Complex Power by Complex Exponential is Analytic

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Lemma

Let $\psi, \eta \in \C$ be constant.

Let $\map f z = z^\psi \, \map \exp {-\eta z}$, where:

$z^\psi$ denotes $z$ to the power of $\psi$, defined on its principal branch
$\map \exp {-\eta z}$ denotes the complex exponential function.


Then $f$ is analytic on any simply connected domain that does not contain the origin nor any points on the negative real axis.


Proof

Let $z$ be written in exponential form:

$z = r \map \exp {i \theta}$

where:

$r > 0$
$\theta \in \hointl {-\pi} \pi$


Let $\psi = a + i b, \eta = c + i d$.

By the definition of $f$:

\(\ds \map f {r \, \map \exp {i \theta} }\) \(=\) \(\ds z^\psi \, \map \exp {-\eta z}\)
\(\ds \) \(=\) \(\ds \paren {r e^{i \theta} }^\psi \, \map \exp {-\eta r \, \map \exp {i \theta} }\)
\(\ds \) \(=\) \(\ds \map \exp {\psi \, \map \Log {r \, \map \exp {i \theta} } } \, \map \exp {-\eta r \paren {\cos \theta + i \sin \theta} }\)
\(\ds \) \(=\) \(\ds \map \exp {\psi \ln r + i \psi \theta - \eta r \cos \theta - i \eta r \sin \theta}\)
\(\ds \) \(=\) \(\ds \map \exp {\paren {a + i b} \ln r + i \paren {a + i b} \theta - \paren {c + i d} r \cos \theta - i \paren {c + i d} r \sin \theta}\)
\(\ds \) \(=\) \(\ds \map \exp {a \ln r + i b \ln r + i a \theta - b \theta - c r \cos \theta - i d r \cos \theta - i c r \sin \theta + d r \sin \theta}\)
\(\ds \) \(=\) \(\ds \map \exp {a \ln r - b \theta - c r \cos \theta + d r \sin \theta} \, \map \exp {i \paren {b \ln r + a \theta - d r \cos \theta - c r \sin \theta} }\)


Define:

\(\ds \map g {r, \theta}\) \(=\) \(\ds a \ln r - b \theta - c r \cos \theta + d r \sin \theta\)
\(\ds \map h {r, \theta}\) \(=\) \(\ds b \ln r + a \theta - d r \cos \theta - c r \sin \theta\)

Then:

\(\ds \map f {r \, \map \exp {i \theta} }\) \(=\) \(\ds \map \exp g \, \map \exp {i h}\)
\(\ds \) \(=\) \(\ds \map \exp g \, \map \cos h + i \map \exp g \, \map \sin h\)


Define:

\(\ds \map u {r, \theta}\) \(=\) \(\ds \map \exp g \, \map \cos h\)
\(\ds \map v {r, \theta}\) \(=\) \(\ds \map \exp g \, \map \sin h\)

Then:

\(\ds \map f {r \, \map \exp {i \theta } }\) \(=\) \(\ds u + iv\)


We check the Polar Form of Cauchy-Riemann Equations.

As a preliminary:

\(\ds \frac {\partial g} {\partial r}\) \(=\) \(\ds \frac a r - c \cos \theta + d \sin \theta\)
\(\ds \frac {\partial h} {\partial r}\) \(=\) \(\ds \frac b r - d \cos \theta - \sin \theta\)
\(\ds \frac {\partial g} {\partial \theta}\) \(=\) \(\ds -b + c r \sin \theta + d r \cos \theta\)
\(\ds \frac {\partial h} {\partial \theta}\) \(=\) \(\ds a + d r \sin \theta - c r \cos \theta\)

Then:

\(\ds \frac {\partial u} {\partial r}\) \(=\) \(\ds \map \exp g \frac {\partial g} {\partial r} \map \cos h - \map \exp g \, \map \sin f \frac {\partial h} {\partial r}\)
\(\ds \) \(=\) \(\ds \frac 1 r \map \exp g \paren {\paren {a - r c \cos \theta + r d \sin \theta} \, \map \cos h + \map \sin h \paren {- b + r d \cos \theta + r c \sin \theta} }\)
\(\ds \) \(=\) \(\ds \frac 1 r \map \exp g \paren {\frac {\partial h} {\partial \theta} \map \cos h + \, \map \sin h \frac {\partial g} {\partial \theta} }\)
\(\ds \) \(=\) \(\ds \frac 1 r \frac {\partial v} {\partial \theta}\)
\(\ds \frac {\partial u} {\partial \theta}\) \(=\) \(\ds \map \exp g \frac {\partial g} {\partial \theta} \, \map \cos h - \map \exp g \, \map \sin h \frac {\partial h} {\partial \theta}\)
\(\ds \) \(=\) \(\ds -r \, \map \exp g \paren {\paren {\frac b r - c \sin \theta - d \cos \theta} \, \map \cos h + \map \sin h \paren {\frac a r + d \sin \theta - c \cos \theta} }\) Sine Function is Odd
\(\ds \) \(=\) \(\ds -r \, \map \exp g \paren {\frac {\partial h} {\partial r} \, \map \cos h + \map \sin h \frac {\partial g} {\partial r} }\)
\(\ds \) \(=\) \(\ds -r \frac {\partial v} {\partial r}\)

$\blacksquare$

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