1 \documentclass{article}
7 \subsection{Definitions}
9 Definitions for power and log:
11 x^a \equiv e^{a \ln x}
14 \ln x \equiv \ln |x| + i \arg(x) \mbox{ where } -\pi < \arg(x) \le \pi
17 \subsection{General rules}
22 for arbitrary complex $x$ and $y$
25 x^{-a} = \frac{1}{x^a}
27 for arbitrary complex $x$ and $a$
29 \subsection{$(ax)^b=a^b x^b$}
31 \subsubsection{$b$ integer, $x$ and $a$ arbitrary complex}
36 (ax)^b & = & \underbrace{(ax) \cdots (ax)}_{b \times}
38 & = & \underbrace{a \cdots a}_{b \times}
39 \underbrace{x \cdots x}_{b \times}
41 & = & a^b x^b \mbox{ q.e.d.}
44 if $b<0$ (so $b=-|b|$)
46 (ax)^b & = & \frac{1}{(ax)^{|b|}}
48 & = & \frac{1}{a^{|b|} x^{|b|}}
50 & = & a^{-|b|} x^{-|b|}
55 \subsubsection{$a>0$, $x$ and $b$ arbitrary complex}
58 (ax)^b & = & e^{b \ln(ax)}
60 & = & e^{b (\ln |ax| + i \arg(ax))}
63 if $a$ is real and positive:
65 \ln |ax| = \ln |a| + \ln |x| = \ln a + \ln |x|
74 e^{b (\ln |ax| + i \arg(ax))} & = &
75 e^{b (\ln a + \ln |x| + i \arg(x))}
77 & = & e^{b (\ln a + \ln x)}
79 & = & e^{b \ln a} e^{b \ln x}
81 & = & a^b x^b \mbox{ q.e.d.}
84 \subsection{$(x^a)^b = x^{ab}$}
86 \subsubsection{$b$ integer, $x$ and $a$ arbitrary complex}
91 (x^a)^b & = & \underbrace{(x^a) \cdots (x^a)}_{b \times}
93 & = & \underbrace{e^{a \ln x} \cdots e^{a \ln x}}_{b \times}
95 & = & e^{\underbrace{\scriptstyle a \ln x + \dots + a \ln x}_{b \times}}
99 & = & x^{ab} \mbox{ q.e.d.}
102 if $b<0$ (so $b=-|b|$)
104 (x^a)^b & = & \frac{1}{(x^a)^{|b|}}
106 & = & \frac{1}{x^{a|b|}}
113 \subsubsection{$-1 < a \le 1$, $x$ and $b$ arbitrary complex}
117 x^a=e^{a \ln|x| + ia\arg(x)}
125 \arg(x^a)-a\arg(x)=2k\pi
127 now if $-1 < a \le 1$, then $-\pi < a\arg(x) \le \pi$,
132 (Note that for $a=-1$ this may not be true, as $-1 \arg(x)$ may be equal to $-\pi$.)
135 \ln(x^a) & = & \ln|x^a| + i\arg(x^a)
137 & = & \ln (e^{a\ln|x|})+ia\arg(x)
139 & = & a \ln |x| + ia\arg(x) \mbox{ (because $a\ln|x|$ is real)}
145 (x^a)^b & = & e^{b\ln x^a}
149 & = & x^{ab} \mbox{ q.e.d.}
152 proof contributed by Adam Strzebonski from Wolfram Research
153 ({\tt adams@wolfram.com}) in newsgroup {\tt sci.math.symbolic}.