Complex Power by Complex Exponential is Analytic
== 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 $f$: {{begin-eqn}} {{eqn | l = \map f {r \, \map \exp {i \theta} }       | r = z^\psi \, \map \exp {-\eta z} }} {{eqn | r = \paren {r e^{i \theta} }^\psi \, \map \exp {-\eta r \, \map \exp {i \theta} } }} {{eqn | r = \map \exp {\psi \, \map \Log {r \, \map \exp {i \theta} } } \, \map \exp {-\eta r \paren {\cos \theta + i \sin \theta} } }} {{eqn | r = \map \exp {\psi \ln r + i \psi \theta - \eta r \cos \theta - i \eta r \sin \theta} }} {{eqn | r = \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} }} {{eqn | r = \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} }} {{eqn | r = \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} } }} {{end-eqn}} Define: {{begin-eqn}} {{eqn | l = \map g {r, \theta}       | r = a \ln r - b \theta - c r \cos \theta + d r \sin \theta }} {{eqn | l = \map h {r, \theta}       | r = b \ln r + a \theta - d r \cos \theta - c r \sin \theta }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \map f {r \, \map \exp {i \theta} }       | r = \map \exp g \, \map \exp {i h} }} {{eqn | r = \map \exp g \, \map \cos h + i \map \exp g \, \map \sin h }} {{end-eqn}} Define: {{begin-eqn}} {{eqn | l = \map u {r, \theta}       | r = \map \exp g \, \map \cos h }} {{eqn | l = \map v {r, \theta}       | r = \map \exp g \, \map \sin h }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \map f {r \, \map \exp {i \theta } }        | r = u + iv }} {{end-eqn}} We check the Polar Form of Cauchy-Riemann Equations. As a preliminary:  {{begin-eqn}} {{eqn | l = \frac {\partial g} {\partial r}       | r = \frac a r - c \cos \theta + d \sin \theta }} {{eqn | l = \frac {\partial h} {\partial r}       | r = \frac b r - d \cos \theta - \sin \theta }} {{eqn | l = \frac {\partial g} {\partial \theta}       | r = -b + c r \sin \theta + d r \cos \theta }} {{eqn | l = \frac {\partial h} {\partial \theta}       | r = a + d r \sin \theta - c r \cos \theta }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \frac {\partial u} {\partial r}       | r = \map \exp g \frac {\partial g} {\partial r} \map \cos h - \map \exp g \, \map \sin f \frac {\partial h} {\partial r} }} {{eqn | r = \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} } }} {{eqn | r = \frac 1 r \map \exp g \paren {\frac {\partial h} {\partial \theta} \map \cos h + \, \map \sin h \frac {\partial g} {\partial \theta} } }} {{eqn | r = \frac 1 r \frac {\partial v} {\partial \theta} }} {{eqn | l = \frac {\partial u} {\partial \theta}       | r = \map \exp g \frac {\partial g} {\partial \theta} \, \map \cos h - \map \exp g \, \map \sin h \frac {\partial h} {\partial \theta} }} {{eqn | r = -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} }       | c = Sine Function is Odd }} {{eqn | r = -r \, \map \exp g \paren {\frac {\partial h} {\partial r} \, \map \cos h + \map \sin h \frac {\partial g} {\partial r} } }} {{eqn | r = -r \frac {\partial v} {\partial r} }} {{end-eqn}}