Failure to see equality in Zagier's proof of the prime number theorem I was looking through Zagier's Newman's Short Proof of the Prime Number Theorem.
I understand the idea of the whole paper and the proof, but I fail to see in the proof the last equality, when it says:
$$ \left| e^{zT} \left(1+\dfrac{z^2}{R^2} \right) \dfrac{1}{z} \right|=e^{\Re (z) T} \cdot \dfrac{2\Re(z)}{R^2} $$
Could someone give me a more elaborated version of this equality?
I can't see how we can take the equality on the semi-circle of radius $R$.
The full paper can be found here .


 A: We have, with $z=Re^{i\theta}$,
\begin{align*}
\left| {e^{zT} \left( {1 + \frac{{z^2 }}{{R^2 }}} \right)\frac{1}{z}} \right| & = \left| {e^{zT} } \right|\left| {1 + \frac{{z^2 }}{{R^2 }}} \right|\frac{1}{{\left| z \right|}} = e^{\Re (z)T} \left| {1 + \frac{{z^2 }}{{R^2 }}} \right|\frac{1}{R} \\ & = e^{\Re (z)T} \left| {1 + \frac{{R^2 e^{2i\theta } }}{{R^2 }}} \right|\frac{1}{R}  = e^{\Re (z)T} \left| {1 + e^{2i\theta } } \right|\frac{1}{R}= e^{\Re (z)T} \frac{{2\left| {\cos \theta } \right|}}{R} \\ &= e^{\Re (z)T} \frac{{2R\left| {\cos \theta } \right|}}{{R^2 }} = e^{\Re (z)T} \frac{{2\left| {\Re (z)} \right|}}{{R^2 }}.
\end{align*}
Now note that on the semi-circle, $\Re(z)\ge 0$, i.e., $|\Re(z)|=\Re(z)$.
A: It suffices to prove that for $z=Re^{i\theta}$ for $|\theta|\le\frac\pi2$ there is
$$
\left|1+{z^2\over R^2}\right|=2\cos\theta.
$$
By definition, we have
$$
\left(1+{z^2\over R^2}\right)={R^2+z^2\over R^2}={R^2(1+\cos2\theta)+iR^2\sin2\theta\over R^2}=(1+\cos\theta)+i\sin\theta,
$$
so when taking absolute values on both side, there is
\begin{aligned}
\left|1+{z^2\over R^2}\right|
&=\sqrt{(1+\cos2\theta)^2+\sin^22\theta}=\sqrt{1+2\cos\theta+\cos^22\theta+\sin^22\theta} \\
&=2\sqrt{(1+\cos2\theta)/2}=2\sqrt{\cos^2\theta}=2\cos\theta.
\end{aligned}
Hence, the derivation is completed.
