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

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 $$|z|<1$$.

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.
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.
• To show that the function is analytic except on the interval $\{-\infty< \Re z \le 0\}$ do we just say that $-\operatorname{Log}|z| - i\pi$ is undefined and hence discontinuous for the principal log function $\operatorname{Log}$? – K.M Apr 6 at 18:20
• Log $is$ defined on the interval, except of course at $0$. But is is not continuous there: after defining Log as the inverse of the exponential restricted to the strip as in my answer, you show directly that Log is discontinuous on the interval: pick any point $\neq 0$ on the interval and see what happens when you approach it from $\Im >0$ and $\Im <0$. – Matematleta Apr 6 at 19:33