Section 3.1 Differentiable and Analytic Functions
Using our imagination, we take our lead from elementary calculus and define the derivative of \(f\) at \(z_0\text{,}\) written \(f\,'(z_0)\text{,}\) by
\begin{equation}
f\,'(z_0) = \lim\limits_{z \to z_0}\frac{f(z) -f(z_0)}{z-z_0}\text{,}\tag{3.1.1}
\end{equation}
provided the limit exists. If it does, we say that the function
\(f\) is differentiable at
\(z_0\text{.}\) If we write
\(\Delta z=z-z_0\text{,}\) then we can express Equation
(3.1.1) in the form
\begin{equation}
f\,'(z_0) = \lim\limits_{\Delta z \to 0}\frac{f(z_0+\Delta z) -f(z_0)}{\Delta z}\text{.}\tag{3.1.2}
\end{equation}
If we let \(w=f(z)\) and \(\Delta w=f(z)-f(z_0)\text{,}\) then we can use the Leibniz notation \(\frac{dw}{dz}\) for the derivative:
\begin{equation}
f\,'(z_0) = \frac{dw}{dz}= \lim\limits_{\Delta z \to 0}\frac{\Delta w}{\Delta z}\text{.}\tag{3.1.3}
\end{equation}
Example 3.1.1.
If \(f(z)=z^3\text{,}\) show that \(f\,'(z) =3z^2\text{.}\)
Solution.
\begin{align*}
f\,'(z_0) \amp = \lim_{z \to z_0}\frac{z^3-z_0^3}{z-z_0}\\
\amp = \lim_{z \to z_0}\frac{(z-z_0) (z^2+z_0z+z_0^2)}{z-z_0}\\
\amp = \lim_{z \to z_0}(z^2+z_0z+z_0^2)\\
\amp = 3z_0^2\text{.}
\end{align*}
We can drop the subscript on \(z_0\) to obtain \(f\,'(z) =3z^2\) as a general formula.
Pay careful attention to the value
\(\Delta z\) in Equation
(3.1.3); the limit must be independent of the manner in which
\(\Delta z \to 0\text{.}\) If we can find two curves that end at
\(z_0\) along which
\(\frac{\Delta w}{\Delta z}\) approaches distinct values, then
\(\frac{ \Delta w}{\Delta z}\) does
not have a limit as
\(\Delta z \to 0\text{,}\) so
\(f\) does
not have a derivative at
\(z_0\text{.}\) The same observation applies to the limits in Equations
(3.1.1) and
(3.1.2).
Example 3.1.2.
Show that the function \(w=f(z) = \overline{z}=x-iy\) is nowhere differentiable.
Solution.
We choose two approaches to the point \(z_0=x_0+iy_0\) and compute limits of the difference quotients. First, we approach \(z_0=x_0+iy_0\) along a line parallel to the \(x\)-axis by forcing \(z\) to be of the form \(z=x+iy_0\text{.}\)
\begin{align*}
\lim\limits_{z \to z_0}\frac{f(z) -f(z_0)}{z-z_0} \amp = \lim\limits_{(x+iy_0) \to (x_0+iy_0)} \frac{f(x+iy_0) - f(x_0+iy_0)}{(x+iy_0)-(x_0+iy_0)}\\
\amp = \lim\limits_{(x+iy_0) \to (x_0+iy_0)}\frac{(x-iy_0) -(x_0-iy_0)}{(x-x_0) +i(y_0-y_0)}\\
\amp = \lim\limits_{(x+iy_0) \to (x_0+iy_0)}\frac{x-x_0}{x-x_0}\\
\amp = 1\text{.}
\end{align*}
Next, we approach \(z_0\) along a line parallel to the \(y\) axis by forcing \(z\) to be of the form \(z=x_0+iy\text{.}\)
\begin{align*}
\lim_{z \to z_0}\frac{f(z) -f(z_0)}{z-z_0} \amp = \lim\limits_{(x_0+iy) \to (x_0+iy_0)}\frac{f(x_0+iy) -f(x_0+iy_0)}{(x_0+iy) -(x_0+iy_0)}\\
\amp = \lim\limits_{(x_0+iy) \to (x_0+iy_0)}\frac{(x_0-iy) -(x_0-iy_0)}{(x_0-x_0) +i(y-y_0)}\\
\amp = \lim\limits_{(x_0+iy) \to (x_0+iy_0)}\frac{-i(y-y_0)}{i(y-y_0)}\\
\amp = -1\text{.}
\end{align*}
The limits along the two paths are different, so there is no possible value for the right side of Equation
(3.1.1). Therefore
\(f( z) = \overline{z}\) is not differentiable at the point
\(z_0\text{,}\) and since
\(z_0\) was arbitrary,
\(f(z)\) is nowhere differentiable.
We seldom are interested in studying functions that aren’t differentiable, or are differentiable at only a single point. Complex functions that have a derivative at all points in a neighborhood of \(z_0\) deserve further study. Indeed, functions that are differentiable in neighborhoods of points are pillars of the complex analysis edifice. We give them a special name, as indicated in the following definition.
Definition 3.1.4. Analytic.
The complex function \(f\) is analytic at the point \(z_0\) provided there is some \(\varepsilon>0\) such that \(f\,'( z)\) exists for all \(z\in D_{\varepsilon}(z_0)\text{.}\) In other words, \(f\) must be differentiable not only at \(z_0\text{,}\) but also at all points in some \(\varepsilon\)-neighborhood of \(z_0\text{.}\)
If \(f\) is analytic at each point in the region \(R\text{,}\) then we say that \(f\) is analytic on \(R\text{.}\) Again, we have a special term if \(f\) is analytic on the whole complex plane.
Definition 3.1.5. Entire.
If \(f\) is analytic on the whole complex plane then \(f\) is said to be entire.
Points of non analyticity for a function are called singular points. They are important for certain applications in physics and engineering.
Our definition of the derivative for complex functions is formally the same as for real functions and is the natural extension from real variables to complex variables. The basic differentiation formulas are identical to those for real functions, and we obtain the same rules for differentiating powers, sums, products, quotients, and compositions of functions. We can easily establish the proof of the differentiation formulas by using the limit theorems.
Suppose that
\(f\) and
\(g\) are differentiable. From Equation
(3.1.1) and the technique exhibited in the solution to
Example 3.1.1, we can establish the following rules, which are virtually identical to those for real-valued functions.
\begin{align}
\frac{d}{dz}C \amp = 0, \text{ where } C \text{ is a constant, and }\tag{3.1.4}\\
\frac{d}{dz}z^n \amp = nz^{n-1}, \text{ where } n \text{ is a positive integer } .\tag{3.1.5}\\
\frac{d}{dz}[Cf(z)] \amp = Cf\,'(z)\tag{3.1.6}\\
\frac{d}{dz}[f(z)+g(z)] \amp = f\,'(z) + g\,'(z)\tag{3.1.7}\\
\frac{d}{dz} [f(z) g(z)] \amp = f(z) g\,'(z) + g(z) f\,'(z)\tag{3.1.8}\\
\frac{d}{dz}\frac{f(z)}{g(z)} \amp = \frac{g(z) f\,'(z) -f(z) g\,'(z)}{[g(z)]^2} \text{ if } g(z) \ne 0, \text{ and }\tag{3.1.9}\\
\frac{d}{dz}\,f\big(g(z)\big) \amp = f\,'\big(g(z)\big) g\,'(z)\text{.}\tag{3.1.10}
\end{align}
Important particular cases of Equations
(3.1.9) and
(3.1.10), respectively, are
\begin{align}
\frac{d}{dz}\frac{1}{z^n} \amp = \frac{-n}{z^{n+1}}\text{ ,\qquad for } z \ne 0, n \text{ a positive integer, and }\tag{3.1.11}\\
\frac{d}{dz}[f(z)]^n \amp =n[f(z)]^{n-1}f\,'(z), n \text{ a positive integer. }\tag{3.1.12}
\end{align}
Example 3.1.6.
If we use Equation
(3.1.12) with
\(f(z)=z^2+i2z+3\) and
\(f\,'(z) =2z+2i\text{,}\) then we get
\begin{equation*}
\frac{d}{dz}(z^2+i2z+3)^4=8(z^2+i2z+3)^3(z+i)
\end{equation*}
The proofs of the rules given in Equations
(3.1.4) through
(3.1.10) depend on the validity of extending theorems for real functions to their complex companions. Equation
(3.1.8), for example, relies on
Theorem 3.1.7.
Theorem 3.1.7.
If \(f\) is differentiable at \(z_0\text{,}\) then \(f\) is continuous at \(z_0\text{.}\)
Proof.
\begin{equation*}
\lim_{z \to z_0}\frac{f(z) -f(z_0)}{z-z_0}=f\,'(z_0)\text{.}
\end{equation*}
Using the multiplicative property of limits given by Formula
(2.3.5), we get
\begin{align*}
\lim\limits_{z \to z_0} [f(z) -f(z_0)] \amp = \lim_{z \to z_0}\frac{f(z) -f(z_0)}{z-z_0}(z-z_0)\\
\amp = \lim_{z \to z_0}\frac{f(z) -f(z_0)}{z-z_0}\lim\limits_{z \to z_0}(z-z_0)\\
\amp = f\,'(z_0) \cdot 0=0\text{.}
\end{align*}
This result implies that \(\lim\limits_{z \to z_0}f(z) =f(z_0)\text{,}\) which is equivalent to showing that \(f\) is continuous at \(z_0\text{.}\)
We can establish Equation
(3.1.8) from
Theorem 3.1.7. Letting
\(h(z)=f(z)g(z)\) and using Definition 3.1, we write
\begin{equation*}
h\,'(z_0) = \lim\limits_{z \to z_0}\frac{h(z) -h(z_0)}{z-z_0} = \lim_{z \to z_0}\frac{f(z) g(z)-f(z_0)g(z_0)}{z-z_0}\text{.}
\end{equation*}
If we subtract and add the term \(f(z_0) g(z)\) in the numerator, we get
\begin{align*}
h\,'(z_0) \amp = \lim_{z \to z_0}\frac{f(z) g(z) - f(z_0) g(z) + f(z_0) g(z) - f(z_0)g(z_0)}{z-z_0}\\
\amp = \lim_{z \to z_0}\frac{f(z)g(z) - f(z_0)g(z)}{z-z_0} + \lim_{z \to z_0}\frac{f(z_0)g(z) - f(z_0)g(z_0)}{z-z_0}\\
\amp = \lim_{z \to z_0}\frac{f(z) -f(z_0)}{z-z_0}\lim_{z \to z_0}g(z) + f(z_0) \lim\limits_{z \to z_0}\frac{g(z) -g(z_0)}{z-z_0}\text{.}
\end{align*}
Using the definition of the derivative given by Equation
(3.1.1) and the continuity of
\(g\text{,}\) we obtain
\(h\,'(z_0) =f\,'(z_0) g(z_0) +f(z_0) g\,'(z_0)\text{,}\) which is what we wanted to establish. We leave the proofs of the other rules as exercises.
The rule for differentiating polynomials carries over to the complex case as well. If we let \(P(z)\) be a polynomial of degree \(n\text{,}\) so
\begin{equation*}
P(z)=a_0 + a_1z + a_2z^2 + \cdots + a_nz^n\text{.}
\end{equation*}
then mathematical induction, along with Equations
(3.1.5) and
(3.1.7), gives
\begin{equation*}
P\,'(z) =a_1 + 2a_2z + 3a_3z^2 + \cdots + na_nz^{n-1}\text{.}
\end{equation*}
Again, we leave the proof of this result as an exercise.
We can use the differentiation rules as aids in determining when functions are analytic. For example, Equation
(3.1.9) tells us that if
\(P(z)\) and
\(Q(z)\) are polynomials, then their quotient
\(\frac{P(z)}{Q(z)}\) is analytic at all points where
\(Q(z) \ne 0\text{.}\) This condition implies that the function
\(f(z) = \frac{1}{z}\) is analytic for all
\(z\ne 0\text{.}\) The square root function is more complicated. If
\(f(z)=z^{\frac{1}{2}} = |z|^{\frac{1}{2}}e^{i\frac{\mathrm{Arg}(z)}{2}}\text{,}\) then
\(f\) is analytic at all points except
\(z=0\) (because
\(\mathrm{Arg}(0)\) is undefined) and at points that lie along the negative
\(x\) axis. The argument function, and therefore the function
\(f\) itself, are not continuous at points that lie along the negative
\(x\) axis.
We close this section with a complex extension of a famous theorem, the proof of which will be given in
Chapter 7.
Theorem 3.1.8. L’Hôpital’s rule.
Assume that \(f\) and \(g\) are analytic at \(z_0\text{.}\) If \(f(z_0) =0\text{,}\) \(g(z_0) =0\text{,}\) and \(g\,'(z_0) \ne 0\text{,}\) then
\begin{equation*}
\lim_{z \to z_0}\frac{f(z)}{g(z)} = \lim_{z \to z_0}\frac{f\,'(z)}{g\,'(z)}
\end{equation*}
Proof.
Exercises Exercises
1.
Find the derivatives of the following functions.
(a)
\(g(z) =(z^2-iz+9)^{5}\text{.}\)
Solution.
\(f\,'(z) =15z^2-8z+7\text{.}\)
(b)
\(h(z) = \frac{2z+1}{z+2}\) for \(z \ne -2\text{.}\)
(c)
\(F(z) = (z^2+(1-3i)z+1)(z^4+3z^2+5i)\text{.}\)
Solution.
\(h\,'(z) =\frac{3s}{(z+2)^2}\) for \(z \ne -2\text{.}\)
2.
Show that the following functions are differentiable nowhere.
(a)
\(f(z) = \mathrm{Re}(z)\text{.}\)
(b)
\(f(z) = \mathrm{Im}(z)\text{.}\)
3.
If \(f\) and \(g\) are entire functions, which of the following are necessarily entire?
(a)
\([f(z)]^3\text{.}\)
Solution.
(b)
\(f(z)g(z)\text{.}\)
Solution.
(c)
\(\frac{f(z)}{g(z)}\text{.}\)
Solution.
Entire, provided that \(g(z) \ne 0\) for all \(z\text{.}\)
(d)
\(f(\frac{1}{z})\text{.}\)
Solution.
(e)
\(f(z-1)\text{.}\)
Solution.
(f)
\(f\big(g(z)\big)\text{.}\)
Solution.
4.
5.
Let \(P(z) =a_0+a_1z+\cdots +a_nz^n\) be a polynomial of degree \(n \ge 1\text{.}\)
(a)
Show that \(P\,'(z) = a_1 + 2a_2z + \cdots + na_nz^{n-1}\text{.}\)
(b)
Show that, for \(k=0,1, \ldots,
n, \; a_k= \frac{P^{(k)}(0)}{k!}\text{,}\) where \(P^{(k)}\) denotes the \(k^{\text{ th } }\) derivative of \(P\text{.}\) (By convention, \(P^{(0)}(z) = P(z)\text{.}\))
Solution.
The result is clearly true when \(n=1\text{.}\) Assume for some \(n>1\) that \(P\,'(z)=a_1+2a_2z+\cdots+na_nz^{n-1}\text{.}\) Consider \(Q(z) =\sum\limits_{k=0}^{n+1}a_kz^k=\sum \limits_{k=0}^na_kz^k+a_{n+1}z^{n+1}\text{.}\) Since the derivative of the sum of two terms is the sum of the derivatives, we have \(Q\,'(z) = \frac{d}{dz}\left(\sum\limits_{k=0}^na_kz^k\right) + \frac{d}{dz}(a_{n+1}z^{n+1})\text{.}\) The induction assumption gives the result.
6.
Let \(P\) be a polynomial of degree 2, given by
\begin{equation*}
P(z) = (z-z_1)(z-z_2)\text{.}
\end{equation*}
where \(z_1\ne z_2\text{.}\) Show that
\begin{equation*}
\frac{P\,'(z)}{P(z)} = \frac{1}{z-z_1} + \frac{1}{z-z_2}\text{.}
\end{equation*}
Note: The quotient \(\frac{P\,'(z)}{P(z)}\) is known as the logarithmic derivative of \(P\text{.}\)
7.
Use L’Hôpital’s rule to find the following limits.
(a)
\(\lim\limits_{z \to i}\frac{z^4-1}{z-i}\text{.}\)
Solution.
(b)
\(\lim\limits_{z \to 1+i}\frac{z^2-iz-1-i}{z^2-2z+2}\text{.}\)
(c)
\(\lim\limits_{z \to -i}\frac{z^{6}+1}{z^2+1}\text{.}\)
Solution.
(d)
\(\lim\limits_{z \to 1+i}\frac{z^4+4}{z^2-2z+2}\text{.}\)
(e)
\(\lim\limits_{z \to 1+i\sqrt{3}}\frac{z^{6}-64}{z^3+8}\text{.}\)
Solution.
(f)
\(\lim\limits_{z \to -1+i\sqrt{3}}\frac{z^{9}-512}{z^3-8}\text{.}\)
8.
Use Equation
(3.1.1) to show that
\(\frac{d}{dz}\frac{1}{z}= -\frac{1}{z^2}\text{.}\)
9.
Show that \(\frac{d}{dz}z^{-n}=-nz^{-n-1}\text{,}\) where \(n\) is a positive integer.
Solution.
\(\frac{d}{dz}z^{-n}=\frac{d}{dz}(\frac{1}{z^n})\text{.}\) Apply the quotient rule: \(\frac{d}{dz}(\frac{1}{z^n}) = \frac{(z^n) \frac{d}{dz}(1) - 1\frac{d}{dz}(z^n)}{(z^n)^2}\text{.}\) Now simplify.
10.
Verify the identity.
\begin{equation*}
\frac{d}{dz} f(z)g(z)h(z) = f\,'(z)g(z)h(z) + f(z)g\,'(z)h(z) + f(z)g(z)h\,'(z)
\end{equation*}
11.
Show that the function \(f(z) =|z|^2\) is differentiable only at the point \(z_0=0\text{.}\) \hint{To show that \(f\) is not differentiable at \(z_0 \ne 0\text{,}\) choose horizontal and vertical lines through the point \(z_0\) and show that \(\frac{\Delta w}{ \Delta z}\) approaches two distinct values as \(\Delta z \to 0\) along those two lines.}
Solution.
Evaluate \(\lim\limits_{z \to 0}\frac{f(z)-f(0)}{z-0}=\lim\limits_{z \to 0}\frac{|z|^2-0}{z-0} = \lim\limits_{z \to 0}\frac{z\bar{z}}{z} = \lim\limits_{z \to 0}\bar{z} = 0\text{.}\) Follow the hint for the rest.
12.
Verify the following identities:
(a)
\(\frac{d}{dz}C = 0\text{,}\) where
\(C\) is a constant: Identity
(3.1.4).
(b)
\(\frac{d}{dz}[f(z)+g(z)] = f\,'(z) + g\,'(z)\text{:}\) Identity
(3.1.7).
(c)
\(\frac{d}{dz}\frac{f(z)}{g(z)} = \frac{g(z) f\,'(z) -f(z) g\,'(z)}{[g(z)]^2}\) if
\(g(z) \ne 0\text{:}\) Identity
(3.1.9).
(d)
\(\frac{d}{dz}\,f\big(g(z)\big) = f\,'\big(g(z)\big) g\,'(z)\text{:}\) Identity
(3.1.10).
(e)
\(\frac{d}{dz}[f(z)]^n = n[f(z)]^{n-1}f\,'(z)\text{,}\) where n is a positive integer: Identity
(3.1.12).
13.
Consider the differentiable function \(f(z)=z^3\) and the two points \(z_1=1\) and \(z_2=i\text{.}\) Show that there does not exist a point \(c\) on the line \(y=1-x\) between 1 and \(i\) such that \(\frac{f(z_2)-f(z_1)}{z_2-z_1} = f\,'(c)\text{.}\) This result shows that the mean value theorem for derivatives does not extend to complex functions.
Solution.
\(\frac{f(z_2)-f(z_1)}{z_2-z_1} = \frac{i^3-1^3}{i-1} = \frac{-i-1}{i-1}=i\text{.}\) The minimum modulus of points on the line \(y=1-x\) is \(\frac{\sqrt{2}}{2}\) (prove this!). But \(f\,'(z) =3z^2\text{,}\) and the only solutions to the equation \(3z^2=i\) have moduli equal to \(\frac{\sqrt{3}}{3}\) (prove), which is less than \(\frac{\sqrt{2}}{2}\) (prove this also).
14.
Let \(f(z)=z^{\frac{1}{n}}\) denote the multivalued “\(n\)th root function,” where \(n\) is a positive integer. Use the chain rule to show that, if \(g(z)\) is any branch of the \(n\)th root function, then
\begin{equation*}
g\,'(z) = \frac{1}{n}\frac{g(z)}{z}
\end{equation*}
in some suitably chosen domain (which you should specify).
15.
Explain why the composition of two entire functions is an entire function.
16.
Let \(f\) be differentiable at \(z_0\text{.}\) Show that there exists a function \(\eta(z)\) such that
\begin{equation*}
f(z) =f(z_0) +f\,'(z_0) (z-z_0) + \eta(z)(z-z_0)\text{,}
\end{equation*}
where \(\eta (z) \to 0\text{ as } z \to z_0\text{.}\)
