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.  
40 Tall Pine Drive, Sudbury, MA 01776  
Tele.  (800) 832-0034, FAX:  (508)  443-8000  
E-mail:  mkt@jbpub.com  
Internet:  http://www.jbpub.com/  
This free software is compliments of the authors.
John H. Mathews,   mathews@fullerton.edu 
Russell W. Howell,  howell@westmont.edu

CHAPTER 2   COMPLEX FUNCTIONS

\(
\(Needs["Graphics`ComplexMap`"];\)\ \)
Needs["Graphics`ImplicitPlot`"]; 
Needs["Graphics`Colors`"]; 
Needs["Algebra`InequalitySolve`"]; 
a = u = v = w = x = y = z = Z = 
                      0; 
Clear[a, ac, al, b, bc, bhc, bl, bverti, circlea, circleb, eq1, eq2, 
        eq3, eq4, eq5, eqn, eqns, f,  F, hc, hl, horiz, Iden, ineq, ineq1, 
        ineq2, ineq3, ineq4, ineq5, ineq6, ineq7, ineq8, ineq9, θ, r, 
        u, u0, u1, u2,  U, U0, v, vc, vl, verti, V, w, wplane, W, x,  X, y, 
        Y, z, Z, zplane];

Section 2.1	Functions of a Complex Variable

Section 2.2	Transformations and Linear Mappings

Section 2.3	The Mappings w = z^n and w = z^1/n

Section 2.4  Limits and Continuity

Let u = u(x,y) be a real-valued function of the two real variables x and y. Recall that u has the limit (Subscript[u, 0]) as (x,y) approaches ((Subscript[x, 0]), (Subscript[y, 0])) provided that the value of u(x,y) can be made to get as close as we please to the value (Subscript[u, 0]) by taking (x,y) to be sufficiently close to ((Subscript[x, 0]), (Subscript[y, 0])). When this happens we write  

			(lim_((x, y) -> ((Subscript[x, 0]), (Subscript[y, 0])))) u (x, y) = 
        (Subscript[u, 0]).

In more technical language, u has the limit (Subscript[u, 0]) as (x,y) approaches ((Subscript[x, 0]), (Subscript[y, 0])) if and only if |u (x, y) - (Subscript[u, 0])| can be made arbitrarily small by making both |x - (Subscript[x, 0])| and |y - (Subscript[y, 0])| small. This is like the definition of a limit for functions of one variable, except that there are two variables instead of one. Since (x,y) is a point in the xy-plane, and the distance between (x,y) and ((Subscript[x, 0]), (Subscript[y, 0])) is (Sqrt[(((x - (Subscript[x, 0])))^2) + (((y - (Subscript[y, 0])))^2)]), we can give a precise definition of a limit as follows.

Definition 2.3 ( limit of u(x,y) ),  Page 69.  The expression (lim_((x, y) -> ((Subscript[x, 0]), (Subscript[y, 0])))) u (x, y) = 
        (Subscript[u, 0]) means that for each number ϵ > 0, there corresponds a number δ > 0 such that  

			|u (x, y) - (Subscript[u, 0])|<ϵ whenever 0 < (Sqrt[(((x - (Subscript[x, 0])))^2) + (((y - (Subscript[y, 0])))^2)]) < δ.

Example 2.14,  Page 69.  The function  u (x, y) = ((2  (x^3))/((x^2)  +  (y^2)))  has the limit  0  as  (x,y)  approaches  (0,0).

Solution 2.14.

Enter the function  u[x, y] = ((2  (x^3))/((x^2)  +  (y^2))).

Remove[x, y]; 
Clear[lim, r, θ, u, u1, u2, U, U0]; 
u[x_, y_] = ((2  (x^3))/((x^2) + (y^2))); 
Print["u[x,y] = ", u[x, y]];

Find the iterated limit,  first let  x -> 0,   second let  y -> 0.

\(
\(u1\  = \ Limit[u[x, y], x -> 0];\)\ \)
u2  =  Limit[u1 ,  y -> 0]; 
Print["u[x,y] = ", u[x, y]]; 
\(\(Print[\*""    \!\(lim\+\(x -> 0\)\) "", 
        u[x, y], " = ", u1];\)\ \)
Print[""]; 
\(\(Print[\*""(lim_(y -> 0)) (lim_(x -> 0)) ("", u[x, y], \*"") = (lim_(y -> 0)) ("", 
        u1, ")"];\)\ \)
\(\(Print[\*""(lim_(y -> 0)) (lim_(x -> 0)) ("", u[x, y], ") = ", u2];\)\ \)

Find the iterated limit,  first let  y -> 0,   second let  x -> 0.

\(
\(u1\  = \ Limit[u[x, y], y -> 0];\)\ \)
u2  =  Limit[u1 ,  x -> 0]; 
Print["u[x,y] = ", u[x, y]]; 
\(\(Print[\*""    \!\(lim\+\(y -> 0\)\) "", 
        u[x, y], " = ", u1];\)\ \)
Print[""]; 
\(\(Print[\*""(lim_(x -> 0)) (lim_(y -> 0)) ("", u[x, y], \*"") = (lim_(x -> 0)) ("", 
        u1, ")"];\)\ \)
\(\(Print[\*""(lim_(x -> 0)) (lim_(y -> 0)) ("", u[x, y], ") = ", u2];\)\ \)

Find the polar limit  fix  θ,   and let  r -> 0.

\(
\(U[r_, θ_]\  = \ 
        Simplify[u[r\ Cos[θ], r\ Sin[θ]]];\)\ \)
lim  =  Limit[U[r, θ], r -> 0]; 
Print["U[r,θ] = ", 
        U[r, θ]]; 
Print[""]; 
\(\(Print[\*""\!\(lim\+\(r -> 0\)\) "", 
        U[r, θ], " = ", lim];\)\ \)

So, along all lines through the origin, the limit is 0.

A more rigorous proof is needed for other approaches to the origin.  

	\(\(\(\(\(0\)\(≤\)\) | \(2  x\^3\)\/\(x\^2 + y\^2\) | \(\(≤\)\(\(2 | x\^3 | \)\/\(\(|\)\(x\^2 + y\^2\)\(|\)\)\)\)\)\(
\)
      \)\)
\(\(\(\(\(0\)\(≤\)\) | \(2  x\^3\)\/\(x\^2 + y\^2\) | \(\(≤\)\(\(2 | 
                x\( | \^3\)\)\/\(\(|\)\(z\)\( | \^2\)\)\)\)\)\(
\)
      \)\)
\(\(\(\(\(0\)\(≤\)\) | \(2  x\^3\)\/\(x\^2 + y\^2\) | \(\(≤\)\(\(2 | 
                z\( | \^3\)\)\/\(\(|\)\(z\)\( | \^2\)\)\)\)\)\(
\)
      \)\)
\(\(\(\(\(0\)\(≤\)\) | \(2  x\^3\)\/\(x\^2 + y\^2\) | \(\(≤\)\(2\)\) | z | 


        0 ≤ 
          lim\+\(\(\(|\)\(z\)\(|\)\)\  -> \ 0\) | \(2  x\^3\)\/\(x\^2 + y\^2\) | \(\(≤\)\(2 
             lim\+\(\(\(|\)\(z\)\(|\)\)\  -> \ 0\)\)\) | 
        z | 


        0 ≤ 
          lim\+\(\(\(|\)\(z\)\(|\)\)\  -> \ 0\) | \(2  x\^3\)\/\(x\^2 + y\^2\) | \(\(≤\)\(0\)\)\)\(
\)
      \)\)
Therefore
(lim_(z  ->  0))((2  (x^3))/((x^2) + (y^2))) = 0   

We can use Mathematica to make a ContourPlot of  u(x,u).

u[x_, 
          y_] = ((2  (x^3))/((x^2) + (y^2))); 
cont  =  {-0.8, -0.4, -0.2, -0.1, -0.05, 0.05, 
          0.1, 0.2, 0.4, 0.8}; 
ContourPlot[u[x,  y], {x, -1, 1}, {y, -1, 1}, 
        Contours -> cont, 
PlotPoints -> 50, 
        ColorFunction -> Hue]; 
Print["u[x,y] = ", u[x, y]]; 
Print[""]; 
Print[
        u[x, y]  ==  
          c, "  where"]; 
Print["c = ", cont];

Notice that the contours with smaller  |c|  values occur closer to the origin.
We see that  u (x, y) = ((x^3)/((x^2)  +  (y^2)))  has the limit  0  as  (x,y)  approaches  (0,0).

The value (Subscript[u, 0]) of the limit must not depend on how (x,y) approaches ((Subscript[x, 0]), (Subscript[y, 0])), so u(x,y) must approach the value (Subscript[u, 0]) when (x,y)  approaches ((Subscript[x, 0]), (Subscript[y, 0])) along any curve that ends at the point ((Subscript[x, 0]), (Subscript[y, 0])). Conversely, if we can find two curves (Subscript[C, 1]) and (Subscript[C, 2]) that end at ((Subscript[x, 0]), (Subscript[y, 0])) along which u(x,y) approaches the two distinct values (Subscript[u, 1]) and (Subscript[u, 2]), respectively, then u(x,y) does not have a limit as (x,y) approaches ((Subscript[x, 0]), (Subscript[y, 0])).

Example 2.15,  Page 70.  The function  u (x, y) = (( x  y)/((x^2)  +  (y^2)))  does NOT have a limit as  (x,y)  approaches  (0,0).

Solution 2.15.

Enter the function  u[x, y]  = (( x y)/((x^2)  +  (y^2))).

Remove[x, y]; 
Clear[lim, r, θ, u, u0, u1, u2, U, U0]; 
u[x_, y_]  = (( x y)/((x^2) + (y^2))); 
Print["u[x,y] = ", u[x, y]];

Find the iterated limit,  first let  x -> 0,   second let  y -> 0.

\(
\(u1\  = \ Limit[u[x, y], x -> 0];\)\ \)
u2  =  Limit[u1, y -> 0]; 
Print["u[x,y] = ", u[x, y]]; 
\(\(Print[\*""    \!\(lim\+\(x -> 0\)\) "", 
        u[x, y], " = ", u1];\)\ \)
Print[""]; 
\(\(Print[\*""(lim_(y -> 0)) (lim_(x -> 0)) ("", u[x, y], \*"") = (lim_(y -> 0)) ("", 
        u1, ")"];\)\ \)
\(\(Print[\*""(lim_(y -> 0)) (lim_(x -> 0)) ("", u[x, y], ") = ", u2];\)\ \)

Find the iterated limit,  first let  y -> 0,   second let  x -> 0.

\(
\(u1\  = \ Limit[u[x, y], y -> 0];\)\ \)
u2  =  Limit[u1, x -> 0]; 
Print["u[x,y] = ", u[x, y]]; 
\(\(Print[\*""    \!\(lim\+\(y -> 0\)\) "", 
        u[x, y], " = ", u1];\)\ \)
Print[""]; 
\(\(Print[\*""(lim_(x -> 0)) (lim_(y -> 0)) ("", u[x, y], \*"") = (lim_(x -> 0)) ("", 
        u1, ")"];\)\ \)
\(\(Print[\*""(lim_(x -> 0)) (lim_(y -> 0)) ("", u[x, y], ") = ", u2];\)\ \)

Find the polar limit  fix  θ,   and let  r -> 0.

U0  =  
        u[r Cos[θ], r Sin[θ]]; 
U[r_, θ_]  =  Simplify[U0]; 
lim  =  
        Limit[U[r, θ], r -> 0]; 
Print["u[x,y] = ", u[x, y]]; 
Print["U[r,θ] = ", U0]; 
Print["U[r,θ] = ", 
        U[r, θ]]; 
Print[""]; 
\(\(Print[\*""\!\(lim\+\(r -> 0\)\) "", 
        U[r, θ], " = ", lim];\)\ \)

The value of this limit is dependent on the angle  θ.  Therefore,   u (x, y) = (( x  y)/((x^2)  +  (y^2)))  does NOT have a limit  as  (x,y)  approaches  (0,0).

We can use Mathematica to make a ContourPlot of  u(x,u).

u[x_, 
          y_] = (( x y)/((x^2) + (y^2))); 
cont  =  Table[c, {c, -1, 1, 0.2}]; 
ContourPlot[u[x,  y], {x, -1, 1}, {y, -1, 1}, 
        Contours -> cont, 
PlotPoints -> 100, 
        ColorFunction -> Hue]; 
Print["u[x,y] = ", u[x, y]]; 
Print[""]; 
Print[
        u[x, y]  ==  
          c, "  where"]; 
Print["c = ", cont];

Notice that the contours with the given "c" values all go through the origin.  Hence "many limits" are possible as one approaches the origin.  We see that  u (x, y) = (( x  y)/((x^2)  +  (y^2)))  does NOT have a (unique) limit as  (x,y)  approaches  (0,0).

Let f(z) be a complex function of the complex variable z that is defined for all values of z in some neighborhood of (Subscript[z, 0]), except perhaps at the point (Subscript[z, 0]). We say that f has the limit (Subscript[w, 0]) as z approaches (Subscript[z, 0]) provided that the value f(z) can be made as close as we please to the value (Subscript[w, 0]) by taking z to be sufficiently close to (Subscript[z, 0]).  When this happens we write 

			(lim_(z -> (Subscript[z, 0])))f (z)=(Subscript[w, 0]).  

Since the distance between the points z and (Subscript[z, 0]) can be expressed by |z-(Subscript[z, 0])|, we can give a precise definition similar to the one for a function of two variables.

Definition 2.3 ( limit of f(z) ),  Page 71.  The expression (lim_(z -> (Subscript[z, 0])))f (z)=(Subscript[w, 0]) means that for each number ϵ > 0, there exists a real number δ > 0 such that  
			|f (z) - (Subscript[w, 0])|<ϵ whenever 0<|z-(Subscript[z, 0])|<δ.  

This can also be expressed as  

			f (z) ∈ (Subscript[D, ϵ]) ((Subscript[w, 0])) whenever \(f \((z)\) ∈ \(D\_δ\%*\) \((w\_0)\)\).

The latter formulation gives a nice context for looking at this definition. It says that for each disk of radius ϵ > 0 about the point (Subscript[w, 0]) (represented by (Subscript[D, ϵ]) ((Subscript[w, 0]))) there is a punctured disk of radius δ > 0 about the point (Subscript[z, 0]) (represented by \(\(D\_δ\%*\) \((w\_0)\)\)) such that the image of each point in the punctured δ > 0 disk lies in the ϵ > 0. The image of the δ > 0 disk does not have to fill up the entire ϵ > 0;  but if z approaches (Subscript[z, 0]) along a curve that ends at (Subscript[z, 0]), then w=f(z) approaches (Subscript[w, 0]).

Example 2.16 on Page 71.  Show if \(f \((z)\) = \(\(z\)\(\ \)\)\&_\), then (lim_(z -> (Subscript[z, 0])))f (z)=(OverBar[(Subscript[z, 0]) ]), where (Subscript[z, 0]) is any complex number.

Solution 2.16.

f[X_, Y_] = 
        ComplexExpand[
          Conjugate[X + I Y] ]; 
Z = x + I y; 
(Subscript[z, 0]) = (Subscript[x, 0]) + I (Subscript[y, 0]); 
lim = 
        Limit[Limit[f[X, Y], X -> (Subscript[x, 0])], 
          Y -> (Subscript[y, 0])]; 
Print["f[x,y] = ", f[x, y] ]; 
\(\(Print[\*""    \!\(z\_0\) = "", 
        z\_0\ ];\)\ \)
\(\(Print[\*""\!\(lim\+\(z -> z\_0\)\) f[x,y] = "", 
        lim\ ];\)\ \)

If we consider w=f(z) as a mapping from the z-plane into the w-plane and think about the previous geometric interpretation of a limit, then we are led to conclude that the limit of a function f should be determined by the limits of its real and imaginary parts u and v. This will also give us a tool for computing limits.

Theorem 2.1,  Page 72.  Let  f (z) = u (x, y) +  i v (x, y)  be a complex function that is defined in some neighborhood of  (Subscript[z, 0]),  except perhaps at  (Subscript[z, 0])=(Subscript[x, 0])+i (Subscript[y, 0]).  Then  

			(lim_(z -> (Subscript[z, 0])))f (z)=(Subscript[w, 0])=(Subscript[u, 0]) + i (Subscript[v, 0])  
		if and only if  
			(lim_((x, y) -> ((Subscript[x, 0]), (Subscript[y, 0]))))u (x, y)=(Subscript[u, 0])  and  (lim_((x, y) -> ((Subscript[x, 0]), (Subscript[y, 0]))))v (x, y)=(Subscript[v, 0]).

Proof of Theorem 2.1,  see text Page 72.

Example 2.17, Page 73.  Show that (lim_(z -> 1 + i))((z^2) - 2  z + 
              1) = -1.

Solution 2.17.

Enter the function f[z] and find the limit.

Clear[f, lim, z]; 
f[z_] = 1 - 2  z + (z^2); 
lim1 = 
        Limit[1, z -> 1 + I]; 
lim2 = 
        Limit[-2 z, 
          z -> 1 + I]; 
lim3 = Limit[(z^2), z -> 1 + I]; 
lim  =  Limit[f[z], z -> 1 + I]; 
Print["f[z] = ", f[z]]; 
Print[""]; 
\(\(Print[\*""\!\(lim\+\(z -> 1 + i\)\) "", 1, " = ", 
        lim1];\)\ \)
\(\(Print[\*""\!\(lim\+\(z -> 1 + i\)\) "", \(-2\) 
          z, " = ", lim2];\)\ \)
\(\(Print[\*""\!\(lim\+\(z -> 1 + i\)\) "", z\^2, " = ", 
        lim3];\)\ \)
Print[""]; 
\(\(Print[\*""\!\(lim\+\(z -> 1 + i\)\) ("", f[z], ") = ", 
        lim1, " + (", lim2, ") + ", lim3];\)\ \)
\(\(Print[\*""\!\(lim\+\(z -> 1 + i\)\) ("", f[z], ") = ", 
        lim];\)\ \)
\(\(Print[\*""\!\(lim\+\(z -> 1 + i\)\) f[z] = "", 
        lim];\)\ \)

We can use Mathematica to make a mapping of a neighborhood of  z = 1 + i  in the z-plane and its image in the w-plane.

\(
\(Clear[Z, W];\)\ \)
Iden[z_]  =  z; 
f[z_]  =  1 - 2  z + (z^2); 
Z[z_]  =  Iden[z + 1 + I]; 
W[z_]  =  f[z + 1 + I]; 
PolarMap[
        Z , {0, 0.1, 0.02}, {0, 
          2  π, (π/12)}, 
PlotRange -> {{0, 
              1.1}, {0, 1.1}}, 
        AspectRatio -> 
          1, 
Ticks -> {Range[0, 1, (1/2)], 
            Range[0, 1, (1/2)]}, AxesLabel -> {"x", "y"}, 
        PlotStyle -> {Pink, Magenta}];  
Print["A neighborhood of z = 1 + \[ImaginaryI]."]; 
\(\(PolarMap[
        W\ , {0, 0.1, 0.02}, {0, 
          2  π, π\/12}, 
PlotRange -> {{\(-1.32\), 0}, {\(-0.22\), 0.22}}, 
        AspectRatio -> 
          1\/3, 
Ticks -> {Range[\(-1\), 0, 1\/2], 
            Range[\(-1\)\/5, 1\/5, 1\/5]}, 
        AxesLabel -> {"u", "v"}, 
        PlotStyle -> {Pink, Magenta}];\)\ \ \)
Print["The image of the neighborhood,"]; 
Print["under the mapping  w = ", f[z], ","]; 
Print["observe that  w -> -1  as  z -> 1 + \
\[ImaginaryI]"];

We see that  (lim_(z  ->  1 + i)) (1 - 2  z + 
              (z^2)) = -1.

Limits of complex functions are formally the same as in the case of real functions, and the sum, difference, product, and quotient of functions have limits given by the sum, difference, product, and quotient of the respective limits. We state this result as a theorem and leave the proof as an exercise.

Theorem 2.2,  Page 73.  Let  (lim_(z -> (Subscript[z, 0])))f (z) = A  and  (lim_(z -> (Subscript[z, 0])))g (z) = B.  Then  

	(lim_(z -> (Subscript[z, 0])))f (z) + g (z) = A + B,  

	(lim_(z -> (Subscript[z, 0])))f (z) - g (z) = A - B,  

	(lim_(z -> (Subscript[z, 0])))f (z) g (z) = A B,  

	(lim_(z -> (Subscript[z, 0])))((f (z))/(g (z)))=(A/B),  provided that  B ≠ 0.

Proof of Theorem 2.2,  see text Page 73.

Definition 2.5 ( continuity of u(x,y) ),  Page 74.  Let u(x,y) be a real-valued function of the two real variables x and y. We say that u is continuous at the point ((Subscript[x, 0]),(Subscript[y, 0])) if the following three conditions are satisfied:

			(lim_((x, y) -> ((Subscript[x, 0]), (Subscript[y, 0])))) u (x, y) exists,  
		and
			u((Subscript[x, 0]),(Subscript[y, 0])) exists,  
		and
			(lim_((x, y) -> ((Subscript[x, 0]), (Subscript[y, 0])))) u (x, y)=u((Subscript[x, 0]),(Subscript[y, 0])).

The third condition actually implies the first two conditions, since the existence of the quantity on each side of equation the third condition is implicitly understood to exist. For example, if u (x, y) = ((x^3)/((x^2) + (y^2))) when (x, y) ≠ (0, 0) and if u (0, 0) = 0, then we have already seen that u (x, y) = ->0 as (x, y) = ->0  so that all three conditions are satisfied. Hence u(x,y) is continuous at (0,0).

	There is a very similar definition for complex valued functions:

Definition 2.6 ( continuity of f(z) ),  Page 74.  Let f(z) be a complex function of the complex variable z that is defined for all values of z in some neighborhood of (Subscript[z, 0]). We say that f is continuous at (Subscript[z, 0])if the following three conditions are satisfied:  

			(lim_(z -> (Subscript[z, 0])))f (z) exists,  
		and
			f((Subscript[z, 0])) exists,  
		and
			(lim_(z -> (Subscript[z, 0])))f (z)=f((Subscript[z, 0])).

Remark 2.3. Example 2.16 shows that the function \(f \((z)\) = \(\(z\)\(\ \)\)\&_\),  is continuous.

A complex function f is continuous if and only if its real and imaginary parts u and v are continuous. The proof of this fact is an immediate consequence of theorem 2.1. Continuity of complex functions is formally the same as in the case of real functions, and sums, differences, and products of continuous functions are continuous;  their quotient is continuous at points where the denominator is not zero. These results are summarized by the following theorems, and the proofs are left as exercises.

Theorem 2.3,  Page 75.  Let  f (z) = u (x, y) +  i v (x, y)  be a defined in some neighborhood of  (Subscript[z, 0]).  Then  f (z)  is continuous at  (Subscript[z, 0])=(Subscript[x, 0])+i (Subscript[y, 0])  if and only if  u (x, y)  and  v (x, y)  are continuous at  ((Subscript[x, 0]),(Subscript[y, 0])).

Proof of Theorem 2.3,  see text Page 75.

Theorem 2.4,  Page 75.  Suppose that  f (z)  and  g (z)  are continuous at the point  (Subscript[z, 0]).  Then the following functions are continuous at (Subscript[z, 0]).  

	Their sum   f (z) + g (z),  

	Their difference   f (z) - g (z),  

	Their product   f (z) g (z),  

	Their quotient   ((f (z))/(g (z))),  provided that  g((Subscript[z, 0]))≠0.

	Their composition  f (g (z))  provided that  f (z)  is continuous in a neighborhood of the point  g((Subscript[z, 0])).

Proof of Theorem 2.4,  see text Page 75.

Example 2.18,  Page 75.  Show that the polynomial function given by  w = P (z) = (Subscript[a, 0]) + (Subscript[a, 1]) z + (Subscript[a, 2]) 
                  (z^2) +  ...  + (Subscript[a, n]) (z^n)  is continuous at each point (Subscript[z, 0])  in the complex plane.

Solution 2.18.

For illustration, we use  n = 5.
Enter the function P[z].

Remove[a, z]; 
Clear[lim, P]; 
P[z_]  =  Sum[(Subscript[a, n]) (z^n),  {n, 0, 5}]; 
\(\(lim\  = \ 
        Expand[Limit[P[z], \ 
            z -> \*""\!\(z\_0\)""]];\)\ \)
Print["     P[z] = ", P[z]]; 
Print[" "]; 
\(\(Print[\*""    P[\!\(z\_0\)] = "", 
        P[\*""\!\(z\_0\)""]];\)\ \)
\(\(Print[\*""\!\(lim\+\(z -> z\_0\)\) P[z] = "", 
        lim];\)\ \)
Print[" "]; 
\(\(Print[\*"" Is P[(Subscript[z, 0])] = (lim_(z -> (Subscript[z, 0]))) P[z] ?  "", lim\ \  == \ \ P[\*""(Subscript[z, 0])""]];\)\ \)

Therefore  P(z)  is a continuous function for all complex numbers.

An Extra Example for Page 75.  Show that the polynomial  p[z]  =  1  - z  - (z^2)  + (z^3)  - (z^4) + (z^5)  is continuous at the point (Subscript[z, 0]) = 1  in the complex plane.

Solution for Extra Example, Page 75.

Enter the function  p[z]  =  1  - z  - (z^2)  + (z^3)  - (z^4) + (z^5)  and find the limit.

Remove[z]; 
Clear[a, lim, p, z0]; 
z0 = 1; 
p[z_]  =  1  - z  - (z^2)  + (z^3)  - (z^4) + (z^5); 
lim  =  Expand[Limit[p[z],  z -> z0]]; 
Print["     p[z] = ", p[z]]; 
Print[" "]; 
Print["    p[", z0, "] = ", p[z0]]; 
\(\(Print[\*""    \!\(lim\+\(z -> z\_0\)\) p[z] = "", 
        lim];\)\ \)
Print[""]; 
\(\(Print[\*""      \!\(z\_0\) = "", 
        z0];\)\ \)
Print[" "]; 
\(\(Print[\*"" Is p[(Subscript[z, 0])] = (lim_(z -> (Subscript[z, 0]))) p[z] ?  "", lim\ \  == \ \ p[z0]];\)\ \)

We can use Mathematica to make a mapping of a neighborhood of  z = 1  in the z-plane and its image in the w-plane.

\(
\(Clear[Z, W];\)\ \)
Iden[z_]  =  z; 
f[z_]  =  1  - z  - (z^2)  + (z^3)  - (z^4) + (z^5); 
Z[z_]  =  Iden[z + 1]; 
W[z_]  =  f[z + 1]; 
PolarMap[
        Z , {0, 0.09, 0.01}, {0, 
          2  π, (π/12)}, 
PlotRange -> {{0, 
              1.1}, {-0.11, 0.11}}, 
        AspectRatio -> 
          (1/5), 
Ticks -> {Range[0, 1, 0.5], 
            Range[0, 1, 0.5]}, AxesLabel -> {"x", "y"}, 
        PlotStyle -> {Green, DarkGreen}]; 
Print["A neighborhood of z = 1."]; 
\(\(PolarMap[
        W\ , {0, 0.09, 0.01}, {0, 
          2  π, π\/12}, 
PlotRange -> {{\(-0.1\), 0.2}, {\(-0.15\), 0.15}}, 
        AspectRatio -> 
          1, 
Ticks -> {Range[0, 0.2, 0.1], 
            Range[\(-0.1\), 0.1, 0.1]}, AxesLabel -> {"u", "v"}, 
        PlotStyle -> {Green, DarkGreen}];\)\ \)
Print["The image of the neighborhood, under"]; 
Print["the mapping  w = ", f[z], ","]; 
Print["observe that  w -> 0  as  z -> 1"];

We see that  (lim_(z  ->  1)) (1 - z - (z^2) + (z^3) - (z^4) + (z^5)) = 
        0.

Another Extra Example for Page 75.  Show that the polynomial  p[z]  =  -1 + 11  z  - 3  (z^2)  + -4 (z^3)  - 3  (z^4) + 
          2  (z^5)  is continuous at the point (Subscript[z, 0]) = -1 + i  in the complex plane.

Solution for Another Extra Example, Page 75.

Enter the function  p[z]  =  -1 + 11  z  - 3  (z^2)  + -4 (z^3)  - 3  (z^4) + 
          2  (z^5)  and find the limit.

Remove[z]; 
Clear[a, lim, p, z0]; 
z0 = -1 + I; 
p[z_]  =  -1 + 11  z  - 3  (z^2)  + -4 (z^3)  - 3  (z^4) + 
          2  (z^5); 
lim  =  Expand[Limit[p[z],  z -> z0]]; 
Print["     p[z] = ", p[z]]; 
Print[" "]; 
Print["    p[", z0, "] = ", p[z0]]; 
\(\(Print[\*""    \!\(lim\+\(z -> z\_0\)\) p[z] = "", 
        lim];\)\ \)
Print[""]; 
\(\(Print[\*""      \!\(z\_0\) = "", 
        z0];\)\ \)
Print[" "]; 
\(\(Print[\*"" Is p[(Subscript[z, 0])] = (lim_(z -> (Subscript[z, 0]))) p[z] ?  "", lim\ \  == \ \ p[z0]];\)\ \)

We can use Mathematica to make a mapping of a neighborhood of  z = -1 + i  in the z-plane and its image in the w-plane.

\(
\(Clear[Z, W];\)\ \)
Iden[z_]  =  z; 
f[z_]  =  -1 + 11  z  - 3  (z^2)  + -4 (z^3)  - 3  (z^4) + 
          2  (z^5); 
Z[z_]  =  Iden[z - 1 + I]; 
W[z_]  =  f[z - 1 + I]; 
\(\(PolarMap[
        Z\ , {0, 0.002, 0.0002}, {0, 
          2  π, π\/12}, 
PlotRange -> {{\(-1.005\), \(-0.995\)}, {0.995, 1.005}}, 
        AspectRatio -> 
          1, 
Ticks -> {Range[\(-1.005\), \(-0.995\), 
              0.005], Range[0.995, 1.005, 0.005]}, 
        AxesLabel -> {"x", "y"}, 
        PlotStyle -> {Cyan, Blue}];\)\ \)
Print["A neighborhood of z = -1 + \[ImaginaryI]."]; 
\(\(PolarMap[
        W\ , {0, 0.002, 0.0002}, {0, 
          2  π, π\/12}, 
PlotRange -> {{\(-0.1\), 0.1}, {0.9, 1.1}}, 
        AspectRatio -> 
          1, 
Ticks -> {Range[\(-0.1\), 0.1, 0.1], 
            Range[0.9, 1.1, 0.1]}, AxesLabel -> {"u", "v"}, 
        PlotStyle -> {Cyan, Blue}];\)\ \)
Print["The image of the neighborhood, under the"]; 
Print["mapping  w = ", f[z], ","]; 
Print["observe that  w -> \[ImaginaryI]  as  z -> -1 + \
\[ImaginaryI]"];

We see that  (lim_(z  ->  -1 + i)) ( -1 + 
              11  z  - 3  (z^2)  + -4 (z^3)  - 3  (z^4) + 
              2  (z^5)) = i.

One technique for computing limits is to apply the previous theorem to quotients. Let P and Q be polynomials. If Q ((Subscript[z, 0])) ≠ 0, then  

			\(lim\+\(z\  -> \ z\_0\)\ \(P \((z)\)\)\/\(Q \((z)\)\) = \(P \((z\_0)\)\)\/\(Q \((z\_0)\)\)\).  

	Another technique, involves factoring polynomials. If both P ((Subscript[z, 0])) = 0 and Q ((Subscript[z, 0])) = 0, then P and Q can be factored as P (z) = (z - (Subscript[z, 0])) (Subscript[P, 1]) (z) and Q (z) = (z - (Subscript[z, 0])) (Subscript[Q, 1]) (z), and if  Q ((Subscript[z, 0])) ≠ 0, then the limit is  

			\(\(\(lim\+\(z\  -> \ z\_0\)\ \(P \((z)\)\)\/\(Q \((z)\)\)\)\(=\)\(\ \)\)\)\(\(\((z - z\_0)\) \(P\_1\) \((z)\)\)\/\(\((z - z\_0)\) \(Q\_1\) \((z)\)\) = \(\(P\_1\) \((z\_0)\)\)\/\(\(Q\_1\) \((z\_0)\)\)\).

Example 2.19,  Page 76.  Show that \(lim\+\(z\  -> \ 1 + i\)\ \(z\^2\  - \ 2  i\)\/\(z\^2\  - \ 2  z\  + \ 2\) = 1 - i\).

Solution 2.19.

Enter the function f[z] and find the limit.

Clear[f, g, lim, z]; 
f[
          z_]  =  (((z^2)  -  2  I)/((z^2)  -  2  z  +  2)); 
Print["f[z] = ", f[z]];

Try to evaluate f[1+i].

Print["f[1 + \[ImaginaryI]] = ", 
        f[1 + I]];

lim  =  
        Limit[f[z], z -> 1 + I]; 
\(\(Print[\*""(lim_(z -> 1 + i)) f[z] = "", \*""(lim_(z -> 1 + i)) "", f[z], " = ", 
        lim];\)\ \)

Factor the numerator and denominator of  f[z].

p = 
        Solve[Numerator[f[z]] == 0]; 
q = 
        Solve[Denominator[f[z]] == 0]; 
n = (z - 
              (Subscript[p, [[1, 1, 2]]])) (z - 
              (Subscript[p, [[2, 1, 2]]])); 
d = (z - 
              (Subscript[q, [[1, 1, 2]]])) (z - 
              (Subscript[q, [[2, 1, 2]]])); 
Print[Numerator[f[z]], " = 0"]; 
Print[p]; 
Print[Denominator[f[z]], " = 0"]; 
Print[q]; 
Print[""]; 
Print["f[z] = ", n, "/", d];

Find the limit of f[z] as z approaches 1 + i.

Clear[p, q]; 
g[z_]  =  Factor[f[z]]; 
p[z_]  =  Numerator[g[z]]; 
q[z_]  =  Denominator[g[z]]; 
Print["       f[z] = ", f[z]]; 
Print["       g[z] = ", g[z]]; 
Print[""]; 
Print["     g[1+\[ImaginaryI]] = (", 
        p[1 + I], ")/(", 
        q[1 + I], ")"]; 
Print["     g[1+\[ImaginaryI]] = ", 
        g[1 + I]]; 
Print[""]; 
\(\(Print[\*""  \!\(lim\+\(z -> 1 + i\)\) f[z] = "", 
        lim];\)\ \)

We can use Mathematica to make a mapping of a neighborhood of  z = 1 + i  in the z-plane and its image in the w-plane.

\(
\(Clear[Z, W];\)\ \)
Iden[z_]  =  z; 
f[
          z_]  =   (((z^2)  -  2  I)/((z^2)  -  2  z  +  2)); 
Z[z_]  =  Iden[z + 1 + I]; 
W[z_]  =  f[z + 1 + I]; 
PolarMap[
        Z , {0.0001, 0.2001, 0.02}, {0, 
          2  π, (π/12)}, 
PlotRange -> {{0, 
              1.21}, {0, 1.21}}, 
        AspectRatio -> 
          1, 
Ticks -> {Range[0, 1, (1/2)], 
            Range[0, 1, (1/2)]}, AxesLabel -> {"x", "y"}, 
        PlotStyle -> {Blue, Cyan}]; 
Print["A neighborhood of z = 1 + \[ImaginaryI]."]; 
PolarMap[
        W , {0.0001, 0.2001, 0.02}, {0, 
          2  π, (π/12)}, 
PlotRange -> {{0, 
              1.12}, {-1.12, 0}}, 
        AspectRatio -> 
          1, 
Ticks -> {Range[0, 1, (1/2)], 
            Range[-1, 0, (1/2)]}, AxesLabel -> {"u", "v"}, 
        PlotStyle -> {Blue, Cyan}]; 
Print["The image of the neighborhood,"]; 
Print["under the mapping  w = ", f[z], ","]; 
Print["observe that w -> 1-\[ImaginaryI] as z -> 1+\
\[ImaginaryI]"];

We see that  \(lim\+\(z\  -> \ 1 + i\)\ \(z\^2\  - \ 2  i\)\/\(z\^2\  - \ 2  z\  + \ 2\) = 1 - i\).

Section 2.4  Exercises for Limits and Continuity
See textbook page 76.

Section 2.5	Branches of Functions

Section 2.6	The Reciprocal Transformation w  = 1/z

GO TO  THE  Table of Contents