KindiakMath

Extending the Logarithm

Recall that we have defined the natural logarithm to be

\ln := \exp|_{\mathbb R}^{-1} : \mathbb R^+ \to \mathbb R,

and that this construction is well-defined since \exp|_{\mathbb R} is injective. The same, however, cannot be said about \exp in general.

Recall that we have e^{it} = \cos(t) + i \sin(t) by Euler’s formula, so that

e^{x+ iy} = e^x (\cos(y) + i \cdot \sin(y)) = e^x \cdot e^{iy}.

Lemma 1. For any z \in \mathbb C, \exp(z) = \exp(z + 2\pi i).

Proof. We observe that

e^{2\pi i} = \cos(2\pi) + i \cdot \sin(2\pi) = 1 + i \cdot 0 = 1.

Hence, writing z = x+ i y,

\begin{aligned} \exp(z) &= \exp(x + iy ) \\ &= e^x \cdot e^{i y} \\ &= e^x \cdot e^{i y} \cdot e^{i(2\pi)} \\ &= e^x \cdot e^{i(y + 2\pi)} \\ &= \exp(x + i(y + 2\pi) ) \\ &= \exp(x + iy + 2\pi i) \\ &= \exp(z + 2\pi i).\end{aligned}

Therefore, \exp(0) = \exp(2\pi i), but 0 \neq 2\pi i, so that \exp is not injective on \mathbb C. This means that whatever our definition of \log such that \log|_{\mathbb R}^+ = \ln, we cannot just define \log := \exp^{-1}; the right-hand side is not well-defined.

As such, we should approach this problem with caution, and take baby steps in establishing \log. Let’s first consider the subset K := \mathbb R + (-\pi, \pi] \cdot i \subseteq \mathbb C.

Lemma 2. \exp|_K : K \to \mathbb C \backslash \{0\} =: \mathbb C^{\times} is bijective.

Proof. Fix w, z \in K and suppose \exp(w) = \exp(z). Write w = r + i s and z = x + i y, so that

e^r \cdot e^{is} = \exp(w) = \exp(z) = e^x \cdot e^{iy} \quad \Rightarrow \quad e^{r-x} = e^{i(y - s)}.

Since the left-hand side is positive and real,

e^{r-x} = |e^{r-x}| = |e^{i(y - s)}| = 1 \quad \Rightarrow \quad r = x.

Hence, since y-s \in (-2\pi, 2\pi),

e^{i(y-s)} = 1 \quad \Rightarrow \quad y-s = 0 \quad \Rightarrow \quad s = y.

Therefore, w = r + is = x + iy = z, proving that \exp|_K is injective.

To prove that \exp_K is surjective, fix w \in \mathbb C^{\times}. Since w / |w| has norm 1, w / |w| = e^{i \theta} for some unique \theta \in (-\pi, \pi]. Therefore, w = |w| \cdot e^{i \theta}. Since \exp|_{\mathbb R} : \mathbb R \to \mathbb R^+ is bijective and |w| > 0, there exists r > 0 such that e^r = |w|. Hence, \exp|_K(r + i \theta) = e^r \cdot e^{i\theta} = w, establishing surjectivity.

In fact, defining K_{y_0} := K + i y_0 for any y_0 \in \mathbb R, \exp|_{K_{y_0}} : K_{y_0} \to \mathbb C^{\times} is injective too.

Definition 1. We define the principal logarithm function by

\mathrm{Log} := \exp|_K^{-1} : \mathbb C^{\times} \backslash \{0\} \to K.

More explicitly,

\mathrm{Log}(r e^{i \theta}) := \exp|_K^{-1}(e^{\ln(r)} \cdot e^{i \theta}) = \ln r + i \cdot \theta.

Furthermore, given z \in \mathbb C^{\times} , we define the principal argument function by

\mathrm{Arg} := \mathrm{Im} \circ \mathrm{Log} : \mathbb C^{\times} \to (-\pi, \pi].

Observe that by construction, \mathrm{Log}|_{\mathbb R^+} = \ln, since

\mathrm{Log}_{\mathbb R^+}(x) = \mathrm{Log}(xe^{i \cdot 0}) = \ln(x) + i \cdot 0 = \ln(x).

In particular, we are justified to make the following claim:

Theorem 1. \mathrm{Log}(-1) = i \pi.

Proof. By construction,

\mathrm{Log}(-1) = \mathrm{Log}(1 \cdot e^{i \pi}) = \ln(1) + i \pi = i \pi.

In a cheeky way, therefore, \ln(-1) = i \pi if we stipulated that \ln|_{ \mathbb R^- } := \mathrm{Log}|_{\mathbb R^-}. Unfortunately, however, this convention still means that \ln(0) is undefined. But this compromise is rather small—we recover \ln defined on \mathbb R \backslash \{0\}.

Theorem 2. \mathrm{Log} is holomorphic on \mathbb C^{\times} \backslash \mathrm{Arg}^{-1}(\pi), and \mathrm{Log}'(z) = 1/z.

Proof. It is clear that K \backslash \mathrm{Arg}^{-1}(\pi) is an open set. To prove the derivative of \mathrm{Log}, recall that \exp' = \exp. Writing u_0 = \mathrm{Log}(z_0) and u = \mathrm{Log}(z_0 + w), since \mathrm{Log} is continuous at u_0, w \to 0 implies u \to u_0, so that

\begin{aligned} \lim_{w \to 0} \frac{ \mathrm{Log}(z_0 + w) - \mathrm{Log}(z_0) }{w} &= \lim_{u \to 0} \frac{ u - u_0 }{ \exp(u) - \exp(u_0) } \\ &= \lim_{u \to 0} \frac 1{ \frac{ \exp(u) - \exp(u_0) }{ u - u_0 } }\\ &= \frac 1{ {\displaystyle \lim_{u \to 0} } \frac{ \exp(u) - \exp(u_0) }{ u - u_0 } } \\ &= \frac 1{ \exp'(u_0) } = \frac 1{ \exp(u_0) } = \frac 1{z_0}. \end{aligned}

What about the general complex logarithm for nonzero z_0 \in \mathbb C? While it is certainly the case that \log(z_0) can take multiple values, one unique value from our construction would be \mathrm{Log}(z), so that \exp(\mathrm{Log}(z_0)) = z_0. That being said, given any other w \in \mathbb C such that \exp(w) = z_0, exponential properties dictate that

\displaystyle \exp(w - \mathrm{Log}(z_0)) = \frac{z_0}{z_0} = 1,

which means w - \mathrm{Log}(z_0) = 2 k \pi i for some k \in \mathbb Z. Therefore,

\exp(w) = z_0 \quad \iff \quad w \in z_0 + 2 \pi i \cdot \mathbb Z.

Therefore, we should think of \log : \mathbb C \to \mathcal P(\mathbb C), rather than \log : \mathbb C \to \mathbb C. In this case, we say that \log is multi-valued or set-valued.

Theorem 3. Define \log : \mathbb C^{\times} \to \mathcal P(\mathbb C) by

\log(z) := \exp^{-1}(z) \subseteq \mathbb C.

Then \log(z) = \mathrm{Log}(z) + 2 \pi i \cdot \mathbb Z and similarly, \mathrm{arg}(z) = \mathrm{Arg}(z) + 2 \pi \cdot \mathbb Z as a set-valued function \mathrm{arg} : \mathbb C^{\times} \to \mathcal P(\mathbb R).

Corollary 1. For any z, w \in \mathbb C^{\times},

\log(z \cdot w) = \log(z) + \log(w).

Consequently, \arg(z \cdot w) = \arg(z) + \arg(w).

Proof. By Theorem 3,

\begin{aligned}\log(z \cdot w) &= \mathrm{Log}(z \cdot w) + 2 \pi i \cdot \mathbb Z \\ &= \mathrm{Log}(z) \pm 2\pi i + \mathrm{Log} (w) \pm 2\pi i + 2 \pi i \cdot \mathbb Z \\ &= (\mathrm{Log}(z) + 2 \pi i \cdot \mathbb Z) + (\mathrm{Log}(w) + 2 \pi i \cdot \mathbb Z) \\ &= \log(z) + \log(w). \end{aligned}

Having extended the logarithm, can we extend the exponential? Namely, what would be a meaningful definition for expressions of the form z^w? Our starting point was expressions of the form e^x, which we can extend to expressions of the form a^x via the identity

\displaystyle a^x := e^{x \ln(a)} \equiv \exp(x \log(a)),

if a > 0. By Theorem 3, expressions of the form \ln(z) are well-defined set-valued functions whenever z \in \mathbb C^{\times}. Fix w \in \mathbb C. Since \exp : \mathbb C \to \mathbb C^{\times} is well-defined too, we can boldly make the definition

\displaystyle z^w := \exp(w \log(z)),\quad z \in \mathbb C^{\times} .

To do this precisely doesn’t take terribly much work—just define the set-valued function f_w : \mathbb C^{\times} \to \mathcal P(\mathbb C) by f_w(z) := \exp(w \log(z)), and subsequently denote z^w = f_w(z) for brevity. Consequently,

\begin{aligned} z^{w_1 + w_2} &= \exp((w_1+w_2) \log(z)) \\ &= \exp(w_1 \log(z))\cdot \exp(w_2 \log(z)) \\ &= z^{w_1} \cdot z^{w_2}. \end{aligned}

In particular, we get a comical answer to evaluating i^i.

Example 1. i^i = e^{-\pi/2 + 2\pi \mathbb Z} \subseteq \mathbb R.

Proof. Using our generalised definition and Theorem 3,

\begin{aligned} i^i = \exp(i \log(i)) &= \exp(i (\mathrm{Log}(i) + 2\pi i \cdot \mathbb Z)) \\ &= \exp(i (i\pi/2 + 2\pi i \cdot \mathbb Z)) \\ &= \exp(- \pi/2 - 2\pi \cdot \mathbb Z) \\ &= e^{-\pi/2 + 2\pi \mathbb Z}. \end{aligned}

Who knew taking the exponential of two purely imaginary numbers could land us in a set of purely real outcomes!

We conclude with a brief chat about n-th roots. Having defined i = \sqrt{-1} in the sense that i^2 = -1, it should come to no surprise that

(-i)^2 = i^2 = -1.

Thus, the equation z^2 = -1 has two roots, namely z = \pm i. Which of these roots should (-1)^{1/2} equal? It turns out that either option work just as well algebraically, and so like the logarithm, it is much more helpful to consider (-1)^{1/2} as set-valued. Define f_n : \mathbb C \to \mathbb C by f_n(w) = w^n. Then define

z^{1/n} := f_n^{-1}(z) \subseteq \mathbb C.

We call the elements of 1^{1/n} the n-th roots of unity.

Example 2. 1^{1/n} = \{ \exp(i \cdot 2k\pi/n) : k = 0, 1,\dots, n-1\}. Denoting \zeta_n := \exp(i \cdot 2\pi/n), 1^{1/n} = \{ 1, \zeta_n, \zeta_n^2, \dots, \zeta_n^{n-1}\} = \zeta_n^{\mathbb Z}.

Proof. Fix z \in 1^{1/n} so that z^n = 1. Then

|z|^n = |z^n| = |1| = 1\quad \Rightarrow \quad |z| = 1

and

\displaystyle n\cdot \mathrm{arg}(z) = \mathrm{arg}(z^n) = \mathrm{arg}(1) = 2\pi \cdot \mathbb Z \quad \Rightarrow \quad \mathrm{arg}(z) = \frac{2\pi}{n} \cdot \mathbb Z,

so that

\displaystyle z = |z| \cdot \exp(i\, \mathrm{arg}(z)) = 1 \cdot \exp\left( \frac{2\pi i}{n} \cdot \mathbb Z\right) = \exp\left( \frac{2k\pi i}{n} \right),

where k varies over 0,1,\dots,n-1 and

\displaystyle \exp\left( \frac{2(k + nd) \pi i}{n} \right) = \exp\left( \frac{2k\pi i}{n} \right) \cdot \underbrace{ \exp(2 d \pi i) }_1 = \exp\left( \frac{2k\pi i}{n} \right).

Example 3. z^{1/n} = |z|^{1/n} \cdot \exp \left(i \cdot \mathrm{Arg}(z) /n \right) \cdot 1^{1/n}.

Proof. Suppose w \in z^{1/n} so that w^n = z. Then |w| = |z|^{1/n} and

\displaystyle \mathrm{arg}(w) = \frac{ \mathrm{arg}(z) }n = \frac{\mathrm{Arg}(z) + 2\pi \cdot \mathbb Z}{n} = \frac{\mathrm{Arg}(z)}{n} + \frac{ \mathrm{arg}(1) }n.

Therefore,

\displaystyle w = |w| \cdot \exp(i \cdot \mathrm{arg}(w)) = |z|^{1/n} \cdot \exp \left(i \cdot \frac{ \mathrm{Arg}(z) }n \right) \cdot 1^{1/n}.

Of course, in the setting that a \in \mathbb R^+, then a^{1/n} = f_n^{-1}(a) = \{ {\sqrt[n]{a}} \} can be viewed as a set-valued function with exactly one element.

Complex analysis lends itself immense power to prove theorems that we otherwise would lack the mathematical resource to establish. We will visit these powerful results in the future. For now, we turn our attention to integration—not as straightforward as in the real case.

Remark 1. We could, in fact, define \log from a group-theoretic perspective. For any w \in \mathbb C^{\times}, define \hat w := w/|w|. Since |\hat w| = 1, there exists \theta \in \mathbb R such that

w = |w| \cdot \hat w = |w| \cdot e^{i \theta}.

Therefore,

w = e^{\ln |w|} \cdot e^{i\theta} = \exp(\ln |w|+i\theta).

Therefore, the map \exp : \mathbb C \to \mathbb C^{\times} is surjective and

\exp(z_1+z_2) = \exp(z_1) \cdot \exp(z_2),\quad z_1,z_2 \in \mathbb C.

Now

\begin{aligned} \ker(\exp) &= \{z \in \mathbb C : \exp(z) = 1\} \\ &= \{ 2k\pi i : k \in \mathbb Z\} = 2\pi i \mathbb Z. \end{aligned}

By the first isomorphism theorem, there exists a bijection

\phi : \mathbb C/{ 2\pi i \mathbb Z } \to \mathbb C^{\times}

such that

\exp(z) = \phi([z]),\quad [z] := z + 2\pi i \mathbb Z.

Furthermore,

\phi([z_1]+[z_2]) = \phi([z_1]) \cdot \phi([z_2]),\quad z_1, z_2 \in \mathbb C.

In this sense, define \log := \phi^{-1}. Since \exp(\ln |w| + i \cdot \theta) = w,

\begin{aligned} \log(w) &= \phi^{-1}(\exp(\ln |w| + i \cdot \theta)) \\ &= [ \ln |w| + i \cdot \theta ] \\ &= (\ln |w| + i \cdot \theta) + 2\pi i \mathbb Z \end{aligned}

and doing some book-keeping yields

\log(w_1 w_2) = \log(w_1) + \log(w_2).

In the special case |w| = 1, w = \hat w and \ln(1) = 0 implies

\log(\hat w) = \log(w) = i \cdot \theta + 2\pi i \mathbb Z \subseteq \mathbb Ri.

Hence, define \arg(w) := \log(\hat w)/i \subseteq \mathbb R, so that

\log(w) = \ln |w| + i \cdot \arg(w),

and \arg(w) = \theta + 2\pi \mathbb Z, and

\begin{aligned} \arg(w_1 w_2) &= \frac{\log(\hat w_1 \hat w_2)}{i} \\ &= \frac{\log(\hat w_1)}{i} + \frac{\log(\hat w_2)}{i} \\ &= \arg(w_1) + \arg(w_2). \end{aligned}

To recover our original definitions, use the Archimedean property to find a unique real number \mathrm{Arg}(w) \in \mathrm{arg}(w) \cap (-\pi, \pi]. Define \mathrm{Log}(w) := \ln |w| + i \cdot \mathrm{Arg}(w). Then

\begin{aligned} \log(w) &= \ln |w| + \log(\hat w) \\ &= \ln |w| + i \cdot \arg(w) \\ &= \ln |w| + i \cdot ( \mathrm{Arg}(w) + 2\pi \mathbb Z) \\ &= (\ln |w| + i \cdot \mathrm{Arg}(w)) + 2\pi i \mathbb Z \\ &= \mathrm{Log}(w) + 2\pi i \mathbb Z. \end{aligned}

—Joel Kindiak, 14 Aug 25, 0033H

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