The Complex Exponential and Logarithm

We prove each part in turn, using standard facts about power series and about holomorphic functions (Cauchy–Riemann equations, the identity theorem) as background machinery, none of which presupposes the present theorem.

(1) Entirety and derivative. The series $\sum z^n/n!$ has radius of convergence $\infty$: by the ratio test, for fixed $z\ne0$,

as $n\to\infty$, so the series converges absolutely for every $z\in\mathbb{C}$. A power series is holomorphic throughout its disk of convergence (here, all of $\mathbb{C}$), and may be differentiated term by term there:

Thus $\exp$ is entire with $\exp'=\exp$.

(2) Addition law. Fix $z_2\in\mathbb{C}$ and define $g(z) = \exp(z)\exp(z_2 - z)$. Then $g$ is entire (product/composition of entire functions), and by the product and chain rules together with (1),

A holomorphic function on the connected set $\mathbb{C}$ with identically zero derivative is constant (its real and imaginary parts have vanishing gradient by the Cauchy–Riemann equations, hence are constant on the connected domain). So $g(z) = g(0) = \exp(0)\exp(z_2) = \exp(z_2)$ for all $z$ (since $\exp(0)=\sum 0^n/n! = 1$). Setting $z = z_1$ gives $\exp(z_1)\exp(z_2-z_1) = \exp(z_2)$, i.e.

More directly: from $g(z)\equiv \exp(z_2)$ with $z=z_1$ we get $\exp(z_1)\exp(z_2-z_1)=\exp(z_2)$. Replacing $z_2$ by $z_1+z_2$ in this identity (valid for all choices of the two free variables) gives $\exp(z_1)\exp(z_2) = \exp(z_1+z_2)$, the addition law. Since $\exp(0)=1$, the addition law with $z_2=-z_1$ gives $\exp(z_1)\exp(-z_1) = \exp(0)=1$, so $\exp(z_1)\ne 0$ and $\exp(-z_1) = 1/\exp(z_1)$, for every $z_1\in\mathbb{C}$.

(3) Euler's formula. For $\theta\in\mathbb{R}$, split the defining series of $\exp(i\theta)$ into real and imaginary parts using $i^{4k}=1, i^{4k+1}=i, i^{4k+2}=-1, i^{4k+3}=-i$:

Rearrangement of the absolutely convergent series is justified (absolute convergence permits regrouping terms). The two real series on the right are exactly the Taylor series of $\cos\theta$ and $\sin\theta$ (this is the standard real-analysis definition/characterization of sine and cosine via their Taylor series, which we take as known from real calculus). Hence $\exp(i\theta) = \cos\theta + i\sin\theta$. For general $z=x+iy$ with $x,y$ real, the addition law gives $e^z = e^x e^{iy} = e^x(\cos y + i\sin y)$, using that $e^x$ is the ordinary real exponential (the restriction of the series to real arguments agrees with the real exponential's Taylor series). Taking modulus, since $\cos^2y+\sin^2y=1$,

and the point $e^z/|e^z| = \cos y + i \sin y$ lies at angle $y$ on the unit circle, so $\arg(e^z) \equiv y \pmod{2\pi}$.

(4) Periodicity and non-injectivity. Since $\cos$ and $\sin$ have period $2\pi$, $e^{i(\theta+2\pi)} = \cos(\theta+2\pi)+i\sin(\theta+2\pi) = \cos\theta+i\sin\theta = e^{i\theta}$. Hence by the addition law, $e^{z+2\pi i} = e^z e^{2\pi i} = e^z\cdot 1 = e^z$. For the "only if" direction of the injectivity claim: suppose $e^{z_1}=e^{z_2}$. By the addition law and part (2), $e^{z_1-z_2} = e^{z_1}e^{-z_2} = e^{z_1}/e^{z_2} = 1$. Write $z_1-z_2 = x+iy$; by part (3), $|e^{z_1-z_2}| = e^x = |1| = 1$, forcing $x=0$ (since $e^x$ is a strictly increasing bijection $\mathbb{R}\to(0,\infty)$, so $e^x=1 \iff x=0$). Then $e^{iy} = \cos y + i\sin y = 1$ forces $\cos y=1,\sin y=0$, which (by the known real-analysis fact that $\cos y=1$ exactly when $y\in2\pi\mathbb{Z}$) means $y\in 2\pi\mathbb{Z}$. So $z_1-z_2 = iy \in 2\pi i\mathbb{Z}$. Conversely if $z_1-z_2=2\pi ik$ for $k\in\mathbb{Z}$, then $e^{z_1}=e^{z_2}e^{2\pi ik}=e^{z_2}\cdot(e^{2\pi i})^k = e^{z_2}\cdot 1^k=e^{z_2}$, using $e^{2\pi i}=\cos2\pi+i\sin2\pi=1$. This proves the "if and only if." Since $\exp(0)=\exp(2\pi i)=1$ but $0\ne 2\pi i$, $\exp$ is not injective on $\mathbb{C}$.

(5) Existence of holomorphic branches on simply connected $U\subseteq\mathbb{C}^*$. Since $U$ is open in $\mathbb{C}^*$, the function $f(z) = 1/z$ is holomorphic on $U$. We use the standard fact (a form of Cauchy's theorem / the primitive existence theorem): on a simply connected open set, every holomorphic function has a holomorphic primitive. Applying this to $f(z)=1/z$ on the simply connected $U$, there exists a holomorphic function $\ell_0: U \to \mathbb{C}$ with $\ell_0'(z) = 1/z$ on $U$. Fix a basepoint $z_0\in U$ and let $w_0$ be any complex number with $e^{w_0} = z_0$ (such $w_0$ exists: writing $z_0 = r_0 e^{i\theta_0}$ in polar form, $w_0=\ln r_0 + i\theta_0$ works by part (3)). Replace $\ell_0$ by $\ell(z) := \ell_0(z) - \ell_0(z_0) + w_0$; this is still a holomorphic primitive of $1/z$ on $U$, now normalized so $\ell(z_0)=w_0$, hence $e^{\ell(z_0)} = e^{w_0}=z_0$. Define $h(z) := z\, e^{-\ell(z)}$ on $U$. Then $h$ is holomorphic (product of holomorphic functions), and by the product rule and part (1),

Since $U$ need not be connected in general, but simple connectivity is usually required to include connectedness (or we apply the argument on each component); assume $U$ connected (the standard convention for "simply connected"). Then $h$ is constant on $U$: $h(z) \equiv h(z_0) = z_0 e^{-w_0} = z_0/e^{w_0} = z_0/z_0 = 1$. So $z\,e^{-\ell(z)} = 1$, i.e. $e^{\ell(z)} = z$ for every $z\in U$. Thus $\ell$ is the desired holomorphic branch, and it is automatically a branch of the logarithm as defined (it is continuous, being holomorphic). Differentiating $e^{\ell(z)}=z$ directly (chain rule) gives $\ell'(z) e^{\ell(z)} = 1$, so $\ell'(z) = 1/e^{\ell(z)} = 1/z$, confirming the derivative formula independently. Uniqueness up to $2\pi i\mathbb{Z}$: if $\ell_1,\ell_2$ are two branches on $U$ (continuous suffices) with $e^{\ell_1(z)}=e^{\ell_2(z)}=z$ for all $z\in U$, then by part (4), for each $z$ there is an integer $k(z)$ with $\ell_1(z)-\ell_2(z) = 2\pi i\, k(z)$. The function $z\mapsto k(z) = (\ell_1(z)-\ell_2(z))/(2\pi i)$ is continuous (difference/quotient of continuous functions) and integer-valued on the connected set $U$, hence constant, since a continuous integer-valued function on a connected set has connected image inside $\mathbb{Z}$ (discrete), forcing the image to be a single point. The principal branch: Let $U = \mathbb{C}\setminus(-\infty,0]$, the plane slit along the non-positive real axis. This $U$ is open; it is simply connected because it is star-shaped with respect to the point $1$ (for any $z\in U$, the segment from $1$ to $z$ stays in $U$ — a short geometric check: crossing the slit would require passing through a non-positive real number, but the straight segment from $1$ to $z=re^{i\theta}$ with $\theta\in(-\pi,\pi)$ only touches the negative real axis if $z$ itself is a non-positive real number, excluded from $U$; star-shaped domains are simply connected). By the general result just proved, a holomorphic branch $\ell$ exists on $U$; one checks directly that $\ell(z) = \ln|z| + i\operatorname{Arg}(z)$ (with $\operatorname{Arg}(z)\in(-\pi,\pi)$ chosen continuously on $U$, which is possible precisely because the slit removes the only obstruction to continuously choosing an angle) satisfies $e^{\ell(z)} = e^{\ln|z|}(\cos(\operatorname{Arg} z) + i \sin(\operatorname{Arg} z)) = |z|\cdot \frac{z}{|z|} = z$ by part (3), and matches the branch produced abstractly up to the additive constant $2\pi i k$ pinned down by evaluating at $z=1$ where $\operatorname{Log}(1) = \ln 1 + i\cdot 0 = 0$, so $k=0$. Hence $\operatorname{Log}$ is exactly this holomorphic branch.

(6) No global continuous branch on $\mathbb{C}^*$. Suppose, for contradiction, that $\ell: \mathbb{C}^*\to\mathbb{C}$ is continuous with $e^{\ell(z)}=z$ for all $z\in\mathbb{C}^*$. Consider the restriction of $\ell$ to the unit circle, parametrized by $\gamma(t) = e^{it}$, $t\in[0,2\pi]$. Define $\phi(t) := \operatorname{Im}(\ell(\gamma(t)))$, a continuous real-valued function of $t\in[0,2\pi]$ (composition of continuous functions). Since $e^{\ell(\gamma(t))} = \gamma(t) = e^{it}$, part (4) gives $\ell(\gamma(t)) - it \in 2\pi i \mathbb{Z}$ for every $t$, i.e. $\phi(t) - t \in 2\pi\mathbb{Z}$. So $n(t):=(\phi(t)-t)/(2\pi)$ is a continuous, integer-valued function on the connected interval $[0,2\pi]$, hence (as in part 5) constant: $n(t) \equiv n$ for some fixed integer $n$. Thus $\phi(t) = t + 2\pi n$ for all $t\in[0,2\pi]$. But $\gamma(0) = \gamma(2\pi) = 1$, so $\phi(0) = \phi(2\pi)$ (both equal $\operatorname{Im}(\ell(1))$, the same value since $\ell$ is a genuine function, single-valued at the point $1$). Yet $\phi(2\pi) - \phi(0) = (2\pi + 2\pi n) - (0+2\pi n) = 2\pi \ne 0$. This contradiction shows no such continuous $\ell$ can exist. $\blacksquare$