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I am working on a complex number problem set and I have had success so far except for the following, which makes me think that I am missing something rather simple in the initial steps that would clear everything up for me. I have made attempts in cartesian, polar, and euler form. I have even cheated by assuming that which is to be proven, and haven't found anything convenient pop out that I could then reverse engineer from the beginning. I have googled but haven't come up with a good search term to even help me find anything useful. I have attempted proof by contradiction, although the curriculum this is from does not generally use contradiction proofs so I am expecting another method to be useful. This is for an International Baccalaureate Higher Level Mathematics problem set and the question comes from the Cambridge textbook, chapter 15 if that's relevant to anyone.

Here's the question.

$$w = \frac{k \,z}{z^2 + 1} \quad \text{with} \;\;z^2 \ne -1\,, \;\operatorname{Im}(w) = \operatorname{Im}(k) \ne 0\,, \;\operatorname{Im}(z) \ne 0$$

Prove $|z| = 1$.

dxiv
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GreaTeacheRopke
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  • Almosty nobody knows MathJax first time here, but we all can learn. It's hard to understand what you meant. – DonAntonio Jun 05 '16 at 14:18
  • The equation and my use of "not =", I hope, should be pretty straightforward. w, k, and z are all complex numbers. "Im(w)" simply means "the imaginary part of w." |z| means the modulus of the complex number z.

    Does that clear things up?

     – GreaTeacheRopke Jun 05 '16 at 14:45
  • Er...no. Your very first equation, with "w". What is that period there? Is that "z^2" multiplying a fraction, or it is in the denominator...or something else?! What is "not =-1 ? The "w", the "z^2", something else...? – DonAntonio Jun 05 '16 at 15:57
  • Hm. Sorry, the period is the end of a sentence. I see why that could have been confusing. So we have the equation

    [w = (kz)/(z² + 1)]

    and the 3 conditions that: (1) z² is not equal to 1. Without coding I was am not sure how to properly type that. (2) the imaginary part of w = the imaginary part of k, both of them are not equal to 0 (ie both w and k have imaginary components). (3) z also has an imaginary component

     – GreaTeacheRopke Jun 05 '16 at 16:37
  • Well, I think I already understood it yet it looks like a horrible, horrible exercise, and after opening up the conditions one gets a terrible equation with second, fourth and more powers...perhaps there's some slick trick to it, but I see it as hopeless. – DonAntonio Jun 05 '16 at 16:53
  • I have also now considered another approach by flipping things to achieve (z^2 + 1)/z = k/w, rewriting (z^2 + 1)/z = z + 1/z = 2cos(theta); I was pretty hopeful when I found this identity hiding in there but it didn't get me anywhere useful in the end. – GreaTeacheRopke Jun 06 '16 at 08:32
  • The equality $;z+\frac1z=2\cos\theta;$ is true iff $;|z|=1;$ and $;z=e^{i\theta};$ , of course...and this is precisely what you're trying to prove! – DonAntonio Jun 06 '16 at 08:48
  • Another thought that has been brought to my attention, but I think has some holes:

    The equation can be rewritten as wz^2 - kz + w = 0. Letting z = a + bi, and this being a quadratic in z, the other root must be a - bi. Now the product of the roots must be w/w = 1 = (a + bi)(a - bi) =a^2 + b^2. But a^2 + b^2 = |z|.

    Some issues I have: the answer does not use the fact that Im(w) = Im(k), and also I don't know if the product of roots theorem can be used for a complex polynomial. Any thoughts on this?

     – GreaTeacheRopke Jun 09 '16 at 07:52
  • That's false, I'm afraid. Only complex roots of real polynomials come in conjugate pairs. The quadratic in $;z;$ you mention is not real as the coefficients aren't. – DonAntonio Jun 09 '16 at 14:19

1 Answers1

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The proposition does not hold true.

Counterexample: $w=1+i\,$, $\;k=2+i$ satisfy the condition $\,\operatorname{Im}(w) = \operatorname{Im}(k) = 1 \ne 0\,$, but the roots of the equation are $\,z_1 = \cfrac{1+i}{2}\,$ and $\,z_2=1-i\,$ neither of which has modulus $1$.

dxiv
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