For example, the square function is defined as $w=f(z)=z^2=r^2e^{i2\theta}$, where r>0 and $-\pi<\theta\leq \pi$. Why is the origin always excluded? Also, why do we use the branch cut on the negative real axis for the square root function when $f$ is also one-to-one on the the larger $D=\{re^{i\theta}:\text{$r>0 \space and -\pi/2<\theta\leq \pi/2$}\}$?
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2The origin need not be excluded, but its excluded often because $\theta$ is undefined for $z=0$. – Simply Beautiful Art Oct 20 '16 at 21:32
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Ah, thank you. I am still wondering about the branch cut. – SihOASHoihd Oct 20 '16 at 21:35
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I don't understand your question about the branch cut for the square root; did you write it wrong? – Oct 20 '16 at 21:53
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I'm defining the square root function as the inverse of $f$. – SihOASHoihd Oct 20 '16 at 22:00
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One can of course define the principal square root function on the negative real axis -- for example by deciding that it will produce a value with positive imaginary part.
In fact one often does that.
However, this produces a function that is not continuous everywhere in its domain -- some some authors prefer to define it with a slightly smaller domain such that the function has an open domain and is continuous and differentiable everywhere in that domain.
We could get an everywhere-continuous function just by excluding the negative reals, but the domain wouldn't be open unless we also exclude the branch point itself. And since most of the theorems in complex analysis are about functions defined on open domains (that are differentiable everywhere in their domain) it is convenient to define and name a function of that shape, rather than a slightly-more-defined function that doesn't match the assumptions of our theorems.
hmakholm left over Monica
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Since $f$ is one-to-one on D we don't need to "decide" what the inverses of f are on the negative x-axis. For example, the $sqrt{-1=i}$. But I haven't thought about the issues with continuity, thanks. – SihOASHoihd Oct 20 '16 at 21:52
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@Michael: The decision may be encoded by whether you take $-\frac\pi2 < \theta \le \frac\pi2$ or $-\frac\pi2 \le\theta < \frac\pi2$ when you select your $D$. – hmakholm left over Monica Oct 20 '16 at 21:59
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In complex analysis, there's a tension between the notion of function and the notion of multi-valued function.
On the one hand, a lot of the things you work with — like the complex square root — are naturally multi-valued functions. On the other hand, one has a background in studying single-valued functions that one would like to apply, and one tends to look for ways to reduce the study of multi-valued functions to the study of single-valued functions.
Excluding the origin from the domain of the square root is an example of something that makes more sense in the multi-valued picture — the multi-valued square root is analytic everywhere except at the origin and at infinity, so that set of points is a good choice of domain.
Although one could continuously extend it to $0$ and to $\infty$, the square root has a nasty singularity at both points, so one tends not to do so.
Furthermore (or alternatively, as a reason why the singularity is so nasty), some of our tools for reducing multi-valued functions to single-valued functions simply don't work at $0$ and $\infty$ — for example, for any point $z$, one often likes to have a whole open set $z \in U$ where each "branch" of the multi-valued function can be expressed as an ordinary single-valued function.