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Show that if $z_0$ is a solution to $(2z-1)^{2014}=(2z+1)^{2014}$, then $\Re(z_0)=0$.

My attempt:

$(2z-1)^{2014}=(2z+1)^{2014}\\ \implies \left(\dfrac{2z-1}{2z+1}\right)^{2014}=1=e^{2k\pi i}, k=0\space \ldots \space 2013$

Let $\omega:=e^{2k\pi i}$

Then $\dfrac{2z-1}{2z+1}=\omega\\ \implies 2z-1=\omega(2z+1)\\ \implies 2z-1=2z\omega+\omega\\ \implies 2z-2z\omega=\omega +1\\ \implies z(2-2\omega)=\omega +1\\ \implies z=\dfrac{\omega +1}{2-2\omega}$

is what I thought would be right, but upon further inspection I noticed that $2-2\omega=0$, so that's not going to work.

Any suggestions as to finding $z$?

Sujaan Kunalan
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    $\omega$ would need to be a $2014$th root of unity, not $e^{2k\pi i}$, which is just $1$. – 2'5 9'2 Feb 26 '14 at 21:56
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    So should it be $\omega:=\operatorname{exp}\left(\dfrac{2k\pi i}{2014}\right)$? – Sujaan Kunalan Feb 26 '14 at 22:03
  • Yes, that also solves your problem with $2-2\omega$... – Thomas Andrews Feb 26 '14 at 22:10
  • Have you studied Möbius transformations? The equation $z=\frac{\omega +1}{2-2\omega}$ is a Möbius transformation, so it has some nice properties. In particular, since it sends $\omega = 1$ to infinity, it sends any circle through $1$ to a line. So you only have to verify the theorem for two points on the unit circle... – Thomas Andrews Feb 26 '14 at 22:14
  • In particular, Möbius transforms are conformal, i.e. they conserve angles. All you need is the angle of $z_0$ to the real axis.. – J.R. Feb 26 '14 at 22:17
  • No we haven't covered Mobius transformatinons yet. It sounds interesting though; I'll look them up. – Sujaan Kunalan Feb 26 '14 at 22:18

2 Answers2

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Since $|2z-1 | = |2z+1|$, this tells us that $2z$ lies on the perpendicular bisector of the line from $-1$ to $1$, IE it lies on the line with real part 0.

Calvin Lin
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If $(2z-1)^{2014}=(2z+1)^{2014}$, then the two sides have the same absolute value, from which we can take a $2014$th root, and $$ \begin{align} \left|2z-1\right|&=\left|2z+1\right|\\ \implies(2z-1)(2\bar{z}-1)&=(2z+1)(2\bar{z}+1)\\ 4z\bar{z}-2z-2\bar{z}+1&=4z\bar{z}+2z+2\bar{z}+1\\ -2\left(z+\bar{z}\right)&=2\left(z+\bar{z}\right)\\ -4\Re z&=4\Re z\\ \implies\Re z&=0 \end{align}$$

2'5 9'2
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