COMPLEX ANALYSIS: Mathematica 4.1 Notebooks
(c) John H. Mathews, and
Russell W. Howell, 2002
Complimentary software to accompany our textbook:

COMPLEX ANALYSIS: for Mathematics and Engineering, 
4th Edition,  2001,  ISBN: 0-7637-1425-9
Jones & Bartlett Publishers, Inc.  
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Tele.  (800) 832-0034, FAX:  (508)  443-8000  
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This free software is compliments of the authors.
John H. Mathews,   mathews@fullerton.edu 
Russell W. Howell,  howell@westmont.edu

CHAPTER 3  ANALYTIC and HARMONIC FUNCTIONS

u = v = w = x = y = z = Z = 
                    0; 
Clear[a, b, difQuotient, dfdt, 
        dfdx, Δf, ΔfΔx, ΔfΔy, ΔfΔz, Δw, Δx, Δy, Δz, E1, 
        E2, E3, E4, E5, f, F, g, g1, g2, g3, g4, g5, lap, lim, L1, L2, sub, 
        u, Ux, Uy, Uθ, v, v1, v2, v3, Vx, Vy, Vθ, w, x, x0, y, 
        y0, z, z0, Z];

Section 3.1	Differentiable Functions

Section 3.2  The Cauchy-Riemann Equations

In Section 3.1we showed that computing the derivative of complex functions written in a nice form such as f (z) = (z^2) is a rather simple task. But life is not so easy, for many times we encounter complex functions written as  f (z) = u (x, y) + i v (x, y). For example, suppose we had  

			f (z) = f (x, 
              y) = u (x, y) + i v (x, y) = ((x^3) - 
                  3 x (y^2)) + i (3 (x^2) y - 
                    (y^3)).

Is there some criterion---perhaps involving the partial derivatives for u, and v---that we can use to determine whether f is differentiable, and if so, to find the value of f (z)?

	The answer to this question is yes, thanks in part to the independent discovery of two important equations relating the partial derivatives of u and v by the French mathematician Augustin Louis Cauchy and the German mathematician Georg Friedrich Bernhard Riemann.

Let  f (z) = u (x, y) + iv (x, y)  be a complex function that is differentiable at the point (Subscript[z, 0]).  Then it is natural to seek a formula for computing  f' ((Subscript[z, 0]))  in terms of the partial derivatives of  u(x,y)  and  v(x,y).  If we investigate this idea, then it is easy to find the required formula;  but we will find that there are special conditions that must be satisfied before it can be used.  In addition, we will discover two important equations relating the partial derivatives of  u  and  v, which were discovered independently by the French mathematician A. L Cauchy and the German mathematician G. F. B. Riemann.

Theorem 3.4,  Page 103. (Cauchy-Riemann Equations)  Suppose that  

			f (z) = f (x + iy) = 
          u (x, y) + i v (x, y)  

is differentiable at the point (Subscript[z, 0]) = (Subscript[x, 0]) + i (Subscript[y, 0]).  Then the partial derivatives of  u  and  v  exist at the point ((Subscript[x, 0]), (Subscript[y, 0])), and we have  

			f' ((Subscript[z, 0])) = (Subscript[u, x]) ((Subscript[x, 0]), 
              (Subscript[y, 0])) + i (Subscript[v, x]) ((Subscript[x, 0]), (Subscript[y, 0])),
		and
			f' ((Subscript[z, 0])) = (Subscript[v, y]) ((Subscript[x, 0]), 
              (Subscript[y, 0])) - i (Subscript[u, y]) ((Subscript[x, 0]), (Subscript[y, 0])).  

Equating the real and imaginary parts of these two versions for the derivatives gives the so-called Cauchy-Riemann Equations  

		(Subscript[u, x]) ((Subscript[x, 0]), (Subscript[y, 0])) = (Subscript[v, y]) ((Subscript[x, 0]), (Subscript[y, 0])) and (Subscript[u, y]) ((Subscript[x, 0]), (Subscript[y, 0])) = -(Subscript[v, x]) ((Subscript[x, 0]), (Subscript[y, 0])).

Proof of Theorem 3.4,  see text Page 103.

Derivation of the Cauchy-Riemann Equations with  Mathematica.

Solution for the Cauchy-Riemann Equations.

Clear[Δx, Δy, E1, E2, 
        E3, E4, E5, f, L1, L2, u, v, x, y, z]; 
TagSet[x, Im[x], 0]; 
TagSet[y, Im[y], 0]; 
TagSet[u, Im[u[x, y]], 0]; 
TagSet[v, Im[v[x, y]], 0]; 
TagSet[x, Re[x], x]; 
TagSet[y, Re[y], y]; 
TagSet[u, Re[u[x, y]], u[x, y]]; 
TagSet[v, Re[v[x, y]], v[x, y]]; 
f[x_  +  I y_]  :=  
        u[x, y]  +  I v[x, y]; 
L1  =  
        Limit[((f[x + Δx + I y] - f[x + I y])/Δ)x, Δx -> 0, 
          Analytic -> True]; 
L2  =  
        Limit[((f[x + I (y + Δy)] - f[x + I y])/(I Δy)), Δy -> 0, Analytic -> True]; 
Print["    f(z) = ", f[x  +  I y]]; 
Print["Re[f(z)] = ", 
        ComplexExpand[Re[f[x + I y]]]]; 
Print["Im[f(z)] = ", 
        ComplexExpand[Im[f[x + I y]]]]; 
Print[" "]; 
\(\(Print[\*""(lim_(Δx -> 0))  Δ(f/Δ)x = "", L1];\)\ \)
\(\(Print[\*""(lim_(Δy -> 0)) \!\(Δf\/\(I\\ Δy\)\) = "", 
        L2];\)\ \)
Print[""]
E1  =  L1  ==  L2; 
Print[E1]; 
Print[" "]; 
Print["The Cauchy-Riemann equations are:"]; 
E2  =  
        ReplaceAll[
          E1, {I -> 0, -I -> 
              0}]; 
Print[E2]; 
E3  =  Thread[-E1, Equal]; 
E4  =  Thread[E2 + E3, Equal]; 
E5  =  
        ReplaceAll[
          E4, {I -> 
              1, -I -> -1}]; 
Print[E5];

We have used Mathematica to derive the Cauchy-Riemann equations  (Subscript[u, x]) ((Subscript[x, 0]), (Subscript[y, 0]))=(Subscript[v, y])((Subscript[x, 0]), (Subscript[y, 0]))  and  (Subscript[u, y])((Subscript[x, 0]), (Subscript[y, 0]))=-(Subscript[v, x]) ((Subscript[x, 0]), (Subscript[y, 0])).

Note carefully some of the implications of this theorem:
	
(i).  If f is differentiable at (Subscript[z, 0]), then we know the Cauchy-Riemann equations will be satisfied at (Subscript[z, 0]), and we can use either equation in Theorem 3.4 to evaluate f' (z). 
	
(ii).  Taking the contrapositive, if the Cauchy-Riemann equations are not satisfied at (Subscript[z, 0]), then we know automatically that f is not differentiable at (Subscript[z, 0]). 

(iii).  On the other hand, just because the Cauchy-Riemann equations are satisfied at (Subscript[z, 0]), we cannot necessarily conclude that f is differentiable at (Subscript[z, 0]).

Example 3.4,  Page 104.  We know that f (z) = (z^2) = (x^2) - (y^2) + i 2 x y  differentiable and that f'[z] = 2  z. We also have  

f (z) = (z^2) = (((x + i y))^2) = ((x^2) - 
                  (y^2)) + i (2 x y).  

As advertised, the Cauchy-Riemann equations are satisfied, because 

	(Subscript[u, x]) (x, y) = 2  x = (Subscript[v, y]) (x, y) and (Subscript[u, y]) (x, y) = -2 y = -(Subscript[v, x]) (x, y).  

Using the formulae in Theorem 3.4 to compute f'[z]  gives

		f'[
          z] = (Subscript[u, x]) (x, y) + i (Subscript[v, x]) (x, 
                y) = 2  x + i 2  
                y = 2  z ,
	and
		f'[
          z] = (Subscript[v, y]) (x, y) - i (Subscript[u, y]) (x, 
                y) = 2  x - i (-2 y) = 2  z,
as expected.

Solution 3.4.

Enter the function f[z] and determine u[x,y] and v[x,y].

Clear[f, x, y, u, v, w, w1, z]; 
f[z_]  =  (z^2); 
w  =  ComplexExpand[f[x + I y]]; 
u[x_, y_] = ComplexExpand[Re[f[x + I y]]]; 
v[x_, y_] = ComplexExpand[Im[f[x + I y]]]; 
Print["  f[z] = ", f[z]]; 
Print["  f[z] = ", w]; 
Print[""]; 
Print["  f[z] = u[x,y] + \[ImaginaryI] v[x,y]"]; 
Print[""]; 
Print["u[x,y] = ", u[x, y]]; 
Print["v[x,y] = ", v[x, y]];

Verify that the Cauchy-Riemann equations hold.

\(
\(Print[\*""Does (∂_x) u[x,y] = (∂_y) v[x,y] ?""];\)\ \)
Print["Does ", (∂_x) 
          u[x, y], " = ", (∂_y) 
          v[x, y], " ?  ", (∂_x) 
            u[x, y] == (∂_y) 
            v[x, y]]; 
Print[""]; 
\(\(Print[\*""Does (∂_y) u[x,y] = -(∂_x) v[x,y] ?""];\)\ \)
Print["Does ", (∂_y) 
          u[x, y], " = ", -(∂_x) 
            v[x, y], " ?  ", (∂_y) 
            u[x, y] == -(∂_x) 
              v[x, y]];

We see that the Cauchy-Riemann equations hold.

Verify that  f'[z] = (Subscript[u, x])[x, y] + i (Subscript[v, x])[x, y].

w1  =  (∂_x) 
            u[x, y]   +   I (∂_x) 
              v[x, y]; 
Print["f'[z] = ", f'[z]]; 
Print["f'[z] = ", f'[x + I y]]; 
Print["f'[z] = ", 
        ComplexExpand[f'[x + I y]]]; 
Print[" "]; 
\(\(Print[\*""f'[z] = (Subscript[u, x])[x,y] + I (Subscript[v, x])[x,y]""];\)\ \)
Print["f'[z] = (", (∂_x) 
          u[x, y], ") + \[ImaginaryI](", (∂_x) 
          v[x, y], ")"]; 
Print["f'[z] = ", w1];

Example 3.5,  Page 105.  Verify that the complex function  \(\(\(f \((z)\)\)\(=\)\(\(\(z\)\(\ \)\)\&_\)\(\ \)\)\)  is NOT analytic for any value of  z. 
Furthermore, the Cauchy-Riemann are not satisfied at any point.

Solution 3.5.

Enter the function f[z].

Clear[f, u, Ux, Ux0, Uy, Uy0, v, Vx, Vx0, Vy, Vy0, 
        w, x, y]; 
TagSet[x, Im[x], 0]; 
TagSet[y, Im[y], 0]; 
TagSet[x, Re[x], x]; 
TagSet[y, Re[y], y]; 
w  =  Conjugate[x + I y]; 
Print["f[z] = ", w]; 
w  =  
        ComplexExpand[w, TargetFunctions -> {Im, Re}]; 
Print["f[z] = ", w];

Find the real and imaginary parts of  f[z].

u[x_, y_]  =  
        Together[ComplexExpand[Re[w]]]; 
v[x_, y_]  =  Together[ComplexExpand[Im[w]]]; 
u[0, 0]  =  0; 
v[0, 0]  =  0; 
f[x_, y_]  =  u[x, y]  +  I v[x, y]; 
f[0, 0]  =  0; 
Print["  f[z] = u[x,y] + \[ImaginaryI] v[x,y]"]; 
Print["  f[z] = ", f[x, y]]; 
Print[""]; 
Print["u[x,y] = ", u[x, y]]; 
Print["v[x,y] = ", v[x, y]];

Determine if the Cauchy-Riemann equations hold.

Ux  =  
        Together[(∂_x) u[x, y]]; 
Uy  =  Together[(∂_y) u[x, y]]; 
Vx  =  Together[(∂_x) v[x, y]]; 
Vy  =  Together[(∂_y) v[x, y]]; 
\(\(Print[\*""\!\(u\_x\)[x,y] = "", Ux];\)\ \)
\(\(Print[\*""\!\(u\_y\)[x,y] = "", Uy];\)\ \)
\(\(Print[\*""\!\(v\_x\)[x,y] = "", Vx];\)\ \)
\(\(Print[\*""\!\(v\_y\)[x,y] = "", 
        Vy];\)\ \)
Print[""]; 
\(\(Print[\*""\!\(u\_x\)[x,y] - \!\(v\_y\)[x,y] = "", 
        Simplify[Ux - Vy]];\)\ \)
\(\(Print[\*""\!\(u\_y\)[x,y] + \!\(v\_x\)[x,y] = "", 
        Simplify[Uy + Vx]];\)\ \)

Hence the Cauchy-Riemann do not hold at any point.

Thus, we have shown that  \(\(\(f \((z)\)\)\(=\)\(\ \)\(\(\(z\)\(\ \)\)\&_\)\(\ \)\)\)  is NOT analytic for any value of  z.

Example 3.6,  Page 105.  Verify that the complex function  \(f \((z)\) = \((\ \(\(z\)\(\ \)\)\&_\ )\)\^2\/z\)  is NOT analytic for any value of  z. 
Furthermore, the Cauchy-Riemann equations hold at (0,0), but f(z) is not differentiable at  z = 0.

Solution 3.6.

Enter the function f[z].

Clear[f, u, Ux, Ux0, Uy, Uy0, v, Vx, Vx0, Vy, Vy0, 
        w, x, y]; 
TagSet[x, Im[x], 0]; 
TagSet[y, Im[y], 0]; 
TagSet[x, Re[x], x]; 
TagSet[y, Re[y], y]; 
\(\(w\  = \ 
        Conjugate[x + I\ y]\^2\/\(x + I\ y\);\)\ \)
Print["f[z] = ", w]; 
w  =  
        ComplexExpand[w, TargetFunctions -> {Im, Re}]; 
Print["f[z] = ", w];

Find the real and imaginary parts of  f[z].

u[x_, y_]  =  
        Together[ComplexExpand[Re[w]]]; 
v[x_, y_]  =  Together[ComplexExpand[Im[w]]]; 
u[0, 0]  =  0; 
v[0, 0]  =  0; 
f[x_, y_]  =  u[x, y]  +  I v[x, y]; 
f[0, 0]  =  0; 
Print["  f[z] = u[x,y] + \[ImaginaryI] v[x,y]"]; 
Print["  f[z] = ", f[x, y]]; 
Print[""]; 
Print["u[x,y] = ", u[x, y]]; 
Print["v[x,y] = ", v[x, y]];

Determine where the Cauchy-Riemann equations hold.

Ux  =  
        Together[(∂_x) u[x, y]]; 
Uy  =  Together[(∂_y) u[x, y]]; 
Vx  =  Together[(∂_x) v[x, y]]; 
Vy  =  Together[(∂_y) v[x, y]]; 
\(\(Print[\*""\!\(u\_x\)[x,y] = "", Ux];\)\ \)
\(\(Print[\*""\!\(u\_y\)[x,y] = "", Uy];\)\ \)
\(\(Print[\*""\!\(v\_x\)[x,y] = "", Vx];\)\ \)
\(\(Print[\*""\!\(v\_y\)[x,y] = "", 
        Vy];\)\ \)
Print[""]; 
\(\(Print[\*""\!\(u\_x\)[x,y] - \!\(v\_y\)[x,y] = 0 = "", 
        Simplify[Ux - Vy]];\)\ \)
\(\(Print[\*""\!\(u\_y\)[x,y] + \!\(v\_x\)[x,y] = 0 = "", 
        Simplify[Uy + Vx]];\)\ \)

Hence the function  f(z)  is differentiable only when  (x^2) - (y^2)=0  and  x y=0  hence only at the origin.   
Let's check to see if the Cauchy-Riemann equations hold at  (0,0).

Ux0  =  
        Limit[((u[x, 0] - u[0, 0])/(x - 0)),  x -> 0,  
          Analytic -> True]; 
Vy0  =  
        Limit[((v[0, y] - v[0, 0])/(y - 0)),  y -> 0,  
          Analytic -> True]; 
Uy0  =  
        Limit[((u[0, y] - u[0, 0])/(y - 0)),  y -> 0,  
          Analytic -> True]; 
Vx0  =  
        Limit[((v[x, 0] - v[0, 0])/(x - 0)),  x -> 0,  
          Analytic -> True]; 
\(\(Print[\*""\!\(u\_x\)[0,0] = "", Ux0];\)\ \)
\(\(Print[\*""\!\(u\_y\)[0,0] = "", Uy0];\)\ \)
\(\(Print[\*""\!\(v\_x\)[0,0] = "", Vx0];\)\ \)
\(\(Print[\*""\!\(v\_y\)[0,0] = "", 
        Vy0];\)\ \)
Print[""]; 
\(\(Print[\*""Is \!\(u\_x\)[0,0] =  \!\(v\_y\)[0,0] ?    "", 
        Ux0 == Vy0];\)\ \)
\(\(Print[\*""Is \!\(u\_y\)[0,0] = -\!\(v\_x\)[0,0] ?    "", 
        Uy0 == \(-Vx0\)];\)\ \)

So,  (Subscript[u, x]) (0, 0)=(Subscript[v, y])(0, 0)  and  (Subscript[u, y])(0, 0)=-(Subscript[v, x]) (0, 0).  
BUT  f(z)  is NOT analytic at (0,0) because f(z) is NOT analytic in a neighborhood of (0,0).

Also,  f(z)  is NOT analytic at (0,0) because the following two limits are distinct.

dfdx  =  
        Limit[((f[x, 0] - f[0, 0])/x), x -> 0, 
          Analytic -> True]; 
dfdt  =  
        Limit[((f[t, t] - f[0, 0])/(t + I t)), t -> 0, 
          Analytic -> True]; 
Print["Along  the  x-axis we have"]; 
\(\(Print[\*""(lim_(x -> 0)) (df/dx) = (lim_(x -> 0)) ((f[x, 0] - f[0, 0])/x)""];\)\ \)
\(\(Print[\*""(lim_(x -> 0)) (df/dx) = (lim_(x -> 0)) ("", f[x, 0] - f[0, 0], ")/(", 
        x, ")"];\)\ \)
\(\(Print[\*""(lim_(x -> 0)) (df/dx) = (lim_(x -> 0)) ("", \((f[x, 0] - f[0, 0])\)/
          x, ")"];\)\ \)
\(\(Print[\*""\!\(lim\+\(x -> 0\)\) \!\(df\/dx\) = "", 
        dfdx];\)\ \)
Print[""]; 
Print["Along the line y=t, y=t we have"]; 
\(\(Print[\*""(lim_(t -> 0)) (df/dt) = (lim_(t -> 0)) \!\(\(f[t, t] - f[0, 0]\)\/\(t + I\\ t\)\)""];\)\ \)
\(\(Print[\*""(lim_(t -> 0)) (df/dt) = (lim_(t -> 0)) ("", f[t, t] - f[0, 0], ")/(", 
        t + I\ t, ")"];\)\ \)
\(\(Print[\*""(lim_(t -> 0)) (df/dt) = (lim_(t -> 0)) ("", \((f[t, t] - 
              f[0, 0])\)/\((t + I\ t)\), ")"];\)\ \)
\(\(Print[\*""\!\(lim\+\(t -> 0\)\) \!\(df\/dt\) = "", 
        dfdt];\)\ \)

Thus, we have shown that  \(f \((z)\) = \((\ \(\(z\)\(\ \)\)\&_\ )\)\^2\/z\)  is NOT analytic for any value of  z.

Example 3.6 reiterates that the mere satisfaction of the Cauchy-Riemann equations is not a sufficient criterion to guarantee the differentiability of a function. However, theorem 3.5  gives us sufficient conditions under which we can use equations (3-15) or (3-16) to compute the derivative f' ((Subscript[z, 0])). They are referred to as the Cauchy-Riemann conditions for differentiability.

Theorem 3.5 (Cauchy-Riemann conditions for differentiability),  Page 106.  Let  f (z) = f (x + i y) = 
          u (x, y) + i v (x, y)  be a continuous function that is defined in some neighborhood of the point  (Subscript[z, 0]) = (Subscript[x, 0]) + i (Subscript[y, 0]). If all the partial derivatives (Subscript[u, x]),(Subscript[u, y]),(Subscript[v, x]) and (Subscript[v, y]) are continuous at the point((Subscript[x, 0]), (Subscript[y, 0])) and if the Cauchy-Riemann equations (Subscript[u, x]) ((Subscript[x, 0]), (Subscript[y, 0])) = (Subscript[v, y]) ((Subscript[x, 0]), (Subscript[y, 0])) and (Subscript[u, y]) ((Subscript[x, 0]), (Subscript[y, 0])) = -(Subscript[v, x]) ((Subscript[x, 0]), (Subscript[y, 0])) hold, then  f(z)  is differentiable at  (Subscript[z, 0]) and the derivative f' ((Subscript[z, 0]))can be computed with either formula 

			f' ((Subscript[z, 0])) = (Subscript[u, x]) ((Subscript[x, 0]), 
              (Subscript[y, 0])) + i (Subscript[v, x]) ((Subscript[x, 0]), (Subscript[y, 0])),
		or
			f' ((Subscript[z, 0])) = (Subscript[v, y]) ((Subscript[x, 0]), 
              (Subscript[y, 0])) - i (Subscript[u, y]) ((Subscript[x, 0]), (Subscript[y, 0])).

Proof of Theorem 3.5,  see text Page 106.

Example 3.7,  Page 108.  Show that the function  f (z) = 
        (x^3) - 3 x (y^2) + i (3 (x^2) y - (y^3))  is differentiable  for all z and find its derivative.  Verify that the Cauchy-Riemann equations hold and that  f'[z] = (Subscript[u, x])[x, y] + i (Subscript[v, x])[x, y].

Solution 3.7.

Enter the function f[z] and determine u[x,y] and v[x,y].

Clear[f, x, y, u, v, w, w1, z]; 
f[z_]  =  (z^3); 
w  =  ComplexExpand[f[x + I y]]; 
u[x_, y_] = ComplexExpand[Re[f[x + I y]]]; 
v[x_, y_] = ComplexExpand[Im[f[x + I y]]]; 
Print["  f[z] = ", f[z]]; 
Print["  f[z] = ", w]; 
Print[""]; 
Print["  f[z] = u[x,y] + \[ImaginaryI] v[x,y]"]; 
Print[""]; 
Print["u[x,y] = ", u[x, y]]; 
Print["v[x,y] = ", v[x, y]];

Verify that the Cauchy-Riemann equations hold.

\(
\(Print[\*""Does (∂_x) u[x,y] = (∂_y) v[x,y] ?""];\)\ \)
Print["Does ", (∂_x) 
          u[x, y], " = ", (∂_y) 
          v[x, y], " ?  ", (∂_x) 
            u[x, y] == (∂_y) 
            v[x, y]]; 
Print[""]; 
\(\(Print[\*""Does (∂_y) u[x,y] = -(∂_x) v[x,y] ?""];\)\ \)
Print["Does ", (∂_y) 
          u[x, y], " = ", -(∂_x) 
            v[x, y], " ?  ", (∂_y) 
            u[x, y] == -(∂_x) 
              v[x, y]];

We see that the Cauchy-Riemann equations hold.

Verify that  f'[z] = (Subscript[u, x])[x, y] + i (Subscript[v, x])[x, y].

w1  =  (∂_x) 
            u[x, y]   +   I (∂_x) 
              v[x, y]; 
Print["f'[z] = ", f'[z]]; 
Print["f'[z] = ", f'[x + I y]]; 
Print["f'[z] = ", 
        ComplexExpand[f'[x + I y]]]; 
Print[" "]; 
\(\(Print[\*""f'[z] = (Subscript[u, x])[x,y] + I (Subscript[v, x])[x,y]""];\)\ \)
Print["f'[z] = (", (∂_x) 
          u[x, y], ") + \[ImaginaryI](", (∂_x) 
          v[x, y], ")"]; 
Print["f'[z] = ", w1];

Example 3.8,  Page 108.  Use the Cauchy-Riemann equations to show that  \(f \((z)\) = e\^\(-y\)\ cos \((x)\) + i\ e\^\(-y\)\ sin \((x)\)\) is differentiable for all  z = x + i y.  Furthermore, f(z) is analytic for all z.

Solution 3.8.

Enter the function f[z] and determine if the Cauchy-Riemann equations hold.

Clear[f, u, Ux, Uy, v, Vx, Vy, w1, x, y, 
        z]; 
u[x_, y_]  =  (E^(-y)) Cos[x]; 
v[x_, y_]  =  (E^(-y)) Sin[x]; 
Ux  =  Together[(∂_x) u[x, y]]; 
Uy  =  Together[(∂_y) u[x, y]]; 
Vx  =  Together[(∂_x) v[x, y]]; 
Vy  =  
        Together[(∂_y) v[x, y]]; 
Print["   f[z] = ", 
        u[x, y] + I v[x, y]]; 
Print["   f[z] = u[x,y] + \[ImaginaryI] v[x,y]"]; 
Print[" u[x,y] = ", u[x, y]]; 
Print[" v[x,y] = ", v[x, y]]; 
Print[""]; 
\(\(Print[\*""\!\(u\_x\)[x,y] = "", Ux];\)\ \)
\(\(Print[\*""\!\(u\_y\)[x,y] = "", Uy];\)\ \)
\(\(Print[\*""\!\(v\_x\)[x,y] = "", Vx];\)\ \)
\(\(Print[\*""\!\(v\_y\)[x,y] = "", 
        Vy];\)\ \)
Print[""]; 
\(\(Print[\*""Is \!\(u\_x\)[x,y] =  \!\(v\_y\)[x,y] ?    "", 
        Ux == Vy];\)\ \)
\(\(Print[\*""Is \!\(u\_y\)[x,y] = -\!\(v\_x\)[x,y] ?    "", 
        Uy == \(-Vx\)];\)\ \)

The Cauchy-Riemann equations hold everywhere, so that  \(f \((z)\) = e\^\(-y\)\ cos \((x)\) + i\ e\^\(-y\)\ sin \((x)\)\) is analytic for all values of  z.

Print[" f[z] =  f[x + \[ImaginaryI] y] = ", 
        u[x, y]  +  I v[x, y]]; 
Print["f'[z] = f'[x + \[ImaginaryI] y] = ", (∂_x) 
            u[x, y]  +  I (∂_x) 
              v[x, y]];

Remark.  We can write this function as  \(f \((z)\) = \(e\^\(i\ z\) = \(e\^\(i\ \((x + i\ y)\)\) = e\^\(\(-y\) + i\ x\)\)\)\),  and investigate  f' (z).

f[
          z_]  =  (E^(I z)); 
w1  =  (∂_x) 
            u[x, y]   +   I (∂_x) 
              v[x, y]; 
Print["    f[z] = ", f[z]]; 
Print[" f[x+\[ImaginaryI]y] = ", 
        f[x + I y]]; 
Print[" f[x+\[ImaginaryI]y] = ", 
        ComplexExpand[
          f[x  +  I y]]]; 
\(\(Print[" f[x+Iy] = u[x,y] + I v[x,y]"];\)\ \)
Print[" "]; 
Print["   f'[z] = ", f'[z]]; 
Print["f'[x+\[ImaginaryI]y] = ", f'[
          x + I y]]; 
Print["f'[x+\[ImaginaryI]y] = ", 
        ComplexExpand[f'[x + I y]]]; 
Print[" "]; 
\(\(Print[\*""f'[x+Iy] = (Subscript[u, x])[x,y]+ I (Subscript[v, x])[x,y]""];\)\ \)
Print["f'[x+\[ImaginaryI]y] = ", 
        w1];

We have shown that  \(f \((z)\) = e\^\(-y\)\ cos \((x)\) + i\ e\^\(-y\)\ sin \((x)\)\) is differentiable and hence analytic for all z.

The Cauchy-Riemann Conditions are particularly useful in determining the set of points for which a function f is differentiable.

Example 3.9,  Page 109.  Show that the complex function  f (z) = 
        (x^3) + 3 x (y^2) + i (3 (x^2) y + (y^3))  is differentiable only at points that lie on the coordinate axes.

Solution 3.9.

Enter the functions u[x,y]  and  v[x,y] and use them to form  f[z].  Then determine where the Cauchy-Riemann equations hold.

Clear[u, Ux, Uy, v, Vx, Vy, x, y]; 
u[x_, y_]  =  (x^3)  +  3 x (y^2); 
v[x_, y_]  =  (y^3)  +  3 (x^2) y; 
Ux  =  Together[(∂_x) u[x, y]]; 
Uy  =  Together[(∂_y) u[x, y]]; 
Vx  =  Together[(∂_x) v[x, y]]; 
Vy  =  
        Together[(∂_y) v[x, y]]; 
Print["   f[z] = ", 
        u[x, y]  +  I v[x, y]]; 
Print["   f[z] = u[x,y] + \[ImaginaryI] v[x,y]"]; 
Print[" u[x,y] = ", u[x, y]]; 
Print[" v[x,y] = ", v[x, y]]; 
Print[""]; 
\(\(Print[\*""\!\(u\_x\)[x,y] = "", Ux];\)\ \)
\(\(Print[\*""\!\(u\_y\)[x,y] = "", Uy];\)\ \)
\(\(Print[\*""\!\(v\_x\)[x,y] = "", Vx];\)\ \)
\(\(Print[\*""\!\(v\_y\)[x,y] = "", 
        Vy];\)\ \)
Print[""]; 
\(\(Print[\*""Is \!\(u\_x\)[x,y] =  \!\(v\_y\)[x,y] ?      "", 
        Ux == Vy];\)\ \)
\(\(Print[\*""Is \!\(u\_y\)[x,y] = -\!\(v\_x\)[x,y] ?      "", 
        Uy == \(-Vx\)];\)\ \)
\(\(Print[\*""Is \!\(u\_y\)[x,y] + \!\(v\_x\)[x,y] = 0 ?   "", 
        Uy + Vx == 0];\)\ \)

The Cauchy-Riemann equations hold only if  x y = 0  which occurs along the coordinate axes.  Furthermore, all of the partial derivatives  (∂_x) u[x, y] = 
        3 (x^2) + 3  (y^2), (∂_y) u[x, y] = 
        6 x y, (∂_x) v[x, y] = 6 x y,  
      and (∂_y) v[x, y] = 3 (x^2) + 3  (y^2)  are continuous.  Hence,  f (z) = 
        (x^3) + 3 x (y^2) + i (3 (x^2) y + (y^3)) is differentiable only when  x=0 or y=0,  which occurs at points that lie on the coordinate axes.  Remark.  In the next section we will learn that  f (z) = 
        (x^3) + 3 x (y^2) + i (3 (x^2) y + (y^3))  is nowhere analytic.

Theorem 3.6  (Polar Form of C-R equations),  Page 109.  Let  \(f \((z)\) = \(f \((r\ e\^\(i\ θ\))\) = U \((r, θ)\) + i\ V \((r, θ)\)\)\)  be a continuous function that is defined in some neighborhood of the point  (Subscript[z, 0]) = (Subscript[r, 0]) (e^(i (Subscript[θ, 0]))).  If all of the first order partial derivatives (Subscript[U, r]),(Subscript[U, θ]),(Subscript[V, x]) and (Subscript[V, θ]) are continuous at the point  ((Subscript[r, 0]), (Subscript[θ, 0]))  and if the Cauchy-Riemann equations 

		\(\(U\_r\) \((r\_0, θ\_0)\) = \(1\/r\_0\) \(V\_θ\) \((r\_0, θ\_0)\)\) and \(\(V\_r\) \((r\_0, θ\_0)\) = \(\(-1\)\/r\_0\) \(U\_θ\) \((r\_0, θ\_0)\)\)  hold, 

then  f(z)  is differentiable at  (Subscript[z, 0]), and we can compute the derivative f ((Subscript[z, 0])) by using either polar formulae  

			\(f' \((z\_0)\) = \(\(f' \((r\_0\ e\^\(i\ θ\_0\))\)\)\(=\)\(\ \)\)\)\(\(\(\ \)\(\(e\^\(\(-i\)\ θ\_0\)\) \((\(U\_r\) \((r\_0, θ\_0)\) + i\ \(V\_r\) \((r\_0, θ\_0)\))\)\)\)\)  
		or  
			\(f' \((z\_0)\) = \(\(f' \((r\_0\ e\^\(i\ θ\_0\))\)\)\(=\)\(\ \)\)\)\(\(1\/r\_0\) \(e\^\(\(-i\)\ θ\_0\)\) \((\(V\_θ\) \((r\_0, θ\_0)\) - i\ \(U\_θ\) \((r\_0, θ\_0)\))\)\).

Proof of Theorem 3.4,  see text Page 109.

Example 3.10,  Page 110.  Show that if  \(f \((z)\) = \(z\^\(1/2\) = 
          r\^\(1/2\)\ \((cos \((θ\/2)\) + i\ sin \((θ\/2)\))\)\)\)  

where the domain is restricted to be  r > 0  and   - π < θ < π,  then the derivative is given by  

		\(f' \((z)\) = \(1\/\(2  z\^\(1/2\)\) = 
          r\^\(\(-1\)/2\)\/2\ \((cos \((θ\/2)\) - i\ sin \((θ\/2)\))\)\)\)  

where  r > 0  and   - π < θ < π.

Solution 3.10.

Enter  u[r, θ]  and   v[r, θ] and use them to form f[z].  Then verify that polar form of the Cauchy-Riemann equations hold.

Clear[r, θ, u, v]; 
u[r_, θ_] = (r^(1/2)) Cos[(θ/2)]; 
v[r_, θ_] = (r^(1/2)) Sin[(θ/2)]; 
Ur  = (∂_r) u[r, θ]; 
Uθ  = (∂_θ) u[r, θ]; 
Vr  = (∂_r) v[r, θ]; 
Vθ  = (∂_θ) 
          v[r, θ]; 
Print["f[r,θ] = ", 
        u[r, θ] + I v[
              r, θ]]; 
Print["f[r,θ] = u[r,θ] + \[ImaginaryI] \
v[r,θ]"]; 
Print["u[r,θ] = ", u[r, θ]]; 
Print["v[r,θ] = ", 
        u[r, θ]]; 
Print[""]; 
\(\(Print[\*""(∂_r) u[r,θ] = "", ∂\_r\ u[r, θ]];\)\ \)
\(\(Print[\*""(∂_θ) u[r,θ] = "", ∂\_θ\ u[r, θ]];\)\ \)
\(\(Print[\*""(∂_r) v[r,θ] = "", ∂\_r\ v[r, θ]];\)\ \)
\(\(Print[\*""(∂_θ) v[r,θ] = "", ∂\_θ\ v[r, θ]];\)\ \)
Print[""]; 
\(\(Print[\*""Is (∂_r) u[r,θ] = (1/r) (∂_θ) v[r,θ] ?   "", ∂\_r\ 
            u[r, θ] == 
          1\/r\ ∂\_θ\ v[r, θ]];\)\ \)
\(\(Print[\*""Is (∂_r) v[r,θ] = ((-1)/r) (∂_θ) u[r,θ] ?  "", ∂\_r\ 
            v[r, θ] == \(-1\)\/r\ ∂\_θ\ 
              u[r, θ]];\)\ \)

Second, find the derivative using polar formulae  \(f' \((z)\) = \(e\^\(\(-i\)\ θ\)\) \((∂\_r\ 
                u[r, θ] + i\ ∂\_r\ 
                  v[r, θ])\)\)  and  \(f' \((z)\) = \(e\^\(\(-i\)\ θ\)\) \((∂\_r\ 
                u[r, θ] + i\ ∂\_r\ 
                  v[r, θ])\)\).

Print[" f(z) = ", 
        u[r, θ] + I v[
              r, θ]]; 
\(\(Print[" f(z) = u[r,θ] + I v[r,θ]"];\)\ \)
Print[""]; 
\(\(Print[\*""f'(z) = (E^(-Iθ))((∂_r) u[r,θ] + I (∂_r) v[r,θ])""];\)\ \)
\(\(Print["f'(z) = ", \(E\^\(\(-I\)\ θ\)\) \((∂\_r\ 
                u[r, θ] + I\ ∂\_r\ 
                  v[r, θ])\)];\)\ \)
\(\(Print["f'(z) = ", 
        ComplexExpand[E\^\(\(-I\)\ θ\)] \((∂\_r\ u[r, θ] + I\ ∂\_r\ 
                  v[r, θ])\)];\)\ \)
\(\(Print["f'(z) = ", 
        Simplify[
          ComplexExpand[E\^\(\(-I\)\ θ\)] \((∂\_r\ 
                  u[r, θ] + I\ ∂\_r\ 
                    v[r, θ])\)]];\)\ \)

We have shown that if  \(f \((z)\) = \(z\^\(1/2\) = 
          r\^\(1/2\)\ \((cos \((θ\/2)\) + i\ sin \((θ\/2)\))\)\)\)  where the domain is restricted to be  r > 0  and   - π < θ < π,  then the derivative is given by  \(f' \((z)\) = \(1\/\(2  z\^\(1/2\)\) = 
          r\^\(\(-1\)/2\)\/2\ \((cos \((θ\/2)\) - i\ sin \((θ\/2)\))\)\)\)  where  r > 0  and   - π < θ < π.

Again, we close this section with an important theoretical result that extends properties of real functions to their complex counterparts.

Theorem 3.7,  Page 111.  Let f be an analytic function in the domain D. If f (z) ≡ 0 for all z in D, then f is constant in D.

Proof of Theorem 3.7,  see text Page 111.

Library Research Experience for Undergraduates

Project I. Write a about analytic functions.

1. Arsove, Maynard G.,''On the Definition of an Analytic Function,'' Am. Math. M., Vol. 62, No. 1. (Jan., 1955), pp. 22-25, Jstor.   

2. Bear, H. S. and G. N. Hile, ''Gradient Characterizations of Analyticity,'' Am. Math. M., Vol. 85, No. 5. (May, 1978), pp. 333-337, Jstor.   

3. Boas, R. P., (1989), ''When is a (C^∞) Function Analytic?'', Math. Intell., V. 11, No. 4, pp. 34-37.

4. Brand, Louis, ''Non-Analytic Functions of a Complex Variable,'' Am. Math. M., Vol. 40, No. 5. (May, 1933), pp. 260-265, Jstor.   

5. Buff, Joseph J., (1973), ''Characterization of an Analytic Function of a Quaternion Variable,'' Pi Mu Epsilon J., Vol. 5, No. 8, pp. 387-392.

6. Buoncristiana, A. M., (1974), ''On Analytic Function of a Quaternion Variable,'' Pi Mu Epsilon J., V. 6, No. 1, pp. 8-11.

7. Cater, F. S., (1984), ''Differentiable, Nowhere Analytic Functions,'' Am. Math. M., Vol. 91, No. 10. (Dec., 1984), pp. 618-624, Jstor.   

8. Folland, G. B., (1986), ''On Characterizations of Analyticity (in Notes),'' Am. Math. M., Vol. 93, No. 8. (Oct., 1986), pp. 640-641, Jstor.   

9. Goodman, A. W., ''On a Characterization of Analytic Functions,'' Am. Math. M., Vol. 71, No. 3. (Mar., 1964), pp. 265-267, Jstor.   

10. Min, Szu-Hoa , ''Non-Analytic Functions,'' Am. Math. M., Vol. 51, No. 9. (Nov., 1944), pp. 510-517, Jstor.   

11. Moppert, C. F., ''On the Notion of Analyticity,'' Proceedings of the American Mathematical Society, Vol. 10, No. 4. (Aug., 1959), pp. 574-576., Jstor.   

12. Snider, Arthur David, (1978), ''On the Definition of Analyticity,'' Int. J. of Math. Ed. in Sci. and Tech., V. 9, No. 3, pp. 373-374.

13. Walsh, Bertram, ''More on Gradient Characterizations of Analyticity (in Classroom Notes),'' Am. Math. M., Vol. 86, No. 6. (Jun. - Jul., 1979), pp. 490-491, Jstor.

Project II. Write a report about the Cauchy-Riemann equations.
	
1. Gray, J. D. and S. A. Morris, ''When is a Function that Satisfies the Cauchy-Riemann Equations Analytic?,'' Am. Math. M., Vol. 85, No. 4. (Apr., 1978), pp. 246-256., Jstor.   

2. Phipps, T. E., ''Cauchy-Riemann Relations in n Dimensions (in Brief Versions),'' Am. Math. M., Vol. 75, No. 8. (Oct., 1968), pp. 881-882, Jstor.   

3. Roy, S., (1979), ''Cauchy-Riemann Differential Equations in Classical Hydrodynamics,'' Int. J. of Math. Ed. in Sci. and Tech., V. 10, No. 2, p. 291.

4. Williams, K. P., ''A Generalization of the Cauchy-Riemann Equations,'' The Annals of Math., 2nd Ser., Vol. 30, No. 1/4. (1928 - 1929), pp. 206-210, Jstor.

Section 3.2  Exercises for The Cauchy-Riemann Equations
See textbook page 111.

Section 3.3	Analytic Functions and Harmonic Functions

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