Lesson 6: Power Series Representations & Analytic Continuation

Lesson 5a built the real-analysis machinery (bounds, rigorous limits, the ML-estimate) specifically so this lesson could use it for real. Here it gets spent on the biggest structural fact in the subject: every holomorphic function is, locally, its own power series. Not approximated by one, the way a real Taylor series merely tracks a smooth function: equal to one, exactly, on an entire disk. Everything else in this lesson (isolated zeros, the identity theorem, analytic continuation, the classification of singularities) falls out of taking that equality seriously.

The radius of convergence, formalized

Lesson 3 used the ratio test informally to find how far a series converges. Here it gets a real derivation: comparing consecutive terms of $\sum c_n(z-z_0)^n$ against a geometric series tail (exactly Lesson 5a's comparison-test machinery) gives

$$R = \frac{1}{L}, \qquad L = \lim_{n\to\infty}\left|\frac{c_{n+1}}{c_n}\right|$$

with $R=\infty$ if $L=0$ and $R=0$ if $L=\infty$. Checked against $e^z=\sum z^n/n!$: $L=0$, so $R=\infty$, matches Lesson 3 exactly. Two new examples confirm the pattern: $\sum nz^n$ has $L=1$, so $R=1$; $\sum z^n/n^2$ has $\frac{n^2}{(n+1)^2}\to1$, so $R=1$ as well, even though the two series look quite different termwise.

Taylor's theorem

Start from the Cauchy Integral Formula and rewrite the kernel $\frac{1}{w-z}$ to expose a geometric series. Setting $r=\frac{z-z_0}{w-z_0}$ (with $z$ interior to a circle $\gamma$ and $w$ on it, so $|r|<1$),

$$\frac{1}{w-z} = \frac{1}{(w-z_0)-(z-z_0)} = \frac{1}{w-z_0}\cdot\frac{1}{1-r} = \frac{1}{w-z_0}\sum_{n=0}^\infty r^n$$

Substituting back into CIF and swapping the sum and integral (legitimate here because the series converges uniformly on $\gamma$, an imported fact, same tier as Lesson 5's differentiation-under-the-integral-sign) turns $f(z)=\frac{1}{2\pi i}\oint_\gamma\frac{f(w)}{w-z}\,dw$ into a power series in $(z-z_0)$, whose $n$-th coefficient is exactly the Cauchy differentiation formula's integral from Lesson 5. That gives Taylor's theorem:

$$f(z) = \sum_{n=0}^\infty \frac{f^{(n)}(z_0)}{n!}(z-z_0)^n$$

for every $z$ inside any circle centered at $z_0$ on which $f$ is holomorphic. This is a genuinely different statement from the real case: holomorphic and analytic turn out to be the same property, with no real-variable analogue at all. A real function can be infinitely differentiable and still fail to equal its own Taylor series (the standard example is $e^{-1/x^2}$); nothing like that can happen for a holomorphic function.

Key takeaway: Taylor's theorem: $f(z)=\displaystyle\sum_{n=0}^\infty \frac{f^{(n)}(z_0)}{n!}(z-z_0)^n$ on any disk where $f$ is holomorphic. Holomorphic $\iff$ analytic: an equivalence with no real-variable counterpart.

A genuine mystery, resolved. $f(z)=\frac{1}{1+z^2}$ is smooth on all of $\mathbb{R}$, so why does its familiar real Taylor series $1-x^2+x^4-\cdots$ mysteriously stop converging at $|x|=1$? Nothing about the real graph explains it. The resolution only appears once $f$ is viewed as a complex function: it fails to be holomorphic exactly at $z=\pm i$, both at distance $1$ from the origin, invisible to anything living on the real axis, but fully determinative of where the series can converge. The same distance-to-singularity logic, checked directly (no series expansion needed): the radius of convergence of $\frac{1}{z-3}$ centered at $0$ is $|3-0|=3$; centered at $1+i$, it's $|3-(1+i)|=|2-i|=\sqrt5$.

Zeros are isolated

Suppose $f$ is holomorphic near $z_0$ and $f(z_0)=0$. Since $f$ equals its Taylor series near $z_0$, and every term but the constant one vanishes at $z=z_0$, $f(z_0)=0$ forces $c_0=0$. Two exhaustive cases follow: either every coefficient is zero ($f\equiv0$ on the whole disk, a dead end), or there's a smallest nonzero coefficient at some index $m$ (the order of the zero). Factoring it out:

$$f(z) = (z-z_0)^m\underbrace{\Big[c_m+c_{m+1}(z-z_0)+c_{m+2}(z-z_0)^2+\cdots\Big]}_{=:\,g(z)} = (z-z_0)^m\,g(z)$$

with $g$ itself holomorphic near $z_0$ and $g(z_0)=c_m\ne0$. (Worked instance to keep in view: $f(z)=z^2+z^3=z^2(1+z)$, so $z_0=0$, $m=2$, $g(z)=1+z$, $g(0)=1$.)

Claim: $g$ is nonzero on some whole disk around $z_0$, which, combined with $(z-z_0)^m$ vanishing only at $z_0$ itself, means $f$'s zero at $z_0$ is isolated: nothing else nearby is a zero.

Proof. $g$ is holomorphic at $z_0$, hence continuous there: for every $\varepsilon>0$ there's a $\delta>0$ such that $|z-z_0|<\delta\implies|g(z)-g(z_0)|<\varepsilon$. Choose $\varepsilon=\frac{|g(z_0)|}{2}$ (legal, since $g(z_0)\ne0$ makes this a genuine positive number). (Why half the distance and not the full distance $|g(z_0)|$: for the toy instance, $\varepsilon=|g(0)|=1$ gives a matching $\delta=1$, but $g$'s actual zero at $z=-1$ sits exactly on that disk's boundary: $|g(z)|\to0$ as $z\to-1$ from inside, so there's no genuine floor. Halving the distance instead guarantees one.)

Now the reverse triangle inequality, in the form that bounds $|g(z)|$ below:

$$|g(z_0)|-|g(z)| \;\le\; |g(z)-g(z_0)| \quad\Longrightarrow\quad |g(z)| \;\ge\; |g(z_0)|-|g(z)-g(z_0)|$$

Since $|g(z)-g(z_0)|<\varepsilon=\frac{|g(z_0)|}{2}$,

$$|g(z)| \;\ge\; |g(z_0)|-|g(z)-g(z_0)| \;>\; |g(z_0)|-\frac{|g(z_0)|}{2} \;=\; \frac{|g(z_0)|}{2} \;>\;0$$

so $g$ has a hard positive floor of $\frac{|g(z_0)|}{2}$ on the whole disk $|z-z_0|<\delta$, never zero there. $\blacksquare$

Isolated Zeros Theorem: if $f$ is holomorphic near $z_0$, $f(z_0)=0$, and $f\not\equiv0$ near $z_0$, there is a $\delta>0$ with $f(z)\ne0$ for all $z$ satisfying $0<|z-z_0|<\delta$.

Play with the toy instance directly: drag $z$, watch $g(z)$ track through the output plane, and see the $\delta$-disk certify a hard floor under $|g|$:

Key takeaway: Order of a zero: if $f(z_0)=0$ and $f\not\equiv0$ nearby, the smallest $m$ with $c_m\ne0$ gives $f(z)=(z-z_0)^m g(z)$ with $g(z_0)\ne0$. Isolated zeros: zeros of a not-identically-zero holomorphic function never accumulate; each one has a punctured neighborhood entirely free of other zeros.

The identity theorem

If two holomorphic functions agree on a set with an accumulation point in the domain (a point every neighborhood of which contains another point of the set), do they agree everywhere? Set $F=f-h$ and ask the same question of $F$'s zeros instead.

Since zeros of $F$ accumulate at some point $p$, there's a sequence $z_n\to p$ (each $z_n\ne p$) with $F(z_n)=0$; $F$ is continuous, so $F(p)=\lim F(z_n)=0$ too. Apply the isolated-zeros dichotomy to $F$ at $p$: either $F\equiv0$ on a whole disk around $p$, or $p$ is an isolated zero. Suppose the second, for contradiction: the Isolated Zeros Theorem then guarantees a punctured disk around $p$ with no other zeros in it. But $p$ being an accumulation point means every neighborhood of $p$, including that exact punctured disk, contains another zero. Contradiction, so $F\equiv0$ on a whole disk around $p$, not merely at the point.

Spreading the zero set to the whole domain. Since the domain is connected, any two points $p,q$ in it can be joined by a continuous path; cover the path with a chain of overlapping disks small enough that $F$ is holomorphic on each. $F\equiv0$ on the first disk, centered at $p$. Pick the next disk's center $z_1$ inside the first disk. Since the whole first disk is zero, $z_1$ is completely surrounded by zeros of $F$, making it an accumulation point of zeros for free. So $z_1$ can't be an isolated zero: the same dichotomy forces $F\equiv0$ on the entire second disk. Repeat along the chain to $q$. Since $q$ was arbitrary, $F\equiv0$ on the whole domain: $f\equiv h$.

Key takeaway: Identity theorem: if $f,h$ are holomorphic on a connected domain and agree on a set with an accumulation point that lies in the domain, $f\equiv h$ on that whole domain (not beyond it, nothing in the argument ever leaves it).

Uniqueness of analytic continuation. If a holomorphic extension of $f$ to a larger connected domain exists, it is the only one. Suppose $h_1,h_2$ both extend $f$ to a larger connected domain $\Omega'\supseteq\Omega$. Let $H=h_1-h_2$; both agree with $f$ on $\Omega$, so $H\equiv0$ there, and $\Omega$, being open, trivially has an accumulation point (every point of an open set is surrounded entirely by other points of it). The identity theorem then forces $H\equiv0$ on all of $\Omega'$: $h_1\equiv h_2$.

A non-example, closing the loop. $\sin(1/z)$ has zeros at $z=\frac{1}{k\pi}$ for every integer $k$, accumulating at $z=0$ itself. That single fact rules out any holomorphic extension of $\sin(1/z)$ to $z=0$: the identity theorem would force it to be identically zero everywhere, contradicting the function's actual behavior away from $0$.

Why $R$ equals the distance to the nearest singularity, derived

Lesson 3 stated the radius-of-convergence rule as a fact to be believed. It's worth deriving, using the same shrinking-circle idea from Lesson 4's proof of CIF: recall that CIF needs $f$ holomorphic on and inside $\gamma$ precisely because the deformation argument sweeps through that whole interior region.

Toy instance: $f(z)=\frac{1}{z-3}$, expanded around $z_0=0$, with its only singularity at $3$.

One series, many circles. For any point $z$ with $|z|<3$, there's an admissible circle certifying convergence there: for $z=2.9$, the circle of radius $2.95$ works (holomorphic on and inside it, since the only singularity is at $3$); for $z=-2.99$, radius $2.999$ works. Crucially, the circle appears in neither coefficient formula: not $c_n=f^{(n)}(z_0)/n!$, nor the Cauchy differentiation integral $\frac{1}{2\pi i}\oint_\gamma\frac{f(w)}{(w-z_0)^{n+1}}\,dw$. It's one fixed series regardless of which admissible circle certifies convergence at a given point. Since every $z$ with $|z|<3$ admits some admissible circle, the series converges on the entire disk $|z|<3$: $R\ge3$.

Closing it from the other side. Suppose instead $R>3$: then the series, and hence $f$ itself, would have to stay bounded on some disk that reaches past the singularity at $z=3$. But $f$ visibly blows up approaching it: $|f(2.7)|\approx3.3$, $|f(2.97)|\approx33$, $|f(2.997)|\approx333$: no finite bound exists. Contradiction, so $R\le3$ as well. Together: $R=3$, exactly.

The honest puzzle. Could $R$ ever exceed the distance to the nearest singularity? Not for $\frac{1}{z-3}$, but consider $g(z)=\frac{\sin z}{z}$ expanded around $z_0=1$: the formula has a singularity at $0$, distance $1$ from the center, yet the series matches $g(-5)$ (distance $6$) to dozens of digits. The vary-the-circle argument above only ever proves $R\ge$ distance to the nearest genuine singularity; whether it's an equality depends on what kind of singularity is actually there. $z=0$ isn't a real obstruction for $\frac{\sin z}{z}$ at all. That's exactly what the next section explains.

Key takeaway: $R\ge$ distance to the nearest singularity, always (the vary-the-circle argument). $R=$ that distance exactly when the blocking singularity is genuine (a pole or an essential singularity); a removable singularity, as the name suggests, doesn't block $R$ at all.

Laurent series and the three kinds of singularity

Three functions, each computed directly at $z_0=0$:

This is a Laurent series: $f(z)=\sum_{n=-\infty}^{\infty}c_n(z-z_0)^n$ on a punctured disk (or more generally an annulus), same coefficient formula as Taylor's, now legal for negative $n$ too. The principal part (the negative-power terms) is exactly what classifies the singularity:

Closing two earlier loops. $\sin(1/z)=\dfrac1z-\dfrac{1}{6z^3}+\dfrac{1}{120z^5}-\cdots$ is essential, which is exactly why its zeros, at $z=\frac{1}{k\pi}$, were free to accumulate at $0$ back in the identity-theorem section: an essential singularity is precisely the kind of local chaos that makes that possible. And $\frac{1}{z-3}$'s pole is exactly the "genuine singularity" the radius-of-convergence section needed: a removable singularity never blows up, so it never blocks $R$ at all.

A general shortcut, for a function shaped as (something)/(power of $z-z_0$): compare orders of vanishing. $\dfrac{\cos z-1}{z^2}$ looks like a pole of order $2$ from the denominator alone, but the numerator $\cos z-1$ also vanishes to order $2$ at $z=0$, exactly canceling: the singularity is removable, with $f(0)=-\frac12$. Contrast $\dfrac{e^z-1}{z^2}$: the numerator vanishes only to order $1$ ($e^z-1=z+\frac{z^2}{2}+\cdots$), leaving one uncanceled factor of $z^{-1}$: a pole of order $1$. In general: pole order = denominator's order of vanishing $-$ numerator's order of vanishing, when that's positive; removable otherwise.

Key takeaway: Laurent series: $f(z)=\sum_{n=-\infty}^\infty c_n(z-z_0)^n$ on an annulus. Classification by the principal part: removable (none), pole of order $m$ (finitely many, deepest $-m$), essential (infinitely many). For $f=\text{numerator}/(z-z_0)^k$: pole order $=k-$(numerator's own order of vanishing), when positive; removable otherwise.

Lesson 5's chain of results all flowed from one integral formula. This lesson's flows from one equality (holomorphic functions are their Taylor series), and that single fact turns out to pin down where a function's zeros can live, how many extensions it can have, and exactly what a singularity is measuring. Next: residues, where the single coefficient $c_{-1}$ of a Laurent series turns out to be worth computing contour integrals for.