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Prove or disprove that

$$\left|\frac{e^{2i\theta} -2e^{i\theta} - 1}{e^{2i\theta} + 2e^{i\theta} -1}\right| = 1$$

This is a step in an attempt to solve a much larger problem, thus I'm fairly sure it's true but not absolutely sure. It looks like it should be simple but it's resisted all my attempts so far.

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If we may know, what is the "larger problem"? –  000 Apr 11 '12 at 19:37
I don't see why the larger problem matters, the only reason I mentioned it was to give the reason why I wasn't sure if it was true or not. –  Thoth Apr 11 '12 at 19:48
@Thoth, it matters because my intellectual curiosity matters to me. That is why I asked. Also, I believe it may have piquanted the curiosity of others. –  000 Apr 11 '12 at 19:49
I'm trying to find a conformal map from the slit open unit disk to the open unit disk which takes boundary to boundary. I have my conformal map but I'm not sure if conformal maps always take boundary to boundary, thus I was trying to prove it did for my particular conformal map, which in this case is $-\frac{z-i2\sqrt{r}e^{i\frac{\theta}{2}} - 1}{z+i2\sqrt{r}e^{i\frac{\theta}{2}} - 1}$ This conformal map wont give exactly the case I'm looking at above, I tweaked it a bit, but hopefully you get the idea. –  Thoth Apr 11 '12 at 19:55

3 Answers 3

up vote 11 down vote accepted

This is true.

$$|z - \frac{1}{z} -2 | = |z - \frac{1}{z} + 2|$$

where $z = e^{i\theta}$, since $\Re(z - \frac{1}{z}) = 0$.

Geometrically, $z - \frac{1}{z}$ lies on the $y$-axis (perpendicular bisector of $(2,0)$ and $(-2,0)$).

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Ah nicely done, thank you. –  Thoth Apr 11 '12 at 20:10
@Thoth: You are welcome! –  Aryabhata Apr 11 '12 at 20:11

Divide by $e^{i\theta}$ the numerator and denominator : $$\left|\frac{e^{2i\theta} -2e^{i\theta} - 1}{e^{2i\theta} + 2e^{i\theta} -1}\right|=\left|\frac{e^{i\theta} -2 - e^{-i\theta}}{e^{i\theta} +2 - e^{-i\theta}}\right|$$ Think at the complex conjuguate of the numerator and conclude!

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Seems like you were getting at the same thing as Aryabhata, +1. –  Thoth Apr 11 '12 at 20:11
@Thoth: yes. I hope that both helped, fine continuation, –  Raymond Manzoni Apr 11 '12 at 20:14

Taking the squared norm of the numerator and denominator separately, $$ \eqalign{ \left|e^{ 2i\theta}\pm e^{ i\theta}-1\right|^2 &= \left(e^{ 2i\theta}\pm e^{ i\theta}-1\right)\cdot \left(e^{-2i\theta}\pm e^{-i\theta}-1\right)\\ & \matrix{=& 1 & \pm2e^{ i\theta} & -e^{2i\theta} \\\\ & \pm2e^{-i\theta} & +4 & \mp2e^{ i\theta} \\\\ & -e^{2i\theta} & \mp2e^{-i\theta} & +1 } \\\\ &= 6 - 2\cos 2\theta \pm 2\cos\theta \mp 2\cos\theta \\\\ &= 6 - 2\cos 2\theta\,. } $$ Notice, however, that this no longer depends on the sign, i.e. it is the same for the numerator and denominator.

But I admit, I like @Raymond's and @Aryabhata's answers much better!

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