Any z\neq 0 can be written in the form \begin{align*} z = {\left\lvert {z} \right\rvert} e^{i\arg(z)} ,\end{align*} where \arg(z) is the multivalued function \arg(z) = \operatorname{Arg}(z) + 2\pi n where \operatorname{Arg}(z)\in (-\pi, \pi] is the principal value of the argument. Note that it’s not necessarily true that if z=x+iy that \operatorname{Arg}(z) = \arctan\qty{y\over x}, because \arctan is also multivalued. There is a formula based on which quadrant z is in:
Note that \operatorname{Arg}(0) is undefined, so that \operatorname{Arg}(z) makes sense on {\mathbb{C}}\setminus(-\infty, 0], where (\infty, 0] is a branch cut. Any choice of interval (\theta_0, \theta_0 + 2\pi] yields a different branch cut.
Since \log(z) and z^\alpha are defined in terms of \operatorname{Arg}, this is where most branching issues come from!
There are two ways to define z^{1\over 2}:
- f_1(z) = \sqrt{{\left\lvert {z} \right\rvert}} e^{i{ \operatorname{Arg}(z) \over 2} } = \sqrt{r}\cos\qty{\theta\over 2} + i\sqrt{r} \sin\qty{\theta\over 2}.
- f_2(z) = \sqrt{{\left\lvert {z} \right\rvert}} e^{i{ \operatorname{Arg}(z) + 2\pi \over 2} } = \sqrt{r}\cos\qty{\theta + 2\pi \over 2} + i\sqrt{r} \sin\qty{\theta + 2\pi \over 2}.
Note that \begin{align*} f_2(z) = \sqrt{r}e^{i{\operatorname{Arg}(z) + 2\pi \over 2} } = \sqrt{r} e^{i{\operatorname{Arg}(z) \over 2}}e^{i\pi} = -\sqrt{r}e^{i{\operatorname{Arg}(z) \over 2}} = -f_1(z) .\end{align*}
Let z_0 = r_0e^{i\pi} \in (-\infty, 0) \subseteq {\mathbb{R}}, and show that z^{1\over 2} is not continuous along (-\infty, 0) by computing \begin{align*} \lim_{z\in \gamma_1} f_1(z) &= i\sqrt{r_0} \\ \lim_{z\in \gamma_2} f_1(z) &= -i\sqrt{r_0} ,\end{align*}
where
- \gamma_1 = \left\{{r_0 e^{it} {~\mathrel{\Big\vert}~}t\in (0, \pi) }\right\},
- \gamma_2 = \left\{{r_0 e^{-it} {~\mathrel{\Big\vert}~}t\in (\pi, 0) }\right\}
solution:
\begin{align*} \lim _{(r, \theta) \rightarrow\left(r_{0}, \pi\right)} f_{1}\left(r e^{i \theta}\right) &=\lim _{(r, \theta) \rightarrow\left(r_{0}, \pi\right)} r^{\frac{1}{2}}\left(\cos \frac{\theta}{2}+i \sin \frac{\theta}{2}\right)=i r_{0}^{\frac{1}{2}}, \quad \text { and } \\ \lim _{(r, \theta) \rightarrow\left(r_{0},-\pi\right)} f_{1}\left(r e^{i \theta}\right) &=\lim _{(r, \theta) \rightarrow\left(r_{0},-\pi\right)} r^{\frac{1}{2}}\left(\cos \frac{\theta}{2}+i \sin \frac{\theta}{2}\right)=-i r_{0}^{\frac{1}{2}} . .\end{align*}
Define \begin{align*} \log(z) \coloneqq\ln\qty{{\left\lvert {z} \right\rvert}} + i\operatorname{Arg}(z) ,\end{align*} where \operatorname{Arg}(z) is the principal argument in (-\pi, \pi]. Note that this is sometimes written \begin{align*} \log(z) \coloneqq\ln\qty{{\left\lvert {r} \right\rvert}} + i\theta .\end{align*}
\begin{align*} \log(1+i) = \ln\qty{\sqrt 2} + \qty{{\pi\over 4} + 2k\pi}i .\end{align*} Note that this assigns an infinite number of complex numbers to 1+i, all on the line \Re(z) = \ln\qty{\sqrt 2} with imaginary parts differing by multiples of 2\pi.
The principal branch of \operatorname{Log} is defined so that \operatorname{Log}(1) = 0, and can be written as \begin{align*} \operatorname{Log}(z) = \int_\gamma {1\over \xi} \,d\xi \end{align*} where \gamma is any piecewise smooth path connecting 1 to z.
Define \begin{align*} z^\alpha \coloneqq e^{\alpha \log(z)} ,\end{align*} where some branch of \log (usually the principal branch) is implicitly chosen.
If \Omega is a connected domain with f\in {\mathcal{O}}^{\times}(\Omega) an invertible regular function with \begin{align*} \int_\gamma {f'\over f} = 0 \end{align*} for all \gamma \subseteq \Omega, then
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There exists a holomorphic g:\Omega\to {\mathbb{C}} such that g = \log(f) and e^g = f.
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g' = {f'\over f}, yielding an explicit formula \begin{align*} g(z) = g(z_0) + \int_{z_0}^z {f'(\xi) \over \xi}\,d\xi .\end{align*}
\operatorname{Log}(f(z)) has branch points at the zeros of f.
Common trick: \begin{align*} f^{1/n} = e^{{1\over n} \log(f)} ,\end{align*} taking (say) a principal branch of \log given by {\mathbb{C}}\setminus(-\infty, 0] \times 0.
Suppose \Omega is a simply-connected region such that 1\in \Omega, 0\not\in\Omega. Then there exists a branch of F(z) \coloneqq\operatorname{Log}(z) such that
- F is holomorphic on \Omega,
- e^{F(z)} = z for all z\in \Omega
- F(x) = \log(x) for x\in {\mathbb{R}} in a neighborhood of 1.
Show that
- \log(zw)\neq \log(z)\log(w)
- \log(\exp(z))\neq z
solution:
Counterexamples:
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Take z=\zeta_4^3 and w=\zeta_4^2, noting that \operatorname{Arg}(z) = {3\pi \over 4} and \operatorname{Arg}(w) = {\pi \over 2} Then zw = e^{5\pi \over 4} but \operatorname{Arg}(zw) = {-3\pi \over 4} because we are forced to use the domain (-\pi, \pi].
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Showing that \log(\exp(z)) is a countably infinite set in {\mathbb{C}}: \begin{align*} \log(\exp(z)) &= \log(\exp(x+iy)) \\ &= \ln\qty{e^x} + (y+2k\pi)i \\ &= (x+iy) + 2k\pi i \\ &= z + 2k\pi .\end{align*}
\begin{align*} \operatorname{Log}(zw) &\neq \operatorname{Log}(z) + \operatorname{Log}(w) \\ \operatorname{Log}(e^z) &\neq z .\end{align*}
For counterexamples, take z=\zeta_4^3=\exp(3\pi i / 4 and w=\zeta_4^2 = \exp(\pi i/2). Then zw= \exp(-3 \pi i /4), using that the domain of \operatorname{Arg} is (-\pi, \pi].
Note the problem: for z\coloneqq x+i0 \in {\mathbb{R}}^{\leq 0}, just above the axis consider z_+ \coloneqq x + i{\varepsilon} and z_- \coloneqq x-i{\varepsilon}. Then
- \log(z_+) = \log{\left\lvert {x} \right\rvert} + i\pi, and
- \log(z_-) = \log{\left\lvert {x} \right\rvert} - i\pi.
So \log can’t even be made continuous if one crosses the branch. The issue is the branch point or branch singularity at z=0.
If f is holomorphic and nonvanishing on a simply-connected region \Omega, then there exists a holomorphic G on \Omega such that
\begin{align*} f(z) = e^{G(z)} .\end{align*}
The complex exponential is 2\pi i periodic, and invertible on any horizontal strip of the form \begin{align*} S_\alpha \coloneqq\left\{{z\in {\mathbb{C}}{~\mathrel{\Big\vert}~}\Im(z) \in [\alpha, \alpha+2\pi] }\right\} .\end{align*}
Generally choose branch cuts that are slits between branch points to prevent monodromy around those points. One can find branch points for \log(f(z)) by noting that \log should be a primitive of {\frac{\partial }{\partial z}\,}\log(f(z)).
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\log(z^2-1) has branch points at z=\pm 1,\ infty. To see \infty is a branch point: by differentiating, the log should be a primitive of {2z\over z^2-1}. Taking a large curve, each term contributes 2\pi i with the same sign, so the value jumps.
So a cut like (-\infty, +1] works, and an explicit formula may be obtained by taking:
- \log(z+1) + \log(z-1) = (r_1 + i\theta_1) + (r_2 + i\theta_2).
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\log\qty{z+1\over z-1}: needs to be a primitive for {1\over z+1} + {1\over z-1}, so \pm 1 are branch points. z=\infty is not a branch point, because integrating over a large curve yields +2\pi i - 2\pi i = 0, so no monodromy.
- So a cut like [-1, 1] works, with a formula \log(z+1) - \log(z-1) = (r_1 + i\theta_1) - (r_2 + i\theta_2).
Prove that if D is a simply connected domain and f(z) is holomorphic and nonzero on D, then f(z)=e^{g(z)}, where g(z) is holomorphic on D.
solution:
