Why do we take $(1+z)^{\alpha}$ as $e^{\alpha \operatorname{Log}(1+z)}$?

Question

Let $\alpha$ be a complex number. Show that if $(1+z)^{\alpha}$ is taken as $e^{\alpha \operatorname{Log}(1+z)}$, then for $|z|< 1$ \begin{equation*} (1+z)^{\alpha} = 1 + \frac{\alpha}{1}z + \frac{\alpha(\alpha -1)}{1\cdot2}z^{2} + \frac{\alpha(\alpha-1)(\alpha-2)}{1\cdot2\cdot3}z^{3} + \dotsb \end{equation*}

Why do we take $(1+z)^{\alpha}$ as $e^{\alpha \operatorname{Log}(1+z)}$? I was thinking that this is to show that $(1+z)^{\alpha}$ is analytic for $|z|<1$. Since $e$ is entire and the composition of analytic functions is analytic, then we can show that $(1+z)^{\alpha}$ is analytic for $|z|<1$. But this requires that we show that $\alpha\operatorname{Log}(1+z)$ is analytic for $|z|<1$. I'm not sure how I would then go about showing that $\alpha\operatorname{Log}(1+z)$ is analytic for $|z|<1$.

Answer 1

That depends upon how you define it, but if you define it as$$z-\frac{z^2}2+\frac{z^3}3-\frac{z^4}4+\cdots,$$then, since the radius of convergence of this power series is $1$, its sum is indeed an analytic function.

Answer 2

To see this, maybe it's best to go back to basics:

Log$z$ is the principal value of the complex logarithm, and is defined as the inverse of the restriction of the analytic many-to-one function $z\mapsto e^z$ to the strip $-\pi<\Im z\le \pi.$

Background: in general, the exponential function sends horizontal lines to open half-lines emanating from the origin; it sends horizontal lines to circles centered at the origin. In fact, it maps bijectively each half-open horizontal strip of width $2\pi$ onto $\mathbb C\setminus \{0\}.$ So, in particular, if we restrict it to $-\pi<\Im z\le \pi$ , we get the principal logarithm Log$:C\setminus \{0\}\to -\pi<\Im z\le \pi$. It is easy to show that this function is analytic $except$ on the interval $\{-\infty<\Re z\le 0\}.$

To finish, note that if $|z|<1$ then $z+1\notin \{-\infty<\Re z\le 0\},$ so Log $(z+1)$ is analytic there.