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Homework Help: Complex numbers: Understanding solutions to tough problems

  1. Oct 17, 2011 #1
    Following are problems from the book "Complex Numbers from A to ...Z" by Titu Andreescu and Dorin Andrica. It's a wonderful book, I'm still adapting to the higher-than-usual level though. My questions/comments are written in bold throughout the problems and solutions.

    Problem 1:
    Prove that for any complex number z,
    |1+z| < \frac{1}{\sqrt{2}} \text{ or } |z^2 + 1| \geq 1.
    Does the "or" in there mean that both of those conditions must be met?

    Suppose by way of contradiction that
    |1+z|<\frac{1}{\sqrt{2}} \text{ and } |1+z^2|<1.
    Setting [itex]z=a+bi[/itex], with [itex]a,b\in \mathbb{R}[/itex] yields [itex]z^{2}=a^{2}-b^{2}+2abi[/itex]. This step I understand.

    We obtain
    (1+a^{2}-b^{2})^{2} + 4a^{2}b^{2}<1 \text{ and } (1+a)^{2}+b^{2}<\frac{1}{2},
    Did this come from just inserting [itex]z^{2}[/itex]? I don't get this when inserting.
    and consequently
    (a^{2}+b^{2})^{2}+2(a^{2}-b^{2})<0 \text{ and } 2(a^{2}+b^{2})+4a+1<0.
    This must just be further reduction, but I need the previous step to understand this one.
    Summing these inequalities implies
    which is a contradiction.

    Problem 2:
    Prove that
    \sqrt{\frac{7}{2}} \leq |1+z| + |1-z+z^{2}| \leq 3\sqrt{\frac{7}{6}}
    for all complex numbers with [itex]|z|=1[/itex].

    Let [itex]t=|1+z|\in [0,2].[/itex] We have
    t^{2}=(1+z)(1+\bar{z})=2+2Re(z), \text{ so } Re(z)=\frac{t^{2}-2}{2}.
    I know that [itex]Re(z)=\frac{z+\bar{z}}{2}[/itex], but isn't the equation missing [itex]z\cdot \bar{z}[/itex]?
    Then [itex]|1-z+z^{2}|=\sqrt{|7-2t^{2}|}[/itex].How did this happen? Inserting what I know about [itex]t[/itex] in [itex]|1-z+z^{2}|[/itex] doesn't get me anywhere.
    It suffices to find the extreme values of the function
    f:[0,2]\rightarrow \mathbb{R}, \, \, \, \, f(t)=t+\sqrt{|7-2t^{2}|}.
    No problemo.
    We obtain
    f\left(\sqrt{\frac{7}{2}}\right)=\sqrt{\frac{7}{2}} \leq t+\sqrt{|7-2t^{2}|} \leq f\left(\sqrt{\frac{7}{6}}\right)=3\sqrt{\frac{7}{6}}
    I understand this as well.

    Any ideas?
  2. jcsd
  3. Oct 18, 2011 #2


    Staff: Mentor

    No - "or" means that one or the other (or possibly both) must be true.
    No. The opposite of the statement with "or" in it is
    |1+z| \geq \frac{1}{\sqrt{2}} \text{ and } |z^2 + 1| < 1.
    It's getting late, so I'm going to turn in, so maybe others will have additional comments on your work.
  4. Oct 18, 2011 #3


    User Avatar
    Science Advisor

    To this point, what Mark44 said: a contradiction would be
    [tex]|1+ z^2|< 1\text{ and } |1+ z|\ge \frac{1}{\sqrt{2}}[/tex]

    No, it's not "just inserting [itex]z^2[/itex]", it is using the definition of "absolute" value of a complex number: [itex]|z|= \sqrt{z\bar{z}}[/itex] so with z= a+ bi, [itex]|a+ bi|= \sqrt{a^2+ b^2}[/itex].

    With z= a+ bi, 1+ z= (a+ 1)+ bi so the absolute value is [itex]\sqrt{(a+1)^2+ b^2}\ge \frac{1}{\sqrt{2}}[/itex]. Squaring both side of the inequality [itex]|1+ z|\ge \frac{1}{\sqrt{2}}[/itex] (which is legal because both sides are positive) we have
    [tex](a+1)^2+ b^2\ge \frac{1}{2}[/tex]

    With z= a+ bi, [itex]z^2= a^2-b^2+ 2abi[/itex], [itex]1+ z^2= a^2- b^2+ 1+ 2abi[/itex] and [itex]|1+ z^2|^2= (1+ a^2- b^2)^2+ 4a^2b^3< 1[/itex].

    The hypothesis is that |z|= 1 so that [itex]z\cdot\bar{z}= 1[/itex]. In particular, with [itex]t^2= |1+ z|^2[/itex], again, taking z= a+ bi, [itex]t^2= (a+1)^2+ b^2= a^2+ 2a+ 1+ b^2= 2a+ 2= 2Re(z)+ 2[/itex] because [itex]|z|^2= a^2+ b^2= 1[/itex].

    Last edited by a moderator: Oct 18, 2011
  5. Oct 18, 2011 #4
    Ah! That makes so much more sense now. Looking at the problem with z=a+bi made much more sense. What happened when you went from [itex]1+z^{2}[/itex] to [itex]|1+z^{2}|^{2}[/itex]? How did [itex]a^{2}-b^{2}+1+2abi[/itex] become [itex](1+a^{2}-b^{2})^{2}+4a^{2}b^{3}[/itex]?

    What about the last part when the function showed up?
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