4.5 Analytic Functions
Definition 4.35. A complex function \(w = f(z)\) is said to be analytic at a point \(z_0\) if \(f\) is differentiable at \(z_0\)
and at every point in some neighbourhood of \(z_0\).
A function \(f\) is analytic in a domain \(D\) if it is analytic at every point in \(D\). A function that is analytic throughout a domain \(D\) is called holomorphic or regular.
Definition 4.36. A function that is analytic at every point \(z\) in the complex plane is said to be an entire function.
- 1.
- A polynomial function \(p(z) = a_nz^n + a_{n - 1}z^{n - 1} + \cdots + a_1z + a_0\,\) where \(n\) is a non negative integer is an entire function.
- 2.
- A rational function \(\frac {p(z)}{q(z)}\) , where \(p\) and \(q\) are polynomial functions is analytic in any domain \(D\)
that contains no point \(z_0\) for which \(q(z_0) = 0\)
Proof.
- 1.
- The constant function and the identity \(f(z)=z\) are differentiable at every point, with derivatives \(0\) and \(1\), directly from the difference quotient. The sum and product rules — proved for complex functions exactly as for real ones, the difference quotient having the same algebra — then make every polynomial differentiable at every point of \(\mathbb {C}\). Differentiable everywhere is what entire means.
- 2.
- Let \(D\) contain no zero of \(q\). At any \(z_0\in D\) both \(p\) and \(q\) are differentiable and \(q(z_0)\neq 0\), so the quotient rule gives \[\left (\frac {p}{q}\right )'(z_0) =\frac {p'(z_0)q(z_0)-p(z_0)q'(z_0)}{q(z_0)^2},\] which exists. So \(p/q\) is differentiable throughout \(D\), that is, analytic there.
Singular Points
Since rational functions \(\, f(z) = \frac {4z}{z^2 - 2z + 2}\,\) is discontinuous at \(z_1 = 1 + i\) and \(z_2 = 1 - i\) and \(f\) fails to be analytic at these points. Thus, we
call these points as singular points.
In general, a point \(z\) at which a complex function \(w = f(z)\) fails to be analytic is called singular point of
\(f\).
Theorem 4.38. If \(f\) is differentiable at a point \(z_0\) in a domain \(D\), then \(f\) is continuous at \(z_0\).
Proof. The limits \(\,\lim \limits _{z \rightarrow z_0} \frac {f(z) - f(z_0)}{z - z_0}\) and \(\lim \limits _{z \rightarrow z_0} (z - z_0)\) exists and equal \(f'(z_0)\) and 0 respectively. Now \begin {align*} \lim \limits _{z \rightarrow z_0} (f(z) - f(z_0)) & = \lim \limits _{z \rightarrow z_0} \Big (\frac {f(z) - f(z_0)}{z - z_0}\Big )(z - z_0)\\\\ & = \lim \limits _{z \rightarrow z_0} \Big (\frac {f(z) - f(z_0)}{z - z_0}\Big )\cdot \lim \limits _{z \rightarrow z_0} (z - z_0)\\\\ & = f'(z_0) \cdot 0 = 0 \end {align*}
From \(\lim \limits _{z \rightarrow z_0} (f(z) - f(z_0)) = 0\), we conclude that \(\lim \limits _{z \rightarrow z_0} f(z) = f(z_0)\). Hence \(f\) is continuous at \(z_0\). □
Theorem 4.39 (L’Hôpital’s Rule). Suppose \(f\) and \(g\) are functions that are analytic at a point \(z_0\) and \(f(z_0) = 0,\quad g(z_0) = 0\) but \(g'(z_0) \neq 0\). Then \[\lim \limits _{z \rightarrow z_0} \frac {f(z)}{g(z)} = \frac {f'(z_0)}{g'(z_0)}\]
Proof. Because \(f(z_0)=g(z_0)=0\), the difference quotients of \(f\) and \(g\) at \(z_0\) are simply \(f(z)/(z-z_0)\) and \(g(z)/(z-z_0)\). So for \(z\neq z_0\), \[\frac {f(z)}{g(z)} =\frac {\dfrac {f(z)-f(z_0)}{z-z_0}}{\dfrac {g(z)-g(z_0)}{z-z_0}} .\] As \(z\rightarrow z_0\) the numerator tends to \(f'(z_0)\) and the denominator to \(g'(z_0)\), both existing because \(f\) and \(g\) are analytic at \(z_0\). Since \(g'(z_0)\neq 0\) the quotient law applies, giving \[\lim _{z\rightarrow z_0}\frac {f(z)}{g(z)}=\frac {f'(z_0)}{g'(z_0)} .\] This is markedly simpler than the real version: no mean value theorem is needed, because the hypothesis \(g'(z_0)\neq 0\) does all the work. □
Example 4.40. Compute \(\,\lim \limits _{z \rightarrow (2 + i)}\frac {z^2 - 4z + 5}{z^3 - z - 10i}\)
Solution
If we identify \(f(z) = z^2 - 4z + 2\) and \(g(z ) = z^3 - z - 10i\), we see that \(f(2 + i) = 0\) and \(g(2 + i) = 0\). The given limit has the indeterminate form of \(\frac {0}{0}\).
Now, since \(f\) and \(g\) are analytic at \(z_0 = 2 + i\), we have
\(f'(z) = 2z - 4\implies f'(2 + i) = 2i,\quad g'(z) = 3z^2 - 1 \implies g'(2 + i) = 8 + 12i,\) we have \begin {align*} \lim \limits _{z \rightarrow (2 + i)} \frac {z^2 - 4z + 5}{z^3 - z - 10i} & = \frac {f'(2 + i)}{g'(2 + i)}= \frac {2i}{8 + 2i}\\\\ & = \frac {3}{26} + \frac {1}{13}i\\ \end {align*}
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