Complex Logarithms
1. Logarithms Defined
Lemma
For any complex number \(z\), let
\[ e^z := \sum_{k=0}^\infty \frac{z^k}{k!}\]
This series converges for all \(z \in {\mathbb C}\) and \(e^{z+w} = e^z e^w\) for all complex numbers \(z\) and \(w\). As a consequence,
\[ e^{x + iy} = e^x \cos y + i e^x \sin y \]
which is infinitely differentiable as a function of \((x,y)\).
Proof
Lemma
Fix any positive \(\rho \in (0,1)\). Then there exists a constant \(C_\rho\) such that for every \(z \in {\mathbb C}\) with \(|z| \leq \rho\), there exists \(w \in {\mathbb C}\) such that
\[ \frac{1}{1-z} = e^{w}\]
with \(|w-z| \leq C_\rho |z|^2\). Moreover \(w\) is a continuous function of \(z\).
Proof
For any \(z \in {\mathbb C}\),
\[{}e^{z} (1 - z){}\]
\[{}= 1 + \sum_{k=1}^\infty \left( \frac{1}{k!} - \frac{1}{(k-1)!} \right) z^k{}\]
\[{}= 1 - \sum_{k=2}^\infty \frac{(k-1) z^k }{k!}.{}\]
Let \(z_0 := z\), and for each \(\ell \geq 1\), let
\[ z_{\ell} := 1 - e^{z_{\ell-1}} (1 - z_{\ell-1}). \]
It follows that
\[ e^{z_{\ell-1}} (1 - z_{\ell-1}) = 1 - z_{\ell}, \]
and by successively substituting these formulas into each other,
\[ e^{z + z_1 + \cdots + z_{\ell-1}} (1-z) = 1 - z_{\ell}\]
for each \(\ell \geq 1\).
By examining the power series, we by virtue of the fact that \(|z| \leq \rho < 1\) that
\[{}|z_{\ell}| \leq \sum_{k=2}^\infty \frac{(k-1)|z_{\ell-1}|^k}{k!}{}\]
\[{}\leq |z_{\ell-1}|^2 \sum_{k=2}^\infty \frac{(k-1)}{k!}{}\]
\[{}= |z_{\ell-1}|^2 \left[ 1 - e^1(1-1) \right]{}\]
\[{}= |z_{\ell-1}|^2.{}\]
Thus \(|z_{\ell}| \leq |z|^{2^{\ell}}\) for each \(\ell \geq 1\). This means that \(|z_{\ell}|\) converges rapidly to zero and the series
\[ z + z_1 + \cdots + z_\ell\]
also converges; if \(w\) is defined to be the limit of this series, then
\[{}|z - w|{}\]
\[{}\leq |z|^2 + |z|^4 + \cdots{}\]
\[{}\leq |z|^2(1 + \rho^2 + \rho^4 + \cdots){}\]
\[{}\leq \frac{|z|^2}{1-\rho^2}.{}\]
Because the exponential function is continuous as a function of the real and imaginary parts of its argument, it follows by taking the limit of the formula
\[ e^{z + z_1 + \cdots + z_{\ell-1}} (1-z) = 1 - z_{\ell}\]
that \(e^{w} (1 - z) = 1\), meaning that \(e^w = 1/(1-z)\).
Continuity of \(w\) follows from the Weierstrass \(M\)-test because each \(z_\ell\) is by itself continuous as a function of \(z\) and \(\sup_{|z| \leq \rho} |z_\ell| \leq \rho^{2^\ell}\).
2. Passing from Products to Sums
Theorem
Suppose that \(L(1,\chi) \neq 0\) for a nontrivial character \(\chi\). Then
\[ \sum_{p \text{ prime}} \frac{\chi(p)}{p^s}\]
is bounded in the limit \(s \rightarrow 1^+\).
The proof is given below in three steps.
Step 1 (Logarithms)
Applying the lemma to \(z = \chi(p)p^{-s}\), which for \(s > 1\) and \(p\) prime has magnitude at most \(1/2\) gives that
\[ \frac{1}{1-\chi(p)p^{-s}} = e^{ \chi(p) p^{-s} + E_p}\]
for \(|E_p| \leq C_{1/2} p^{-2s}\). For \(s > 1\), we can take a finite product over primes less than \(N\) and then take the limit. It will follow that if
\[ S(s,\chi) := \sum_{p \text{ prime}} \frac{\chi(p)}{p^s}, \]
then
\[ L(s,\chi) = e^{S(s,\chi) + E(s)}\]
for some \(E(s)\) satisfying
\[ |E(s)| \leq C_{1/2} \sum_{p \text{ prime}} \frac{1}{p^{2s}} \leq C_{1/2} \sum_{n=1}^\infty \frac{1}{n^2}\]
and which is a continuous function of \(s\) on \(s > 1\) (again, by the Weierstrass \(M\)-test).
Step 2 (Understanding Unboundedness)
Suppose \(S(s,\chi)\) is unbounded as \(s \rightarrow 1^+\). This means there is a decreasing sequence \(s_k \rightarrow 1^+\) such that the maximum absolute value of real and imaginary parts of \(S(s_k,\chi)\) is at least \(k\). We can sort terms into four bins: real part \(\geq k\), real part \(\leq -k\), imaginary part \(\geq k\) and imaginary part \(\leq -k\). Since the sequence is infinite and the bins are finite, there is a subsequence in which all the bins are the same. In other words, we may assume without loss of generality that along the sequence \(s_k \rightarrow 1^+\), the real part of \(S(s_k,\chi)\) diverges to \(\infty\) or to \(-\infty\) or the imaginary part diverges to \(\infty\) or \(-\infty\). Because \(E(s)\) is continuous and bounded on \(s > 1\), the same sort of divergence must occur for the sequence \(S(s_k,\chi) + E(s_k)\). If the real parts tend to \(-\infty\), exponentiating would give that \(L(s_k,\chi) \rightarrow 0\), which cannot occur because \(L(s,\chi)\) is continuous at \(s = 1\). Similarly divergence of the real part to \(\infty\) would mean that \(|L(s_k,\chi)|\) tends to infinity as \(k \rightarrow \infty\), which is also ruled out by continuity.
Step 3 (Imaginary Unboundedness)
Therefore, it suffices to consider the situation when the imaginary part of \(S(s_k,\chi) + E(s_k)\) tends to either \(+\infty\) or \(-\infty\) as \(k \rightarrow \infty\). But since \(S(s,\chi) + E(s)\) is continuous in \(s\), this would imply the existence of two sequences \(s'_k\) and \(s''_k\), both tending to \(1\) from above, such that the imaginary part of \(S(s'_k,\chi) + E(s'_k)\) is an even multiple of \(\pi\) and the imaginary part of \(S(s''_k,\chi) + E(s''_k)\) is an odd multiple of \(\pi\). Exponentiating would give that \(L(s'_k,\chi)\) is real and positive and \(L(s''_k,\chi)\) is real and negative. Because both sequences converge to \(L(1,\chi)\) as \(k \rightarrow \infty\), this would again imply that \(L(1,\chi) = 0\), which have hypothesized is not the case.
3. To-Do List
To complete the proof of Dirichlet's Theorem, we need to:
- Establish that \(L(1,\chi) \neq 0\) when \(\chi\) is a real nontrivial character.