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Say we have two complex numbers:

$re^{i\theta}$ and $se^{i\phi}$

Is there a straightforward way to find the polar form of the midpoint of these two complex numbers? I think I'm correct in saying that the argument of the midpoint will be:

$ \frac{\theta+\phi}{2} $

I am asking this question as I had to find the midpoint between two "consecutive" (consecutive as in the sense of adjacent) roots of unity; in this case the moduli of the two complex numbers in question will be the same, but I was wondering if there was any method for the general case. I know we could do it by first converting to Cartesian, but this seems unsatisfactory. Thanks!

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    $\begingroup$ it is not true that argument of the midpoint of two number is the average of their arguments. take for example $1$ and $2i.$ $\endgroup$
    – abel
    May 25 '15 at 20:02
  • $\begingroup$ Where is the problem in trying to find a polar form of $$\frac12 (re^{i\theta} + se^{i\phi})$$ $\endgroup$
    – AlexR
    May 25 '15 at 20:03
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    $\begingroup$ Not true take the two points $z_1=1$ and $z_2=2i$. The argument of the mid point is not $\pi/4$ as claimed. There is no straightforward way to express the midpoint in polar form $\endgroup$
    – marwalix
    May 25 '15 at 20:03
  • $\begingroup$ @marwalix, why did we come up with the same example? $\endgroup$
    – abel
    May 25 '15 at 20:04
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    $\begingroup$ @abel I'd have chosen $-1$ and $2$. The argument of $\frac12$ is none of $\pm\frac\pi2$. $\endgroup$
    – AlexR
    May 25 '15 at 20:04
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IF you have $r=s$, then the midpoint $te^{i\psi}$ has

$$ \psi = \frac{\theta+\phi}{2}$$

but not, in general, otherwise.

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    $\begingroup$ No in general $t\neq r=s$. $z_1=1$ and $z_2=i$ you have $r=s=1$ and $t=\sqrt 2/2$ $\endgroup$
    – marwalix
    May 25 '15 at 20:09
  • $\begingroup$ @marwalix This is what I thought too, this implies that the midpoint lies on a circle along with the other two points. $\endgroup$
    – talfred
    May 25 '15 at 20:10
  • $\begingroup$ @marwalix Ha, true, thanks - corrected, I'm not sure where that mistake came from. $\endgroup$
    – Joffan
    May 25 '15 at 20:10

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