Holomorphic Functions

From a Derivative at One Point to a Property on an Open Set

Complex Differentiability focused on the limit defining a derivative at one point. Holomorphicity is the corresponding property on an open set: the function must be complex differentiable at every point of that set. The distinction matters because a derivative at one point gives only pointwise information, while holomorphicity describes how a function behaves throughout a region.

The open-set requirement is built into the definition. At each point where a complex derivative is taken, sufficiently small complex increments must remain in the set. An open set guarantees this at every one of its points. We will use “holomorphic” for functions on open sets, whether or not those sets are connected. Some texts reserve “domain” for a connected open set; connectedness is not needed for the results here.

Definition (Holomorphic Function): Let \(U\subseteq\mathbb{C}\) be open and let \(f:U\to\mathbb{C}\). The function \(f\) is holomorphic on \(U\) if it is complex differentiable at every point \(z_0\in U\). It is holomorphic at \(z_0\) if it is complex differentiable at that point.

Definition (Entire Function): A function \(f:\mathbb{C}\to\mathbb{C}\) is entire if it is holomorphic on all of \(\mathbb{C}\).

Thus, “holomorphic at \(z_0\)” is a pointwise statement, whereas “holomorphic on \(U\)” requires the derivative to exist at every point of \(U\). A function may be differentiable at one point without being holomorphic on any open neighborhood of that point. The example \(f(z)=|z|^2\) from Complex Differentiability illustrates this distinction: it is complex differentiable at the origin but not at any nonzero point.

Holomorphicity Is a Local Property

To decide whether a function is holomorphic on a set, it is enough to study it near each point. Conversely, if the function is holomorphic on the members of an open cover, then it is holomorphic on their union. This follows because a derivative at a point depends only on function values in an arbitrarily small neighborhood of that point.

Theorem (Locality of Holomorphicity): Let \(U\subseteq\mathbb{C}\) be open, and suppose \(U\) is covered by open sets \(V_\alpha\), so that \(U=\bigcup_\alpha V_\alpha\). A function \(f:U\to\mathbb{C}\) is holomorphic on \(U\) if and only if its restriction to each \(V_\alpha\) is holomorphic on \(V_\alpha\).

Proof. Suppose first that \(f\) is holomorphic on \(U\). Fix an index \(\alpha\). Every point of \(V_\alpha\) lies in \(U\), and the difference quotient for the restriction \(f|_{V_\alpha}\) is the same as the difference quotient for \(f\). Since \(f\) is complex differentiable at each such point, \(f|_{V_\alpha}\) is holomorphic on \(V_\alpha\).

For the reverse implication, suppose \(f|_{V_\alpha}\) is holomorphic on \(V_\alpha\) for every \(\alpha\). Fix any \(z_0\in U\). Because the sets \(V_\alpha\) cover \(U\), there is an index \(\alpha\) with \(z_0\in V_\alpha\). The set \(V_\alpha\) is open, so all sufficiently small increments \(h\) satisfy \(z_0+h\in V_\alpha\). On those increments, the difference quotient for \(f\) on \(U\) is exactly the difference quotient for \(f|_{V_\alpha}\). The latter has a limit as \(h\to0\), since \(f|_{V_\alpha}\) is holomorphic there. Thus \(f\) is complex differentiable at \(z_0\). As \(z_0\) was arbitrary, \(f\) is holomorphic on \(U\). \(\square\)

The theorem lets us use convenient neighborhoods, such as disks, when checking holomorphicity. It also explains why a function’s domain is part of the claim: a formula might define a holomorphic function on one open set but fail to define a function at all points of a larger set.

Worked Example: A Polynomial Is Entire

Consider \(p(z)=z^3-2z+5\). Fix an arbitrary \(z_0\in\mathbb{C}\). For \(h\neq0\), expand the numerator of the difference quotient:

\begin{aligned} \frac{p(z_0+h)-p(z_0)}{h} &=\frac{(z_0+h)^3-2(z_0+h)+5-(z_0^3-2z_0+5)}{h}\\ &=\frac{3z_0^2h+3z_0h^2+h^3-2h}{h}\\ &=3z_0^2+3z_0h+h^2-2. \end{aligned}

As \(h\to0\) through complex values, this expression tends to \(3z_0^2-2\). The calculation holds for every \(z_0\in\mathbb{C}\), so \(p\) is complex differentiable everywhere and hence entire, with \(p'(z)=3z^2-2\). The same reasoning applies to any polynomial: its difference quotient reduces to a polynomial in \(h\) whose limit exists at every point.

Taking Reciprocals

A reciprocal is defined only where the original function is nonzero. On that set, taking a reciprocal preserves holomorphicity. The derivative formula can be obtained directly from the difference quotient; the continuity of a complex-differentiable function, established in Complex Differentiability, supplies the limit needed in the denominator.

Theorem (Reciprocal of a Holomorphic Function): Let \(U\subseteq\mathbb{C}\) be open, and let \(f:U\to\mathbb{C}\) be holomorphic with \(f(z)\neq0\) for every \(z\in U\). Then \(1/f\) is holomorphic on \(U\), and \[ \left(\frac{1}{f}\right)'(z)=-\frac{f'(z)}{f(z)^2} \] for every \(z\in U\).

Proof. Fix \(z_0\in U\), and write \(f_0=f(z_0)\), which is nonzero by hypothesis. For nonzero \(h\) sufficiently small that \(z_0+h\in U\), the reciprocal difference quotient is

\frac{\dfrac{1}{f(z_0+h)}-\dfrac{1}{f_0}}{h} = -\frac{f(z_0+h)-f_0}{h\,f(z_0+h)f_0}.

Since \(f\) is complex differentiable at \(z_0\), it is continuous there, so \(f(z_0+h)\to f_0\) as \(h\to0\). The difference quotient \(\bigl(f(z_0+h)-f_0\bigr)/h\) tends to \(f'(z_0)\), and the product \(f(z_0+h)f_0\) tends to \(f_0^2\neq0\). Taking the limit gives

\left(\frac{1}{f}\right)'(z_0)=-\frac{f'(z_0)}{f_0^2}.

Thus \(1/f\) is complex differentiable at the arbitrary point \(z_0\), and therefore holomorphic on \(U\). \(\square\)

Worked Example: A Reciprocal on a Punctured Plane

Let \(f(z)=1/z\) on \(U=\mathbb{C}\setminus\{0\}\). The set \(U\) is open, and \(f\) has no zeros there. Apply the reciprocal theorem to the function \(g(z)=z\), which is complex differentiable everywhere with \(g'(z)=1\). At every \(z\neq0\),

f'(z)=\left(\frac{1}{g(z)}\right)' =-\frac{g'(z)}{g(z)^2} =-\frac{1}{z^2}.

This proves that \(1/z\) is holomorphic on \(\mathbb{C}\setminus\{0\}\). It is not entire: its formula is undefined at zero, so it is not a function from all of \(\mathbb{C}\) to \(\mathbb{C}\). The derivative calculation also shows why the excluded point matters: \(-1/z^2\) is not defined there.

Rational Functions and Their Domains

The reciprocal theorem combines with the sum and product rules for complex derivatives, proved in Complex Differentiability. In particular, a quotient of holomorphic functions is holomorphic wherever the denominator does not vanish. A rational expression should therefore always be paired with its domain, rather than judged solely by its algebraic appearance.

Worked Example: A Rational Function Away from Its Pole

Consider

r(z)=\frac{z+1}{z-3}.

Its natural domain is \(U=\mathbb{C}\setminus\{3\}\). Rewrite the numerator as \(z+1=(z-3)+4\). For \(z\neq3\),

r(z)=1+\frac{4}{z-3}.

The function \(z\mapsto z-3\) is holomorphic and nonzero on \(U\), so its reciprocal is holomorphic there by the reciprocal theorem. Multiplication by \(4\) and addition of the constant \(1\) preserve holomorphicity by the sum and product rules. The derivative is

r'(z)=4\left(-\frac{1}{(z-3)^2}\right) =-\frac{4}{(z-3)^2},\qquad z\neq3.

For a direct algebra check, differentiating the quotient by the algebraic expression gives

\frac{(z-3)-(z+1)}{(z-3)^2} =\frac{-4}{(z-3)^2},

which agrees with the derivative obtained from the reciprocal theorem. The point \(z=3\) must remain excluded: the original quotient has zero denominator there.

Worked Example: Restricting a Holomorphic Function to a Smaller Open Set

The function \(f(z)=1/z\) is holomorphic on \(U=\mathbb{C}\setminus\{0\}\), as shown above. Now let \(V=\{z\in\mathbb{C}:|z-2|<1\}\). This is an open disk contained in \(U\): if \(z\in V\), then the triangle inequality gives

2=|2|\leq |2-z|+|z|<1+|z|,

so \(|z|>1\), in particular \(z\neq0\). The locality theorem implies that the restriction of \(f\) to \(V\) is holomorphic. Its derivative there is \(f'(z)=-1/z^2\). This example emphasizes that holomorphicity is assessed relative to an open set: the same formula is holomorphic on the punctured plane and on this disk, but it does not define an entire function.

What Holomorphicity Does—and Does Not—Say

The definition requires a complex derivative at every point in an open set. It does not merely require the function to be continuous, and it does not follow from having a derivative at one chosen point. Complex differentiability at a point implies continuity at that point, as established earlier, so a holomorphic function is continuous throughout its open domain. The converse is not part of the definition and is false in general.

It is also important not to confuse holomorphicity with the Cauchy–Riemann equations at a single point. Complex Differentiability established those equations as necessary conditions for a derivative at that point, but also showed that they can hold without a complex derivative. The next tutorial studies the equations further and develops conditions under which they can be used as a test.

For now, a reliable first check is to identify the open set and then verify complex differentiability at every point in it. Polynomial functions are entire; reciprocals are holomorphic where their denominators do not vanish; and rational functions are holomorphic away from their poles. The locality theorem permits these checks to be made on smaller open neighborhoods whenever that is more convenient.

Check Your Understanding

Use the definitions and results in this tutorial to answer the following questions.

What additional requirement distinguishes a function holomorphic on an open set from one complex differentiable at a single point?
State the locality theorem for holomorphicity and explain why openness is needed in its proof.
Under what condition on a holomorphic function \(f\) is \(1/f\) holomorphic, and what is its derivative?
On what set is \(1/(z+2i)\) holomorphic? Use the reciprocal theorem to compute its derivative there.
Why is \((z+1)/(z-3)\) holomorphic on \(\mathbb{C}\setminus\{3\}\) but not an entire function?

What You'll Learn