z_2) | |z_3| - |z_4| | \leq (|z_1| + |z_2|) |z_3 + z_4|) [/tex]

    which holds since you know [itex] Re(z_1 + z_2) \leq |z_1| + |z_2|[/itex] and [itex]| |z_3| - |z_4| _ \leq |z_3 + z_4|[/itex]

    Then divide both sides by the appropriate quantities and note that everything you're dividing by is positive
  6. Jun 27, 2009 #5
    Yes that makes sense but how do I show that graphically? I tried changing it to [tex]\left.|z| + i = 3[/tex] is simply because I thought that would help make graphing it easier? Could I also say it is a circle with the radius of 3-i?

    So should I be transforming these z's into their components such that [tex]\left.|z| = \sqrt{x^{2}+y^{2}}[/tex]? I guess one of my problem is failing to understand where this is going to lead? Will dividing eventually lead me to a point that shows z3 is not equal to z4? I guess my last question is when I divide these quantities, their relationships always seem to be in terms of inequalities, for example:
    [tex]\left.Re(z_{1} + z_{2}) \leq |z_{1} + z_{2}| [/tex]
    [tex]\left.|z_{1} + z_{2}| \leq |z_{1}| + |z_{2}|[/tex]
    so If I were to say divide our expression by [tex]\left. |z_{1}| + |z_{2}|[/tex]
    I haven't done inequalities in years but we can't just simply divide the terms like that right because they're not always equal.

    Thanks for all the help!!!!
  7. Jun 27, 2009 #6


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    Graphically? Draw the circle |z+i| = 3. Shade in all the points on the inside of that circle. What's confusing you about that?

    Don't use the triangle inequality when it's not needed. Keep it simple:

    Fact: [tex] Re(z_1 + z_2) \leq |z_1| + |z_2| [/tex]

    Fact: [tex] ||z_3|-|z_4|| \leq |z_3 + z_4|[/tex]

    So if [tex]A = Re(z_1 + z_2); B = |z_1| + |z_2|; C= ||z_3|-|z_4||; D = |z_3 + z_4|[/tex]

    We have

    [tex] A \leq B[/tex] and [tex] C \leq D[/tex]

    Also A, B, C and D are positive so

    [tex] AC \leq BD[/tex]

    This is a simple property of inequalities

    As [tex]|z_3| =/= |z_4|[/tex] C must be non-zero. Also, it must be [tex] z_3 =/= -z_4[/tex] so D must be non-zero. Hence
    [tex] \frac{A}{D} \leq \frac{B}{C}[/tex]
    by dividing both sides by the positive, non-zero quantity CD. Compare this with the desired result
  8. Jun 27, 2009 #7
    Well what confused me is the inequality. I understood the expression when it has the equal sign, but the problem that was given was [tex] |z+i| \leq 3 [/tex] so I was trying to find a way to change it into simply an equals sign.

    Thanks, I see what you mean now. Do you have any advice on approaching a proof problem? I always have the hardest time with them because in a way it is so vague and it seems like there are so many ways to go that I eventually have no idea what to do next.
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