Sums of complex numbers - proof in Rudin's book I have one question about a proof in Rudin's book:
If $z_1 ..., z_N$ are complex numbers then there is a subset $S$ of 
$ \{1,..., N \}$ for which $|\sum_{k \in S} z_k| \ge \frac{1}{\pi} \sum_{1}^N |z_k|$
Proof:
Write $z_k = |z_k| e^{i \alpha_k}$. 
For $\phi \in [- \pi, \pi] $ let  $\ \ S(\phi):= \{ k | \cos (\alpha_k - \phi)>0 \}$. 
Then $|\sum _{S(\phi)} z_k| = |\sum_{S(\phi) } e^{-i \phi} z_k| \ge Re(\sum_{S(\phi) } e^{-i \phi} z_k) = \sum_1^N |z_k| \cos ^+ (\alpha_k - \phi) $.
Could you tell me why the two equations hold (the inequality is quite obvious) and what $\cos ^+$ stands for?
 A: $\cos^+$ is the positive part of the cosine. Generally, for a real-valued function $f$, we have the positive and negative parts of $f$:
\begin{align}
f^+(x) &= \max \{ 0, f(x)\} = \begin{cases} f(x) &, f(x) \geqslant 0 \\ 0 &, f(x) < 0,\end{cases}\\
f^-(x) &= \max \{ 0, -f(x)\} = \begin{cases} -f(x) &, f(x) < 0 \\ 0 &, f(x) \geqslant 0. \end{cases}
\end{align}
Note that both, the psoitive part and the negative part are non-negative (somewhat confusing at first), and we have $f(x) = f^+(x) - f^-(x)$ as well as $\lvert f(x)\rvert = f^+(x) + f^-(x)$.
We have the first equality
$$\left\lvert \sum_{k\in S(\phi)} z_k\right\rvert = \left\lvert \sum_{k\in S(\phi)} e^{-i\phi}z_k\right\rvert$$
since $\lvert e^{-i\phi}\rvert = 1$, and every term in the sum is multiplied with the same factor, so that means the entire sum is multiplied with $e^{-i\phi}$, which has modulus $1$. Since $\lvert ab\rvert = \lvert a\rvert\cdot \lvert b\rvert$, multiplying with a number of modulus $1$ doesn't change the modulus.
The second equality comes from writing $z_k$ in polar form, so
$$e^{-i\phi}z_k = e^{-i\phi} \lvert z_k\rvert e^{i\alpha_k} = \lvert z_k\rvert e^{i(\alpha_k-\phi)},$$
and taking the real part gives
$$\operatorname{Re} \left(e^{-i\phi}z_k\right) = \lvert z_k\rvert\cos (\alpha_k-\phi).$$
Since for $k\in S(\phi)$ we have $\cos (\alpha_k-\phi) = \cos^+(\alpha_k-\phi)$, and $\cos^+(\alpha_k-\phi) = 0$ for $k\notin S(\phi)$, we have
$$\sum_{k\in S(\phi)} \lvert z_k\rvert \cos (\alpha_k-\phi) = \sum_{k\in S(\phi)} \lvert z_k\rvert \cos^+ (\alpha_k-\phi) = \sum_{k=1}^N  \lvert z_k\rvert \cos^+ (\alpha_k-\phi).$$
