Proof that the principal logarithm is holomorphic off the branch cut

Builds on Principal branch Branch cut Principal value of the argument Argument of a complex number cos pi = -1 Product of exponentials Power series of the exponential function Logarithm is the inverse of the exponential Euler's formula Power series of cosine Power series of sine Angle addition formulas Pythagorean trigonometric identity Tangent function Arctangent Derivative of arctangent Differentiable functions are continuous Composition of continuous functions is continuous Derivative of the complex exponential Complex exponential Holomorphic function

Proof

We show that $\operatorname{Log}$ is continuous on $S$ and then invert the derivative of $\exp$ .

The set $(-\infty, 0]$ is closed in $\C$ , so the slit plane $S$ is open. For $z \in S$ , the principal branch is $\operatorname{Log} z = \ln \abs{z} + i \operatorname{Arg} z$ , with $\operatorname{Arg} z \in (-\pi, \pi]$ the principal argument.
Let $z \in S$ and $\theta = \operatorname{Arg} z$ . If $\theta = \pi$ , then by the definition of an argument and $\cos \pi = -1$ , $\sin \pi = 0$ , $z = -\abs{z} \in (-\infty, 0]$ . This is false, so $\theta \in (-\pi, \pi)$ .
Let $z \in S$ . By the product of exponentials, $e^{\operatorname{Log} z} = e^{\ln \abs{z}} e^{i \operatorname{Arg} z}$ . Here $e^{\ln \abs{z}} = \abs{z}$ , because the exponential series is the real exponential at real points and $e^{\ln t} = t$ .
By Euler's formula, with the real cosine and sine at real points, $e^{i\theta} = \cos \theta + i \sin \theta$ . So $e^{\operatorname{Log} z} = \abs{z} (\cos \theta + i \sin \theta) = z$ .
Write $z = x + iy \in S$ with $x, y$ real, and let $\varphi = \theta / 2 \in (-\pi/2, \pi/2)$ . By the angle addition formulas and the Pythagorean identity, $\sin \theta = 2 \sin \varphi \cos \varphi$ and $\cos \theta = 2 \cos^2 \varphi - 1$ .
Since $x = \abs{z} \cos \theta$ and $y = \abs{z} \sin \theta$ , step 5 gives $\abs{z} + x = 2 \abs{z} \cos^2 \varphi$ and $y = 2 \abs{z} \sin \varphi \cos \varphi$ .
We have $\abs{z} + x > 0$ on $S$ . Indeed, if $y \ne 0$ then $\abs{z} > \abs{x} \ge -x$ . If $y = 0$ then $x > 0$ , because $z \in S$ .
By steps 6 and 7, $\cos \varphi \ne 0$ and $\frac{y}{\abs{z} + x} = \frac{\sin \varphi}{\cos \varphi} = \tan \varphi$ , by the definition of the tangent.
Since $\varphi \in (-\pi/2, \pi/2)$ , the definition of the arctangent gives $\varphi = \arctan \frac{y}{\abs{z} + x}$ . So $\operatorname{Arg} z = 2 \arctan \frac{y}{\abs{z} + x}$ for each $z \in S$ .
The functions $x$ , $y$ and $\abs{z}$ are continuous in $z$ , and the denominator in step 9 is positive on $S$ . The arctangent is differentiable, hence continuous. The function $\ln$ is continuous on $(0, \infty)$ .
By step 10 and the continuity of compositions, $\operatorname{Arg}$ and $\ln \abs{\cdot}$ are continuous on $S$ . So $\operatorname{Log}$ is continuous on $S$ .
Fix $z_0 \in S$ and put $w_0 = \operatorname{Log} z_0$ . For $z \in S$ with $z \ne z_0$ , put $w = \operatorname{Log} z$ . By step 4, $e^w = z \ne z_0 = e^{w_0}$ , so $w \ne w_0$ .
Then $\frac{\operatorname{Log} z - \operatorname{Log} z_0}{z - z_0} = \frac{w - w_0}{e^w - e^{w_0}} = \frac{1}{Q(w)}$ , where $Q(w) = \frac{e^w - e^{w_0}}{w - w_0}$ for $w \ne w_0$ .
By the derivative of the complex exponential, $Q(w) \to e^{w_0} = z_0$ as $w \to w_0$ with $w \ne w_0$ .
Let $\varepsilon > 0$ . Choose $\eta > 0$ with $\abs{Q(w) - z_0} < \varepsilon$ for $0 < \abs{w - w_0} < \eta$ . By step 11, choose $\delta > 0$ with $\abs{\operatorname{Log} z - w_0} < \eta$ for $z \in S$ , $\abs{z - z_0} < \delta$ .
By steps 12 and 15, $\abs{Q(\operatorname{Log} z) - z_0} < \varepsilon$ for $z \in S$ with $0 < \abs{z - z_0} < \delta$ . So $Q(\operatorname{Log} z) \to z_0$ as $z \to z_0$ .
We have $z_0 \ne 0$ . When $\abs{Q - z_0} < \abs{z_0}/2$ , we get $\abs{Q} > \abs{z_0}/2$ and $\abs{1/Q - 1/z_0} = \abs{Q - z_0} / (\abs{Q} \abs{z_0}) < 2 \abs{Q - z_0} / \abs{z_0}^2$ .
By steps 13, 16 and 17, the difference quotient of $\operatorname{Log}$ at $z_0$ tends to $1/z_0$ . So $\operatorname{Log}'(z_0) = 1/z_0$ .
This holds at each point of the open set $S$ , so $\operatorname{Log}$ is holomorphic on $S$ with $\operatorname{Log}'(z) = 1/z$ .

Idea

On the slit plane, the principal argument has the explicit formula $2 \arctan(y/(\abs{z} + x))$ , which shows that $\operatorname{Log}$ is continuous. Since $\operatorname{Log}$ inverts the exponential, its difference quotient is the reciprocal of a difference quotient of $\exp$ , which tends to $e^{w_0} = z_0$ .

Reading path

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Proof that the principal logarithm is holomorphic off the branch cut proof

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3 pages build on this one.

Tag	`ESYG79Z43Y`
URI	`m2://ESYG79Z43Y`
Kind	proof
Writing	W0 stub
MSC	33B10
Proves	The principal logarithm is holomorphic off the branch cut
Uses	Principal branch, Branch cut, Principal value of the argument, Argument of a complex number, cos pi = -1, Product of exponentials, Power series of the exponential function, Logarithm is the inverse of the exponential, Euler's formula, Power series of cosine, Power series of sine, Angle addition formulas, Pythagorean trigonometric identity, Tangent function, Arctangent, Derivative of arctangent, Differentiable functions are continuous, Composition of continuous functions is continuous, Derivative of the complex exponential, Complex exponential, Holomorphic function
Changed	4 November 2026

Proof that the principal logarithm is holomorphic off the branch cut

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