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Complex Analysis: an Open Source Textbook

Russell W. Howell, John H. Mathews

Section 3.2 The Cauchy-Riemann Equations

In Section 3.1 we showed that computing the derivative of complex functions written in a form such as \(f(z)=z^2\) is a rather simple task. But life isn’t always so easy. Many times we encounter complex functions written in the form of \(f(z) = f(x, y) = u(x, y) +iv(x, y)\text{.}\) For example, suppose we had
\begin{equation} f(z) = f(x, y) = u(x, y) +iv(x, y) = (x^3-3xy^2) +i(3x^2y-y^3)\text{.}\tag{3.2.1} \end{equation}
Is there some criterion that we can use to determine whether \(f\) is differentiable, and if so, to find the value of \(f\,'(z)\text{?}\)
The answer to this question is yes, thanks to the independent discovery of two important equations by the French mathematician Augustin-Louis Cauchy
 1 
A.L. Cauchy (1789–1857) played a prominent role in the development of complex analysis, and his name appears often throughout this text. The last name is not pronounced as “kaushee.” The first syllable has a long “o” sound, like the word kosher, but with the second syllable having a long “e” instead of “er” at the end. Thus, we pronounce Cauchy as “kōsh\={e}.” In mathematical circles it is not kosher to mispronounce Cauchy!
and the German mathematician Georg Friedrich Bernhard Riemann.
First, let’s reconsider the derivative of \(f(z)=z^2\text{.}\) As we have stated, the limit given in Equation (3.1.1) must not depend on how \(z\) approaches \(z_0\text{.}\) We investigate two such approaches: a horizontal approach and a vertical approach to \(z_0\text{.}\) Recall from our graphical analysis of \(w=z^2\) that the image of a square is a “curvilinear quadrilateral.” For convenience, we let the square have vertices \(z_0=2+i\text{,}\) \(z_1=2.01+i\text{,}\) \(z_2=2+1.01i\text{,}\) and \(z_3=2.01+1.01i\text{.}\) Then the image points are \(w_0=3+4i\text{,}\) \(w_1=3.0401+4.02i\text{,}\) \(w_2=2.9799+4.04i\text{,}\) and \(w_3=3.02+4.0602i\text{,}\) as shown in Figure 3.2.1.
Figure 3.2.1. The image of a small square with vertex {\(z_0=2+i\)}, using \(w=z^2\)
We know that \(f\) is differentiable, so the limit of the difference quotient \(\frac{f(z) -f(z_0)}{z-\,z_0}\) exists no matter how we approach \(z_0=2+i\text{.}\) Thus we can approximate \(f\,'(2+i)\) by using horizontal or vertical increments in \(z\text{:}\)
\begin{equation*} f\,'(2+i) \approx \frac{f(2.01+i) -f(2+i)}{(2.01+i) -(2+i)}= \frac{0.0401+0.02i}{0.01}=4.01+2i \end{equation*}
and
\begin{equation*} f\,'(2+i) \approx \frac{f(2+1.01i) - f(2+i)}{(2+1.01i) - (2+i)} = \frac{-0.0201+0.04i}{0.01i}=4+2.01i\text{.} \end{equation*}
These computations lead to the idea of taking limits along the horizontal and vertical directions. When we do so, we get
\begin{equation*} f\,'(2+i) = \lim\limits_{h \to 0}\frac{f(2+h+i) -f(2+i)}{h}= \lim\limits_{h \to 0}\frac{4h+h^2+i2h}{h}=4+2i \end{equation*}
and
\begin{equation*} f\,'(2+i) = \lim\limits_{h \to 0}\frac{f(2+i+ih) -f(2+i)}{ih}= \lim\limits_{h \to 0}\frac{-2h-h^2+i4h}{ih}=4+2i \end{equation*}
We now generalize this idea by taking limits of an arbitrary differentiable complex function and obtain an important result.

Proof.

Because \(f\) is differentiable, we know that \(\lim\limits_{z \to z_0}(\frac{f(z) -f(z_0)}{z-z_0})\) exists regardless of the path we take as \(z \to z_0\text{.}\) We will choose horizontal and vertical lines that pass through the point \(z_0=(x_0,y_0)\) and compute the limiting values of \(\frac{f(z) -f(z_0)}{(z-z_0)}\) along these lines. Equating the two resulting limits will yield Equations (3.2.4). For the horizontal approach to \(z_0\text{,}\) we set \(z=x+iy_0\) and obtain
\begin{align*} f\,'(z_0) \amp = \lim_{(x,y_0) \to (x_0,y_0)}\frac{f(x+iy_0) -f(x_0+iy_0)}{x+iy_0-(x_0+iy_0)}\\ \amp = \lim_{x \to x_0}\frac{u(x,y_0) -u(x_0,y_0) +i[v(x,y_0) -v(x_0,y_0)]}{x-x_0}\\ \amp = \lim_{x \to x_0}\frac{u(x,y_0) -u(x_0,y_0)}{x-x_0}+i\lim_{x \to x_0}\frac{v(x,y_0) -v(x_0,y_0)}{x-x_0}\text{.} \end{align*}
The last two limits are the partial derivatives of \(u\) and \(v\) with respect to \(x\text{,}\) so
\begin{equation*} f\,'(z_0) = u_x(x_0,y_0) + iv_x(x_0,y_0)\text{.} \end{equation*}
giving us Equation (3.2.2).
Along the vertical approach to \(z_0\text{,}\) we have \(z=x_0+iy\text{,}\) so
\begin{align*} f\,'(z_0) \amp = \lim_{(x_0,y) \to (x_0,y_0)}\frac{f(x_0+iy) - f(x_0+iy_0)}{x_0+iy-(x_0+iy_0)}\\ \amp = \lim_{y \to y_0}\frac{u(x_0,y) -u(x_0,y_0) + i[v(x_0,y) -v(x_0,y_0)]}{i(y-y_0)}\\ \amp = \lim_{y \to y_0}\frac{v(x_0,y) -v(x_0,y_0)}{y-y_0} - i\lim_{y \to y_0}\frac{u(x_0,y) -u(x_0,y_0)}{y-y_0}\text{.} \end{align*}
The last two limits are the partial derivatives of \(u\) and \(v\) with respect to \(y\text{,}\) so
\begin{equation*} f\,'(z_0) =v_y(x_0,y_0) -iu_y(x_0,y_0)\text{.} \end{equation*}
giving us Equation (3.2.3).
Since \(f\) is differentiable at \(z_0\text{,}\) the limits given by Equations (3.2.2) and (3.2.3) must be equal. If we equate the real and imaginary parts in those equations, the result is Equations (3.2.4), and the proof is complete.
Note some of the important implications of this theorem.
  • If \(f\) is differentiable at \(z_0\text{,}\) then the Cauchy-Riemann Equations (3.2.4) will be satisfied at \(z_0\text{,}\) and we can use either Equation (3.2.2) or (3.2.3) to evaluate \(f\,'(z_0)\text{.}\)
  • Taking the contrapositive, if Equations (3.2.4) are not satisfied at \(z_0\text{,}\) then we know automatically that \(f\) is not differentiable at \(z_0\text{.}\)
  • If Equations (3.2.4) are satisfied at \(z_0\text{,}\) however, we cannot necessarily conclude that \(f\) is differentiable at \(z_0\text{.}\)
We now illustrate each of these points.

Example 3.2.3.

We know that \(f(z)=z^2\) is differentiable and that \(f\,'(z) =2z\text{.}\) We also have
\begin{equation*} f(z)=z^2=(x+iy)^2=(x^2-y^2) + i(2xy) =u(x,y) +iv(x,y)\text{.} \end{equation*}
It is easy to verify that Equations (3.2.4) are indeed satisfied:
\begin{equation*} u_x(x,y) =2x=v_y(x,y) \text{ and } u_y(x,y) =-2y=-v_x(x,y)\text{.} \end{equation*}
Using Equations (3.2.2) and (3.2.3), respectively, to compute \(f\,'(z)\) gives
\begin{align*} f\,'(z) \amp = u_x(x,y) +iv_x(x,y) =2x+i2y=2z, \text{ and }\\ f\,'(z) \amp = v_y(x,y) -iu_y(x,y) =2x-i(-2y)=2x+i2y=2z\text{,} \end{align*}
as expected.

Example 3.2.4.

Show that \(f(z) = \overline{z}\) is nowhere differentiable.
Solution.
We have \(f(z) =f(x+iy) =x-iy=u(x,y) + iv(x,y)\text{,}\) where \(u(x,y) =x\) and \(v(x,y) =-y\text{.}\) Thus, for any point \((x,y)\text{,}\) \(u_x(x,y) = 1\) and \(v_y(x,y) = -1\text{.}\) The Cauchy-Riemann equations are not satisfied at any point \(z=(x,y)\text{,}\) so we conclude that \(f\) is nowhere differentiable.

Example 3.2.5.

Show that the function defined by
\begin{equation*} f(x)= \begin{cases}\frac{(\overline{z})^2}{z}=\frac{x^3-3xy^2}{x^2+y^2}+i \frac{y^3-3x^2y}{x^2+y^2} \text{ when } z \ne 0, \text{ and } \\ 0 \text{ when } z=0 \end{cases} \end{equation*}
is not differentiable at the point \(z_0=0\) even though the Cauchy-Riemann equations are satisfied at \((0,0)\text{.}\)
Solution.
We must use limits to calculate the partial derivatives at \((0,0)\text{.}\)
\begin{equation*} u_x(0,0) = \lim_{x \to 0}\frac{u(x,0)-u(0,0)}{x-0} = \lim_{x \to 0}\frac{\frac{x^3-0}{x^2+0}}{x}=1\text{.} \end{equation*}
Similarly, we can show that
\begin{equation*} u_y(0,0) =0\text{,} v_x(0,0) = 0 \text{ and } v_y(0,0) =1\text{.} \end{equation*}
Hence the Cauchy-Riemann equations hold at the point \((0,0)\text{.}\)
We now show that \(f\) is not differentiable at \(z_0=0\text{.}\) Letting \(z\) approach \(0\) along the \(x\) axis gives
\begin{equation*} \lim_{(x,0) \to (0,0)}\frac{f(x+0i) -f(0)}{x+0i-0} = \lim_{x \to 0}\frac{\frac{x^2}{x}-0}{x-0}= \lim\limits_{x \to 0}\frac{x-0}{x-0}=1\text{.} \end{equation*}
But if we let \(z\) approach \(0\) along the line \(y=x\) given by the parametric equations \(x=t\) and \(y=t\text{,}\) then
\begin{equation*} \lim_{(t,t) \to (0,0)}\frac{f(t+it) -f(0)}{t+it-0} = \lim_{t \to 0}\frac{-\frac{2t^3}{2t^2}+i\big(\!-\frac{2t^3}{2t^2}\big)}{t+it} = \lim\limits_{t \to 0}\frac{-t-it}{t+it}=-1\text{.} \end{equation*}
The two limits are distinct, so \(f\) is not differentiable at the origin.
Example 3.2.5 reiterates that the mere satisfaction of the Cauchy-Riemann equations is not sufficient to guarantee the differentiability of a function. The following theorem, however, gives conditions that guarantee the differentiability of \(f\) at \(z_0\text{,}\) so that which we can use either Equation (3.2.2) or (3.2.3) to compute \(f\,'(z_0)\text{.}\) They are referred to as the Cauchy-Riemann conditions for differentiability.

Proof.

Let \(\Delta z = \Delta x+i\Delta y\) and \(\Delta w= \Delta u+i\Delta v\text{,}\) and let \(\Delta z\) be small enough so that \(z\) lies in the \(\varepsilon\)-neighborhood of \(z_0\) in which the hypotheses hold. We need to show that \(\frac{\Delta w}{\Delta z}\) approaches the limit given in Equation (3.2.3) as \(\Delta z\) approaches zero. We write the difference, \(\Delta u\text{,}\) as
\begin{equation*} \Delta u=u(x_0+\Delta x,y_0+\Delta y) -u(x_0,y_0) \end{equation*}
If we subtract and add the term \(u(x_0,y_0+\Delta y)\text{,}\) then we get
\begin{align} \Delta u \amp = [u(x_0+\Delta x,y_0+\Delta y) - u(x_0,y_0+\Delta y)]\notag\\ \amp + [u(x_0,y_0+\Delta y) -u(x_0,y_0)]\text{.}\tag{3.2.5} \end{align}
The partial derivatives \(u_x\) and \(u_y\) exist, so the mean value theorem for real functions of two variables implies that a value \(x^*\) exists between \(x_0\) and \(x_0+\Delta x\) such that we can write the first term in brackets on the right side of Equation (3.2.5) as
\begin{equation*} u(x_0+\Delta x, y_0+\Delta y) - u(x_0,y_0 + \Delta y) = u_x(x^*, y_0+\Delta y) \Delta x\text{.} \end{equation*}
Furthermore, as \(u_x\) and \(u_y\) are continuous at \((x_0, y_0)\text{,}\) there exists a quantity \(\varepsilon_1\) such that
\begin{equation*} u_x(x^*,y_0+\Delta y) =u_x(x_0,y_0) + \varepsilon_1\text{.} \end{equation*}
where \(\varepsilon_1 \to 0\) as \(x^* \to x_0\) and \(\Delta y \to 0\text{.}\) Because \(\Delta x \to 0\) forces \(x^* \to x_0\text{,}\) we can use the equation
\begin{equation} u(x_0+\Delta x,y_0+\Delta y) -u(x_0,y_0+\Delta y) = [u_x(x_0,y_0) + \varepsilon_1] \Delta x\text{.}\tag{3.2.6} \end{equation}
where \(\varepsilon_1 \to 0\) as \(\Delta x \to 0\) and \(\Delta y \to 0\text{.}\) Similarly, there exists a quantity \(\varepsilon_2\) such that the second term in brackets on the right side of Equation (3.2.5) satisfies the equation
\begin{equation} u(x_0,y_0+\Delta y) - u(x_0,y_0) = [u_y(x_0,y_0) + \varepsilon_2] \Delta y\text{.}\tag{3.2.7} \end{equation}
where \(\varepsilon_2 \to 0\) as \(\Delta x \to 0\) and \(\Delta y \to 0\text{.}\) Combining Equations (3.2.6) and (3.2.7) gives
\begin{equation*} \Delta u=(u_x+\varepsilon_1) \Delta x+(u_y+\varepsilon_2) \Delta y\text{.} \end{equation*}
where partial derivatives \(u_x\) and \(u_y\) are evaluated at the point \((x_0,y_0)\) and \(\varepsilon_1\) and \(\varepsilon_2\) tend to zero as \(\Delta x\) and \(\Delta y\) both tend to zero. Similarly, the change \(\Delta v\) is related to the changes \(\Delta x\) and \(\Delta y\) by the equation
\begin{equation*} \Delta v=(v_x+\varepsilon_3) \Delta x+(v_y+\varepsilon_4) \Delta y\text{,} \end{equation*}
where the partial derivatives \(v_x\) and \(v_y\) are evaluated at the point \((x_0,y_0)\) and \(\varepsilon_3\) and \(\varepsilon_4\) tend to zero as \(\Delta x\) and \(\Delta y\) both tend to zero. Combining these last two equations gives
\begin{equation} \Delta w=u_x\Delta x+u_y\Delta y+i(v_x\Delta x+v_y\Delta y) + \varepsilon_1\Delta x + \varepsilon_2\Delta y+i(\varepsilon_3\Delta x+\varepsilon_4\Delta y)\text{.}\tag{3.2.8} \end{equation}
We can use the Cauchy-Riemann equations in Equation (3.2.8) to obtain
\begin{equation*} \Delta w=u_x\Delta x-v_x\Delta y+i(v_x\Delta x+u_x\Delta y) + \varepsilon_1\Delta x + \varepsilon_2\Delta y+i(\varepsilon_3\Delta x+\varepsilon_4\Delta y) \end{equation*}
Now we rearrange the terms and get
\begin{equation*} \Delta w=u_x\left[ \Delta x+i\Delta y\right] +iv_x[\Delta x+i\Delta y] + \varepsilon_1\Delta x+\varepsilon_2\Delta y+i(\varepsilon_3\Delta x+\varepsilon_4\Delta y) \end{equation*}
Since \(\Delta z= \Delta x+i\Delta y\text{,}\) we can divide both sides of this equation by \(\Delta z\) and take the limit as \(\Delta z \to 0\text{:}\)
\begin{equation} \lim_{\Delta z \to 0}\frac{\Delta w}{\Delta z} = u_x+iv_x+\lim_{\Delta z \to 0}\left[\frac{\varepsilon_1\Delta x}{\Delta z} + \frac{\varepsilon_2\Delta y}{\Delta z} + i\frac{\varepsilon_3\Delta x}{\Delta z}+i\frac{\varepsilon_4\Delta y}{\Delta z}\right]\text{.}\tag{3.2.9} \end{equation}
Because \(\varepsilon_1\) tends to zero as \(\Delta x\) and \(\Delta y\) both tend to zero, we have
\begin{equation*} \lim_{\Delta z \to 0}\left|\frac{\varepsilon_1\Delta x}{\Delta z}\right| = \lim_{\Delta z \to 0}\big|\varepsilon_1\big|\left|\frac{\Delta x}{\Delta z}\right| \le \lim_{\Delta z \to 0}\big|\varepsilon_1\big|=0\text{.} \end{equation*}
Similarly, the limits of the other quantities in Equation (3.2.9) involving \(\varepsilon_2\text{,}\) \(\varepsilon_3\text{,}\) \(\varepsilon_4\) are zero. Therefore the limit in Equation (3.2.9) becomes
\begin{equation*} \lim_{\Delta z \to 0}\frac{\Delta w}{\Delta z}=f\,'(z_0) =u_x(x_0,y_0) +iv_x(x_0,y_0)\text{,} \end{equation*}
which completes the proof of the theorem.

Example 3.2.7.

At the beginning of this section (Equation (3.2.1)) we defined the function \(f(z)=u(x,y)+iv(x,y) = x^3-3xy^2+i(3x^2y-y^3)\text{.}\) Show that this function is differentiable for all \(z\text{,}\) and find its derivative.
Solution.
We compute \(u_x(x,y) =v_y(x,y) =3x^2-3y^2\) and \(u_y(x,y) =-6xy=-v_x(x,y)\text{,}\) so the Cauchy-Riemann Equations (3.2.4) are satisfied. Moreover, \(u, \, v, \, u_x, \, u_y, \, v_x\text{,}\) and \(v_y\) are continuous everywhere. By Theorem 3.2.6, \(f\) is differentiable everywhere, and Equation (3.2.2) gives
\begin{equation*} f\,'(z) = u_x(x,y) +iv_x(x,y) = (3x^2-3y^2) +i6xy = 3(x^2-y^2+i2xy) =3z^2\text{.} \end{equation*}
Alternatively, from Equation (3.2.3),
\begin{equation*} f\,'(z) =v_y(x,y) - iu_y(x,y) =(3x^2-3y^2) - i(-6xy) = 3(x^2-y^2+i2xy) = 3z^2\text{.} \end{equation*}
This result isn’t surprising because \((x+iy)^3 = x^3-3xy^2+i(3x^2y-y^3)\) and so the function \(f\) is really our old friend \(f(z)=z^3\text{.}\)

Example 3.2.8.

Show that the function \(f(z) =e^{-y}\cos x+ie^{-y}\sin x\) is differentiable for all \(z\) and find its derivative.
Solution.
We first write \(u(x,y) =e^{-y}\cos x\) and \(v(x,y) = e^{-y}\sin x\) and then compute the partial derivatives.
\begin{align*} u_x(x,y) \amp = v_y(x,y) =-e^{-y}\sin x\ \text{ and }\\ v_x(x,y) \amp =- u_y(x,y) =e^{-y}\cos x\text{.} \end{align*}
We note that \(u, \, v, \, u_x, \, u_v, \, v_x\text{,}\) and \(v_y\) are continuous functions and that the Cauchy-Riemann equations hold for all values of \((x,y)\text{.}\) Hence, using Equation (3.2.2), we write
\begin{equation*} f\,'(z) =f\,'(x,y) =u_x(x,y) +iv_x(x,y) =-e^{-y}\sin x+ie^{-y}\cos x \end{equation*}
The Cauchy-Riemann conditions are particularly useful in determining the set of points for which a function \(f\) is differentiable.

Example 3.2.9.

Show that the function \(f(z) =x^3+3xy^2+i(y^3+3x^2y)\) is differentiable on the \(x\) and \(y\) axes, but analytic nowhere.
Solution.
Recall (Definition 3.1.4) that when we say a function is analytic at a point \(z_0\) we mean that the function is differentiable not only at \(z_0\text{,}\) but also at every point in some \(\varepsilon\)-neighborhood of \(z_0\text{.}\) With this in mind, we proceed to determine where the Cauchy-Riemann equations are satisfied. We write \(u(x,y) =x^3+3xy^2\) and \(v(x,y) =y^3+3x^2y\) and compute the partial derivatives:
\begin{align*} u_x(x,y) \amp = 3x^2+3y^2, \; v_y(x,y) = 3x^2+3y^2, \text{ and }\\ u_y(x,y) \amp = 6xy, \; v_x(x,y) = 6xy\text{.} \end{align*}
Here \(u, \, v, \, u_x, \, u_y\text{,}\) and \(v\) are continuous, and \(u_x(x,y) =v_y(x,y)\) holds for all \((x,y)\text{.}\) But \(u_y(x,y) =-v_x(x,y)\) iff \(6xy=-6xy\text{,}\) which is equivalent to \(12xy=0\text{.}\) The Cauchy-Riemann equations hold only when \(x=0\) or \(y=0\text{,}\) and according to Theorem 3.2.6, \(f\) is differentiable only at points that lie on the coordinate axes. But this means that \(f\) is nowhere analytic because any \(\varepsilon\)-neighborhood about a point on either axis contains points that are not on those axes.
When polar coordinates \((r, \theta)\) are used to locate points in the plane, we use Expression (2.1.2) for a complex function for convenience; that is,
\begin{align*} f(z) \amp = u(x,y) +iv(x,y)\\ f(re^{i\theta}) \amp = u(re^{i\theta}) +iv(re^{i\theta})\\ \amp = U(r, \theta) +iV(r, \theta)\text{.} \end{align*}
where \(U\) and \(V\) are real functions of the real variables \(r\) and \(\theta\text{.}\) The polar form of the Cauchy-Riemann equations and a formula for finding \(f\,'(z)\) in terms of the partial derivatives of \(U(r,\theta )\) and \(V(r,\theta )\) are given in Theorem 3.2.10, which we ask you to prove in Exercise 3.2.10. This theorem makes use of the validity of the Cauchy-Riemann equations for the functions \(u\) and \(v\text{,}\) so the relation between them and the functions \(U\) and \(V\)—namely, \(u(x,y) =u(re^{i\theta}) =U(r,\theta )\) and \(v(x,y) =v(re^{i\theta}) =V(r,\theta )\)—is important.

Example 3.2.11.

Show that, if \(f\) is given by
\begin{equation*} f(re^{i\theta}) =f(z)=z^{\frac{1}{2}}=r^{\frac{1}{2}}\cos \frac{\theta}{2}+ir^{\frac{1}{2}}\sin \frac{\theta}{2}\text{.} \end{equation*}
where the domain is restricted to be \(\{re^{i\theta} : r>0 \text{ and } -\pi \lt \theta \lt \pi\}\text{,}\) then the derivative is given by
\begin{equation*} f\,'(z) = \frac{1}{2z^{\frac{1}{2}}}= \frac{1}{2}r^{-\frac{1}{2}}\cos \frac{\theta}{2} - i\frac{1}{2}r^{-\frac{1}{2}}\sin \frac{\theta}{2}\text{,} \end{equation*}
for every point in the domain.
Solution.
Write
\begin{equation*} U(r,\theta) =r^{\frac{1}{2}}\cos \frac{\theta}{2} \text{ and } V(r,\theta ) = r^{\frac{1}{2}}\sin \frac{\theta}{2}\text{.} \end{equation*}
Then
\begin{align*} U_{r}(r,\theta ) \amp = \frac{1}{r}V_{\theta}(r,\theta) = \frac{1}{2}r^{-\frac{1}{2}}\cos \frac{\theta}{2}, \text{ and }\\ V_{r}(r,\theta ) \amp = -\frac{1}{r}U_{\theta}(r,\theta) = \frac{1}{2}r^{-\frac{1}{2}}\sin \frac{\theta}{2}\text{.} \end{align*}
Since \(U, \, V, \, U_r, \, U_\theta, \, V_r\text{,}\) and \(V_\theta\) are continuous at every point in the domain (note the strict inequality in \(-\pi \lt \theta \lt \pi\)), we use Theorem 3.2.10 and Equation (3.2.11) to get
\begin{align*} f\,'(z) \amp = e^{-i\theta}\left(\frac{1}{2}r^{-\frac{1}{2}}\cos \frac{\theta}{2}+i\frac{1}{2}r^{-\frac{1}{2}}\sin \frac{\theta}{2}\right)\\ \amp = e^{-i\theta}\left(\frac{1}{2}r^{-\frac{1}{2}}e^{i\frac{\theta}{2}}\right) = \frac{1}{2}r^{-\frac{1}{2}}e^{-i\frac{\theta}{2}}= \frac{1}{2z^{\frac{1}{2}}}\text{.} \end{align*}
Note that \(f(z)\) is discontinuous on the negative real axis and is undefined at the origin. Using the terminology of Section 2.4, the negative real axis is a branch cut, and the origin is a branch point for this function.
Two important consequences of the Cauchy-Riemann equations close this section.

Proof.

The equation \(|f(z)|=K\) implies that, for all \(z=(x,y) \in D\text{,}\)
\begin{equation} u(x,y)^2 + v(x,y)^2 = K^2\text{.}\tag{3.2.13} \end{equation}
If \(K=0\text{,}\) then it must be that \(u(x,y)^2=0\) and \(v(x,y)^2=0\) for all \((x,y)\in D\text{,}\) so \(f\) is identically zero on \(D\text{.}\) If \(K \ne 0\text{,}\) then we take the partial derivative of both sides of Equation (3.2.13) with respect to both \(x\) and \(y\text{,}\) resulting in
\begin{equation*} 2uu_x+2vv_x=0 \text{ and } 2uu_y+2vv_y=0\text{.} \end{equation*}
where for brevity we write \(u\) in place of \(u(x,y)\text{,}\) and so on. We can now use the Cauchy-Riemann equations to rewrite this system as
\begin{equation*} uu_x-vu_y=0 \text{ and } vu_x+uu_y=0\text{.} \end{equation*}
Treating \(u\) and \(v\) as coefficients, we have two equations with two unknowns, \(u_x\) and \(u_y\text{.}\) Solving for \(u_x\) and \(u_y\) gives
\begin{equation*} u_x= \frac{0}{u^2+v^2}=0 \text{ and } u_y= \frac{0}{u^2+v^2}=0\text{.} \end{equation*}
Note that it is important here for \(K \ne 0\) in Equation (3.2.13).
A theorem from the calculus of real functions states that, if for all \((x,y) \in D\) we have both \(u_x(x,y) = 0\) and \(u_y(x,y) = 0\text{,}\) then for all \((x,y) \in D, \, u(x,y)=c_1\text{,}\) where \(c_1\) is a constant. Using a similar argument, we find that \(v(x,y)=c_2\text{,}\) for all \((x,y) \in D\text{,}\) and therefore \(f(z)=f(x,y) = c_1+ic_2\text{,}\) for all \((x,y) \in D\text{.}\) In other words, \(f\) is constant on \(D\text{.}\)

Proof.

By the Cauchy-Riemann equations, \(f\,'(z) =u_x(z)+iv_x(z) = v_y(z) -iu_y(z)\) for all \(z\in D\text{.}\) By hypothesis \(f\,'(z) =0\) for all \(z \in D\text{,}\) so for all \(z \in D\) the functions \(u_x, \, u_y, \, v_x\text{,}\) and \(v_y\) are identically zero. As with the conclusion to the proof of Theorem 3.2.12, this situation means both \(u\) and \(v\) are constant functions, from whence the result follows.

Exercises Exercises

1.

Use the Cauchy Riemann conditions to determine where the following functions are differentiable, and evaluate the derivatives at those points where they exist.
(a)
\(f(z)=iz+4i\text{.}\)
Solution.
\(u(x,y)=-y, \; v(x,y)=x+4; u_x=v_y=0, \; u_y=-v_x=-1\text{.}\) The partials are continuous everywhere, so \(f\,'(z)=u_x+iv_x=i\) for all \(z\text{.}\)
(b)
\(f(z) = f(x,y) = \frac{y+ix}{x^2+y^2}\text{.}\)
(c)
\(f(z) = -2(xy+x) + i(x^2-2y-y^2)\text{.}\)
Solution.
\(u_x=v_y=-2(y+1); u_y=-v_x=-2x\text{.}\) The partials are continuous everywhere, so \(f\,'(z) = u_x+iv_x = -2(y+1)+i2x\) for all \(z\text{.}\)
(d)
\(f(z) = x^3-3x^2-3xy^2+3y^2 + i(3x^2y-6xy-y^3)\text{.}\)
(e)
\(f(z) = x^3 + i(1-y)^3\text{.}\)
Solution.
\(f\) is differentiable only at \(z=i\text{,}\) and \(f\,''(i) = 0\text{.}\)
(f)
\(f(z) = z^2+z\text{.}\)
(g)
\(f(z) = x^2+y^2 + i2xy\text{.}\)
Solution.
\(u_x=v_y=2x, \; u_y=2y, \; v_x=2y\text{.}\) The conditions necessary for Theorem Theorem 3.2.6 are satisfied if and only if \(y=0\text{,}\) and for \(z=(x,0), \; f\,'(z)=2x\text{.}\)
(h)
\(f(z) = |z-(2+i)|^2\text{.}\)

2.

Let \(f\) be a differentiable function. Verify the identity \(|f\,'(z)|^2 = u_x^2+v_x^2=u_y^2+v_y^2\text{.}\)

3.

Find the constants \(a\) and \(b\) so that \(f(z) =(2x-y) + i(ax+by)\) is differentiable for all \(z\text{.}\)
Solution.
\(a=1\) and \(b=2\text{.}\)

4.

Let \(f\) be differentiable at \(z_0=r_0e^{i\theta_0}\text{.}\) Let \(z\) approach \(z_0\) along the ray \(r>0\text{,}\) \(\theta = \theta_0\) and use Equation (3.1.1) to show that Equation (3.2.2) holds.

5.

Let \(f(z) =e^x\cos y+ie^x\sin y\text{.}\) Show that both \(f(z)\) and \(f\,'(z)\) are differentiable for all \(z\text{.}\)
Solution.
\(f\,'(z)=f\,''(z)=e^x\cos y+ie^x\sin y\) by Theorem Theorem 3.2.6.

6.

A vector field \(\mathbf{F}(z) =U(x,y) +iV(x,y)\) is said to be irrotational if \(U_y(x,y)=V_x(x,y)\text{.}\) It is said to be solenoidal if \(U_x(x,y)=-V_y(x,y)\text{.}\) If \(f(z)\) is an analytic function, show that \(\mathbf{F}(z)=f(z)\) is both irrotational and solenoidal.

7.

Use any method to show that the following functions are nowhere differentiable.
(a)
\(h(z) =e^y\cos x+ie^y\sin x\text{.}\)
Solution.
\(u_x=-e^y\sin x, \; v_y=e^y\sin x, \; u_y=e^y\cos x\text{,}\) and \(-v_x=-e^y\cos x\text{.}\) The Cauchy-Riemann equations hold if and only if both \(\sin x=0\) and \(\cos x=0\text{,}\) which is impossible.
(b)
\(g(z)=z+\overline{z}\text{.}\)

8.

Use Theorem 3.2.10 with regard to the following.
(a)
Let \(f(z) =f(re^{i\theta}) = \ln r+i\theta\text{,}\) where \(r>0\) and \(-\pi \lt \theta \lt \pi\text{.}\) Show that \(f\) is analytic in the domain indicated and that \(f\,'(z) = \frac{1}{z}\text{.}\)
(b)
Let \(f(z) =(\ln r)^2-\theta^2+i2\theta \ln r\) where, \(r>0\) and \(-\pi \lt \theta \le \pi\text{.}\) Show that \(f\) is analytic for \(r>0, \, -\pi \lt \theta \lt \pi\text{,}\) and find \(f\,'(z)\text{.}\)

9.

Show that the following functions are entire (see Definition 3.1.5).
(a)
\(f(z) = \cosh x\sin y-i\sinh x\cos y.p\)
Solution.
\(u_x=\sinh x\sin y=v_y\) and \(u_y=\cosh x\sin y=-v_x\text{.}\) The partials are continuous everywhere, so \(f\) is entire.
(b)
\(g(z) = \cosh x\cos y+i\sinh x\sin y\text{.}\)

10.

To prove Theorem 3.2.10, the polar form of the Cauchy-Riemann equations,
(a)
Let \(f(z) =f(x,y) =f(re^{i\theta}) =u(re^{i\theta}) +iv(re^{i\theta}) = U(r,\theta )+iV(r,\theta )\text{.}\) Use the transformation \(x=r\cos \theta\) and \(y=r\sin \theta\) \big(i.e., \((x, y) =re^{i\theta}\)\big) and the chain rules
\begin{equation*} U_r = u_x\frac{\partial x}{\partial r}+u_y\frac{\partial y}{\partial r} \text{ and } U_\theta = u_x\frac{\partial x}{\partial \theta} + u_y\frac{\partial y}{\partial \theta} \text{ (similarly for } V)\text{.} \end{equation*}
to prove that
\begin{align*} U_{r} \amp = u_x\cos \theta +u_y\sin \theta, U_{\theta}=-u_xr\sin \theta +u_yr\cos \theta; \text{ and }\\ V_{r} \amp = v_x\cos \theta +v_y\sin \theta, V_{\theta} = -v_xr\sin \theta +v_yr\cos \theta\text{.} \end{align*}
(b)
Use the original Cauchy-Riemann equations for \(u\) and \(v\) and the results of part (a) to prove that \(rU_{r}=V_{\theta}\) and \(rV_{r}=-U_{\theta}\text{,}\) thus verifying Equation (3.2.10)

11.

Determine where the following functions are differentiable and where they are analytic. Explain!
(a)
\(f(z) =x^3+3xy^2+i(y^3+3x^2y)\text{.}\)
Solution.
\(f\) is differentiable only at points on the coordinate axes, so \(f\) is nowhere analytic.
(b)
\(f(z) =8x-x^3-xy^2+i(x^2y+y^3-8y)\text{.}\)
(c)
\(f(z) =x^2-y^2+i2|xy|\text{.}\)
Solution.
\(f\) is differentiable and analytic inside quadrants I and III.

12.

Let \(f\) and \(g\) be analytic functions in the domain \(D\text{.}\) If \(f\,'(z) = g\,'(z)\) for all \(z\) in \(D\text{,}\) then show that \(f(z) =g(z) +C\text{,}\) where \(C\) is a complex constant.

13.

Explain how the limit definition for the derivative in complex analysis and the limit definition for the derivative in calculus are different. How are they similar?
Solution.
The form of the definition is identical, but the meaning is more subtle in the complex case. For starters, the limit must exist when \(z \to z_0\) from any direction in the complex case. The real case is limited to two directions.

14.

Let \(f\) be an analytic function in the domain \(D\text{.}\) Show that if \(\mathrm{Re}[f(z)]=0\) at all points in \(D\text{,}\) then \(f\) is constant in \(D\text{.}\)

15.

Let \(f\) be a nonconstant analytic function in the domain \(D\text{.}\) Show that the function \(g(z) = \overline{f(z)}\) is not analytic in \(D\text{.}\)
Solution.
Since \(f=u+iv\) is analytic, \(u\) and \(v\) must satisfy the Cauchy-Riemann equations. Since \(f\) is not constant, this means the functions \(u\) and \(-v\) do not satisfy the Cauchy-Riemann equations. Explain why this is the case, and then use Theorem Theorem 3.2.2 to conclude that \(g=u-iv\) is not analytic.

16.

Recall that, for \(z=x+iy\text{,}\) \(x= \frac{z+\overline{z}}{2}\) and \(y= \frac{z-\overline{z}}{2i}\text{.}\)
(a)
Temporarily, think of \(z\) and \(\overline{z}\) as dummy symbols for real variables. With this perspective, \(x\) and \(y\) can be viewed as functions of \(z\) and \(\overline{z}\text{.}\) Use the chain rule for a function \(h\) of two variables to show that
\begin{equation*} \frac{\partial h}{\partial\overline{z}} = \frac{\partial h}{\partial x}\frac{\partial x}{\partial \overline{z}} + \frac{\partial h}{\partial y}\frac{\partial y}{\partial \overline{z}} = \frac{1}{2}\left(\frac{\partial h}{\partial x} + i\frac{\partial h}{\partial y}\right)\text{.} \end{equation*}
(b)
Now define the operator \(\frac{\partial}{\partial \overline{z}} = \frac{ 1}{2}(\frac{\partial}{\partial x} +i\frac{\partial}{\partial y})\) that is suggested by the previous equation. With this construct, show that if \(f=u+iv\) is differentiable at \(z=(x ,y)\text{,}\) then, at the point \((x ,y), \, \frac{\partial f}{\partial \overline{z}} = \frac{1}{2}\left[ u_x-v_y+i(y_x+u_y) \right] = 0\text{.}\) Equating real and imaginary parts thus gives the complex form of the Cauchy-Riemann equations: \(\frac{\partial f}{\partial \overline{z}} = 0\text{.}\)
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