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  
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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 1  COMPLEX NUMBERS

Load Mathematica's  "Arrow"  and  "Colors"  packages.

Needs["Graphics`Arrow`"]; 
Needs["Graphics`Colors`"]; 
a = b = c = e = P = r = t = θ = v = V = w = x = y = z = Z = 0; 
Clear[a, b, bdd, c, disk, dot, e, eqn, ext, int, k, list, P, r, r1, 
        r2, set, solset, t, θ, θ1, θ2, v, V, values, w, 
        x, x0, x1, x2, x3, xx, y, y0, y1, y2, y3, yy, z, z1, z2, z3, z4, 
        Z];

Section 1.1	The Origin of Complex Numbers

Section 1.2  The Algebra of Complex Numbers

We have seen that complex numbers came to be viewed as ordered pairs of real numbers. That is, a complex number z is defined to be 
			z = (x, y), 
where x and y are both real numbers.
	The reason we say ordered pair is because we are thinking of a point in the plane. The point (2, 3), for example, is not the same as (3, 2). The order in which we write x and y in the equation makes a difference. Clearly, then, two complex numbers are equal if and only if their x coordinates are and their y coordinates are equal. In other words,
		(x, y)=(u, v) iff x=u and y=v.
	(Throughout this text, iff means if and only if.)

	If we are to have a meaningful number system, there needs to be a method for combining these ordered pairs. We need to define algebraic operations in a consistent way so that the sum, difference, product, and quotient of any two ordered pairs will again be an ordered pair. The key to defining how these numbers should be manipulated is to follow Gauss' lead and equate (x, y) with x+iy. Then, by letting (Subscript[z, 1]) = ((Subscript[x, 1]), (Subscript[y, 1])) and (Subscript[z, 2]) = ((Subscript[x, 2]), (Subscript[y, 2])) be arbitrary complex numbers, we see that  

		(Subscript[z, 1]) + (Subscript[z, 2]) = ((Subscript[x, 1]), (Subscript[y, 1])) + ((Subscript[x, 2]), (Subscript[y, 2]))  
		\(\(\(\ \ \ \ \ \ \ \ \ \ \ \ \ \)\(\(=\)\(\((x\_1 + i\ y\_1)\) + \((x\_2 + i\ y\_2)\)\)\)\)\)  
		             =((Subscript[x, 1]) + 
              (Subscript[x, 2])) + i ((Subscript[y, 1]) + (Subscript[y, 2]))  
		             =((Subscript[x, 1]) + (Subscript[x, 2]), 
          (Subscript[y, 1]) + (Subscript[y, 2]))  

	Thus, if (Subscript[z, 1]) = ((Subscript[x, 1]), (Subscript[y, 1])) and (Subscript[z, 2]) = ((Subscript[x, 2]), (Subscript[y, 2])), are arbitrary complex numbers, the following should certainly make sense:

Definition 1.1, (Addition),  Page 7.  
Formula (1.6)  We can derive the general formula  (Subscript[z, 1])  + (Subscript[z, 2])  = ((Subscript[x, 1]) + (Subscript[y, 1])) + ((Subscript[x, 2]) + (Subscript[y, 2]))=((Subscript[x, 1]) + (Subscript[x, 2]), (Subscript[y, 1]) + (Subscript[y, 2])).

Derivation of  Formula (1.6).

Clear[x, y, z]; 
(Subscript[z, 1]) =  (Subscript[x, 1])  +  I (Subscript[y, 1]); 
(Subscript[z, 2]) =  (Subscript[x, 2])  +  I (Subscript[y, 2]); 
\(\(Print[\*""   \!\(z\_1\) = ""\ , \ z\_1];\)\ \)
\(\(Print[\*""   \!\(z\_2\) = ""\ , \ 
        z\_2];\)\ \)
Print[""]; 
\(\(Print[\*""\!\(z\_1\) + \!\(z\_2\) = ""\ , \ 
        z\_1 + z\_2];\)\ \)
Print[""]; 
\(\(Print[\*""\!\(z\_1\) + \!\(z\_2\) = ""\ , \ 
        ComplexExpand[z\_1 + z\_2]];\)\ \)

Definition 1.2, (Subtraction),  Page 7.  
Formula (1.7)  We can derive the general formula  (Subscript[z, 1])  - (Subscript[z, 2])  = ((Subscript[x, 1]) + (Subscript[y, 1])) - ((Subscript[x, 2]) + (Subscript[y, 2]))=((Subscript[x, 1]) - (Subscript[x, 2]), (Subscript[y, 1]) - (Subscript[y, 2])).

Derivation of  Formula (1.7).

Clear[x, y, z]; 
(Subscript[z, 1]) =  (Subscript[x, 1])  +  I (Subscript[y, 1]); 
(Subscript[z, 2]) =  (Subscript[x, 2])  +  I (Subscript[y, 2]); 
\(\(Print[\*""   \!\(z\_1\) = ""\ , \ z\_1];\)\ \)
\(\(Print[\*""   \!\(z\_2\) = ""\ , \ 
        z\_2];\)\ \)
Print[""]; 
\(\(Print[\*""\!\(z\_1\) - \!\(z\_2\) = ""\ , \ 
        z\_1 - z\_2];\)\ \)
Print[""]; 
\(\(Print[\*""\!\(z\_1\) - \!\(z\_2\) = ""\ , \ 
        ComplexExpand[z\_1 - z\_2]];\)\ \)

Example 1.1,  Page 7.  Given  (Subscript[z, 1]) = 3 + 7  i   and   (Subscript[z, 2]) = 
          5 - 6  i.  (a)  Find  (Subscript[z, 1]) + (Subscript[z, 2])  and  (b)  (Subscript[z, 1]) - (Subscript[z, 2]).

Solution 1.1. (a).

(a)  Compute  (Subscript[z, 1]) + (Subscript[z, 2]).

Clear[V, w, z]; 
(Subscript[z, 1]) = 3 + 7  I; 
(Subscript[z, 2]) = 5 - 6  I; 
(Subscript[w, 1]) = (Subscript[z, 1]) + (Subscript[z, 2]); 
(Subscript[V, 1])  =  
        Graphics[{Red, Arrow[{0,  0}, {Re[(Subscript[z, 1])], Im[(Subscript[z, 1])]}]}]; 
(Subscript[V, 2])  =  
        Graphics[{Green, 
            Arrow[{Re[(Subscript[z, 1])], Im[(Subscript[z, 1])]}, {Re[(Subscript[w, 1])], 
                Im[(Subscript[w, 1])]}]}]; 
(Subscript[V, 3])  =  
        Graphics[{Blue, 
            Arrow[{0,  0}, {Re[(Subscript[w, 1])], Im[(Subscript[w, 1])]}]}]; 
Show[(Subscript[V, 1]), (Subscript[V, 2]), (Subscript[V, 3]), 
        PlotRange -> {{0, 8}, {0, 
              7}}, 
Ticks -> {Range[0, 8, 1], 
            Range[0, 7, 1]}, 
AspectRatio -> (7/8), 
        Axes -> True, 
        AxesLabel -> {"x", "y"}]; 
\(\(Print[\*""     \!\(z\_1\) = ""\ , \ z\_1];\)\ \)
\(\(Print[\*""     \!\(z\_2\) = ""\ , \ z\_2];\)\ \)
\(\(Print[\*""\!\(z\_1\) + \!\(z\_2\) = ""\ , \ 
        z\_1\  + \ z\_2];\)\ \)

Solution 1.1. (b).

(b) Compute (Subscript[z, 1]) - (Subscript[z, 2]).

Clear[w, z]; 
(Subscript[z, 1]) = 3 + 7  I; 
(Subscript[z, 2]) = 5 - 6  I; 
(Subscript[w, 2]) = (Subscript[z, 1]) - (Subscript[z, 2]); 
v1  =  
        Graphics[{Red, Arrow[{0,  0}, {Re[(Subscript[z, 1])], Im[(Subscript[z, 1])]}]}]; 
v2  =  
        Graphics[{Green, 
            Arrow[{Re[(Subscript[w, 2])], Im[(Subscript[w, 2])]}, {Re[(Subscript[z, 1])], 
                Im[(Subscript[z, 1])]}]}]; 
v3  =  
        Graphics[{Blue, 
            Arrow[{0,  0}, {Re[(Subscript[w, 2])], Im[(Subscript[w, 2])]}]}]; 
Show[v1, v2, v3, PlotRange -> {{-2, 3}, {0, 13}}, 
        Ticks -> {Range[-2, 3, 1], 
            Range[0, 13, 1]}, 
AspectRatio -> (13/5), 
        Axes -> True, 
        AxesLabel -> {"x", "y"}]; 
\(\(Print[\*""     \!\(z\_1\) = ""\ , \ z\_1];\)\ \)
\(\(Print[\*""     \!\(z\_2\) = ""\ , \ z\_2];\)\ \)
\(\(Print[\*""\!\(z\_1\) - \!\(z\_2\) = ""\ , \ 
        z\_1\  - \ z\_2];\)\ \)

At this point, it is tempting to define the product (Subscript[z, 1]) (Subscript[z, 2]) as (Subscript[z, 1]) (Subscript[z, 2])  = ((Subscript[x, 1]) (Subscript[x, 2]), (Subscript[y, 1]) (Subscript[y, 2])). It turns out, however, that this is not a good definition, and you will be asked in the problem set for this section to explain why. How, then, should products be defined? Again, if we equate (x,y) with x+iy and assume, for the moment, that i = (Sqrt[-1]) makes sense (so that (i^2) = -1), we have  

			(Subscript[z, 1]) (Subscript[z, 2])  = ((Subscript[x, 1]),  (Subscript[y, 1])) ((Subscript[x, 2]), (Subscript[y, 2]))  
			\(\(\(\ \ \ \ \ \ \ \ \ \ \)\(\(=\)\(\((x\_1 + i\ y\_1)\) \((x\_2 + iy\_2)\)\)\)\)\)  
			\(\(\(\ \ \ \ \ \ \ \ \ \ \)\(\(=\)\(x\_1\ x\_2 + i\ x\_1\ y\_2 + i\ x\_2\ y\_1 + i\^2\ y\_1\ y\_2\)\)\)\)  
			          =(Subscript[x, 1]) (Subscript[x, 2]) - 
          (Subscript[y, 1]) (Subscript[y, 2]) + i ((Subscript[x, 1]) (Subscript[y, 2]) + 
                (Subscript[x, 2]) (Subscript[y, 1])) 
			          =((Subscript[x, 1]) (Subscript[x, 2]) - (Subscript[y, 1]) (Subscript[y, 2]), 
          (Subscript[x, 1]) (Subscript[y, 2]) + (Subscript[x, 2]) (Subscript[y, 1]))  

	Thus, it appears we are forced into the following definition.

Definition 1.3, (Complex Multiplication),  Page 8.  
Formula (1.8)  We can derive the general formula  (Subscript[z, 1]) (Subscript[z, 2])  = ((Subscript[x, 1]),  (Subscript[y, 1])) ((Subscript[x, 2]), (Subscript[y, 2]))=((Subscript[x, 1]) (Subscript[x, 2]) - (Subscript[y, 1]) (Subscript[y, 2]), (Subscript[x, 1]) (Subscript[y, 2]) + (Subscript[x, 2]) (Subscript[y, 1])).

Derivation of  Formula (1.8).

Clear[x, y, z]; 
(Subscript[z, 1]) =  (Subscript[x, 1])  +  I (Subscript[y, 1]); 
(Subscript[z, 2]) =  (Subscript[x, 2])  +  I (Subscript[y, 2]); 
\(\(Print[\*""   \!\(z\_1\) = ""\ , \ z\_1];\)\ \)
\(\(Print[\*""   \!\(z\_2\) = ""\ , \ 
        z\_2];\)\ \)
Print[""]; 
\(\(Print[\*""\!\(z\_1\) \!\(z\_2\) = ""\ , \ 
        z\_1\ z\_2];\)\ \)
Print[""]; 
\(\(Print[\*""\!\(z\_1\) \!\(z\_2\) = ""\ , \ 
        ComplexExpand[z\_1\ z\_2]];\)\ \)

Example 1.2,  Page 8.  Given  (Subscript[z, 1]) = 3 + 7  i   and   (Subscript[z, 2]) = 
          5 - 6  i.  Find  (Subscript[z, 1]) (Subscript[z, 2]).

Solution 1.2.

Compute (Subscript[z, 1]) (Subscript[z, 2]).

Clear[V, w, z]; 
(Subscript[z, 1]) = 3 + 7  I; 
(Subscript[z, 2]) = 5 - 6  I; 
(Subscript[w, 3]) = (Subscript[z, 1]) (Subscript[z, 2]); 
(Subscript[V, 1])  =  
        Graphics[{Red, Arrow[{0,  0}, {Re[(Subscript[z, 1])], Im[(Subscript[z, 1])]}]}]; 
(Subscript[V, 2])  =  
        Graphics[{Green, Arrow[{0, 0}, {Re[(Subscript[z, 2])], Im[(Subscript[z, 2])]}]}]; 
(Subscript[V, 3]) =  
        Graphics[{Blue, 
            Arrow[{0,  0}, {Re[(Subscript[w, 3])], Im[(Subscript[w, 3])]}]}]; 
Show[(Subscript[V, 1]), (Subscript[V, 2]), (Subscript[V, 3]), PlotRange -> {{0, 57}, {-6, 17}}, 
        Ticks -> {Range[0, 57, 5], 
            Range[-6, 17, 2]}, 
AspectRatio -> 1, 
        Axes -> True, 
        AxesLabel -> {"x", "y"}]; 
\(\(Print[\*""   \!\(z\_1\) = ""\ , \ z\_1];\)\ \)
\(\(Print[\*""   \!\(z\_2\) = ""\ , \ z\_2];\)\ \)
\(\(Print[\*""\!\(z\_1\) \!\(z\_2\) = ""\ , \ \(z\_1\) 
          z\_2];\)\ \)

To motivate our definition for division, we will proceed along the same lines as we did for multiplication, assuming (Subscript[z, 2]) ≠ 0:  

			\(z\_1\/z\_2 = \(\((x\_1, \ y\_1)\)\/\((x\_2, y\_2)\) = \(x\_1 + i\ y\_1\)\/\(x\_2 + iy\_2\)\)\)  
	
	At this point we need to figure out a way to be able to write the preceding quantity in the form x+iy. To do this, we use a standard trick and multiply the numerator and denominator by (Subscript[x, 2]) + i(Subscript[y, 2]). This gives 

			\(z\_1\/z\_2 = \(\((x\_1 + i\ y\_1)\) \((x\_2 - i\ y\_2)\)\)\/\(\((x\_2 + i\ y\_2)\) \((x\_2 - i\ y\_2)\)\)\)  
			      =(((Subscript[x, 1]) (Subscript[x, 2]) + 
              (Subscript[y, 1]) (Subscript[y, 2]) + i (-(Subscript[x, 1]) (Subscript[y, 2]) + 
                    (Subscript[x, 2]) (Subscript[y, 1])))/((Subscript[x, 2]^2) + (Subscript[y, 2]^2)))  
			\(\(\(\ \ \ \ \ \ \)\(\(=\)\(\(x\_1\ x\_2 + y\_1\ y\_2\)\/\(x\_2\%2 + y\_2\%2\) + i \(\(\ \)\(\(-x\_1\)\ y\_2 + x\_2\ y\_1\)\)\/\(x\_2\%2 + y\_2\%2\)\)\)\)\)
			  \(\(\(\ \ \ \ \)\(\(=\)\((\(x\_1\ x\_2 + y\_1\ y\_2\)\/\(x\_2\%2 + y\_2\%2\), \(\(\ \)\(\(-x\_1\)\ y\_2 + x\_2\ y\_1\)\)\/\(x\_2\%2 + y\_2\%2\))\)\)\)\).  

Thus, we finally arrive at a rather odd definition.

Definition 1.4, (Complex Division),  Page 9.  
Formula (1.9)  We can derive the general formula  ((Subscript[z, 1])/(Subscript[z, 2])) = ((((Subscript[x, 1]),  (Subscript[y, 1])))/(((Subscript[x, 2]), (Subscript[y, 2]))))=\(\((\(x\_1\ x\_2 + y\_1\ y\_2\)\/\(x\_2\%2 + y\_2\%2\), \(\(\ \)\(\(-x\_1\)\ y\_2 + x\_2\ y\_1\)\)\/\(x\_2\%2 + y\_2\%2\))\)\), for (Subscript[z, 2]) ≠ 0.

Derivation of  Formula (1.9).

Clear[x, y, z]; 
(Subscript[z, 1]) =  (Subscript[x, 1])  +  I (Subscript[y, 1]); 
(Subscript[z, 2]) =  
        (Subscript[x, 2])  +  I (Subscript[y, 2]); 
n = 
        Expand[(Subscript[z, 1]) ((Subscript[x, 2]) - I (Subscript[y, 2]))]; 
d = Expand[(Subscript[z, 2]) ((Subscript[x, 2]) - I (Subscript[y, 2]))]; 
\(\(Print[\*"" \!\(z\_1\) = ""\ , \ z\_1];\)\ \)
\(\(Print[\*"" \!\(z\_2\) = ""\ , \ 
        z\_2];\)\ \)
Print[""]; 
\(\(Print[\*""\!\(z\_1\/z\_2\) = ""\ , \ z\_1\/z\_2];\)\ \)
\(\(Print[\*""\!\(z\_1\/z\_2\) = ""\ , 
        n\/d];\)\ \)
Print[""]; 
\(\(Print[\*""((Subscript[z, 1])/(Subscript[z, 2])) = ""\ , \(ComplexExpand[Re[n]] + I\ ComplexExpand[Im[n]]\)\/d];\)\ \)

Example 1.3,  Page 9.  Given  (Subscript[z, 1]) = 3 + 7  i   and   (Subscript[z, 2]) = 
          5 - 6  i.  Find  ((Subscript[z, 1])/(Subscript[z, 2])).

Solution 1.3.

Compute ((Subscript[z, 1])/(Subscript[z, 2])).

Clear[V, w, z]; 
(Subscript[z, 1]) = 3 + 7  I; 
(Subscript[z, 2]) = 5 - 6  I; 
(Subscript[w, 4]) = ((Subscript[z, 1])/(Subscript[z, 2])); 
(Subscript[V, 1]) =  
        Graphics[{Red, Arrow[{0,  0}, {Re[(Subscript[z, 1])], Im[(Subscript[z, 1])]}]}]; 
(Subscript[V, 2])  =  
        Graphics[{Green, Arrow[{0, 0}, {Re[(Subscript[z, 2])], Im[(Subscript[z, 2])]}]}]; 
(Subscript[V, 3])  =  
        Graphics[{Blue, 
            Arrow[{0,  0}, {Re[(Subscript[w, 4])], Im[(Subscript[w, 4])]}]}]; 
Show[(Subscript[V, 1]), (Subscript[V, 2]), (Subscript[V, 3]), PlotRange -> {{-1, 5}, {-6, 7}}, 
        Ticks -> {Range[-1, 5, 1], 
            Range[-6, 7, 1]}, 
AspectRatio -> 
          (13/6), Axes -> True, 
        AxesLabel -> {"x", "y"}]; 
\(\(Print[\*"" \!\(z\_1\) = ""\ , \ 
        z\_1];\)\ \)
\(\(Print[\*"" \!\(z\_2\) = ""\ , \ z\_2];\)\ \)
\(\(Print[\*""\!\(z\_1\/z\_2\) = ""\ , \ 
        z\_1\/z\_2];\)\ \)

The technique most mathematicians would use to perform operations on complex numbers is to appeal to the notation x+iy and perform the algebraic manipulations, as we did here, rather than to apply the complicated looking definitions we gave for those operations on ordered pairs. This is a valid procedure since the x+iy notation was used as a guide to see how we should define the operations in the first place. It is important to remember, however, that the x+iy notation is nothing more than a convenient bookkeeping device for keeping track of how to manipulate ordered pairs. It is the ordered pair algebraic definitions that really form the foundation on which our complex number system is based. In fact, if you were to program a computer to do arithmetic on complex numbers, your program would perform calculations on ordered pairs, using exactly the definitions that we gave.

	It turns out that our algebraic definitions give complex numbers all the properties we normally ascribe to the real number system. Taken together, they describe what algebraists call a Field. In formal terms, a field is a set (in this case, the complex numbers) together with two binary operations (in this case, addition and multiplication) with the following properties:

(P1) Commutative Law for Addition.  
Property (P1), Page 10.  We can derive the general formula  (Subscript[z, 1]) + (Subscript[z, 2])  =  (Subscript[z, 2]) + (Subscript[z, 1]).

Derivation of  Property (P1).

Clear[x, y, z]; 
(Subscript[z, 1])  =  (Subscript[x, 1])  +  I (Subscript[y, 1]); 
(Subscript[z, 2])  =  (Subscript[x, 2])  +  I (Subscript[y, 2]); 
\(\(Print[\*""     \!\(z\_1\) = ""\ , z\_1];\)\ \)
\(\(Print[\*""     \!\(z\_2\) = ""\ , \ 
        z\_2];\)\ \)
Print[""]; 
\(\(Print[\*""\!\(z\_1\) + \!\(z\_2\) = ""\ , \ 
        z\_1 + z\_2];\)\ \)
\(\(Print[\*""\!\(z\_2\) + \!\(z\_1\) = ""\ , \ 
        z\_2 + z\_1];\)\ \)
Print[""]; 
\(\(Print[\*""Is (Subscript[z, 1]) + (Subscript[z, 2]) = (Subscript[z, 2]) + (Subscript[z, 1]) ? "", z\_1 + z\_2\  == \ z\_2 + z\_1];\)\ \)

(P2) Associative Law for Addition.  
Property (P2), Page 10.  We can derive the general formula  (Subscript[z, 1]) + ((Subscript[z, 2]) + (Subscript[z, 3])) = ((Subscript[z, 1]) + (Subscript[z, 2])) +  (Subscript[z, 3]).

(P3) Additive Identity.
Property (P3), Page 10.  There is a complex number ω such that z + ω = z for all complex numbers z. The number ω is obviously the ordered pair (0,0).

(P4) Additive Inverses.
Property (P4), Page 10.  Given any complex number z, there is a complex number η (depending on z) with the property that z + η = (0, 0). Obviously, if z = (x, y) = x + i y, the number η will be \(η = \(\((\(-x\), \(-y\))\) = \(\(-x\) - i\ y = \(-z\)\)\)\).

(P5) Commutative Law for Multiplication.   
Property (P5), Page 10.  We can derive the general formula  (Subscript[z, 1]) (Subscript[z, 2])  =  (Subscript[z, 2]) (Subscript[z, 1]).

(P6) Associative Law for Addition.  
Property (P6), Page 10.  We can derive the general formula  (Subscript[z, 1]) + ((Subscript[z, 2]) (Subscript[z, 3])) = ((Subscript[z, 1]) (Subscript[z, 2])) +  (Subscript[z, 3]).

(P7) Multiplicative Identity.
Property (P7), Page 10.  There is a complex number ζ such that ζz = z for all complex numbers z. As one might expect, it turns out that (1,0) is the unique complex number ζ with this property. You will be asked to verify this in the problem set for this section.

(P8) Multiplicative Inverses.
Property (P8), Page 10.  Given any number z other than the number (0, 0), there is a complex number (depending on z) which we shall denote by (z^(-1)) with the property that z (z^(-1)) = (1, 0) = 1. Given our definition for division, it seems reasonable that the number (z^(-1)) would be (z^(-1)) = (((1, 0))/z) = (1/z). You will be asked to confirm this in the problem set for this section.

(P9) The Distributive Law.  
Property (P 9), Page 10.  We can derive the general formula (Subscript[z, 1]) ((Subscript[z, 2]) + (Subscript[z, 3])) = (Subscript[z, 1]) ((Subscript[z, 2]) + (Subscript[z, 3])).

Derivation of  Property (P 9).

Clear[x, y, z]; 
(Subscript[z, 1]) = (Subscript[x, 1]) + I (Subscript[y, 1]); 
(Subscript[z, 2]) = (Subscript[x, 2]) + I (Subscript[y, 2]); 
(Subscript[z, 3]) = (Subscript[x, 3]) + I (Subscript[y, 3]); 
\(\(Print[\*""          \!\(z\_1\) = ""\ , \ z\_1];\)\ \)
\(\(Print[\*""          \!\(z\_2\) = ""\ , \ z\_2];\)\ \)
\(\(Print[\*""          \!\(z\_3\) = ""\ , \ z\_3];\)\ \)
Print[""]; 
\(\(Print[\*""  (Subscript[z, 1])((Subscript[z, 2]) + (Subscript[z, 3])) = ""\ , \(z\_1\) \((z\_2 + z\_3)\)];\)\ \)
\(\(Print[\*""(Subscript[z, 1]) (Subscript[z, 2]) + (Subscript[z, 1]) (Subscript[z, 3]) = ""\ , \(z\_1\) z\_2 + z\_1\ z\_3];\)\ \)
Print[""]; 
\(\(Print[\*""  \!\(z\_1\)(\!\(z\_2\) + \!\(z\_3\)) = ""\ , \ 
        Expand[\(z\_1\) \((z\_2 + z\_3)\)]];\)\ \)
\(\(Print[\*""(Subscript[z, 1]) (Subscript[z, 2]) + (Subscript[z, 1]) (Subscript[z, 3]) = ""\ , \ Expand[z\_1\ z\_2 + z\_1\ z\_3]];\)\ \)
Print[""]; 
\(\(Print[\*""Is (Subscript[z, 1])((Subscript[z, 2]) + (Subscript[z, 3])) =  (Subscript[z, 1]) (Subscript[z, 2]) + (Subscript[z, 1]) (Subscript[z, 3]) ?  "", 
        Expand[\(z\_1\) \((z\_2 + z\_3)\)]\ \  == \ \ Expand[
            z\_1\ z\_2 + z\_1\ z\_3]];\)\ \)

None of these properties is difficult to prove. Most of the proofs make use of corresponding facts in the real number system. 
Thus, we may move freely between the notations x+iy and (x,y), depending on which is more convenient for the context in which we are working.

Definition 1.5, (Real Part of z), Page 12.  The real part of z=x+iy, denoted by Re(z), is the real number x.

Definition 1.6, (Imaginary Part of z), Page 12.  The imaginary part of z=x+iy, denoted by Im(z), is the real number y.

Definition 1.7, (Conjugate of z), Page 12.  The conjugate of z=x+iy, denoted by \(\(\(z\)\(\ \)\)\&_\), is the complex number (x,-y) = x - iy.

The following theorem gives some important facts relating to these operations. You will be asked for a proof in the exercises

Example 1.4,  Page 12.  Given  (Subscript[z, 1]) = -3 + 7  i   and   (Subscript[z, 2]) = 
          9 + 4  i.   
1.4 (a)  Find  Re[(Subscript[z, 1])]  and   Re[(Subscript[z, 2])].  
1.4 (b)  Find  Im[(Subscript[z, 1])]  and   Im[(Subscript[z, 2])].  
1.4 (c)  Find  \(\(z\_1\)\&_\ \ \ and\ \ \ \(z\_2\)\&_\).

Solution 1.4 (a).

(a) Compute Re[(Subscript[z, 1])].

Clear[V, z]; 
(Subscript[z, 1])  =  -3  +  7  I; 
(Subscript[V, 1]) =  
        Graphics[{Red, Arrow[{0,  0}, {Re[(Subscript[z, 1])], Im[(Subscript[z, 1])]}]}]; 
(Subscript[V, 2])  =  
        Graphics[{Blue, Arrow[{0,  0}, {Re[(Subscript[z, 1])], 0}]}]; 
Show[(Subscript[V, 1]), (Subscript[V, 2]), 
        PlotRange -> {{-3, 0}, {0, 
              7}}, 
Ticks -> {Range[-3, 0, 1], 
            Range[0, 7, 1]}, 
AspectRatio -> (7/3), 
        Axes -> True, 
        AxesLabel -> {"x", "y"}]; 
\(\(Print[\*""    \!\(z\_1\) = ""\ , \ z\_1];\)\ \)
\(\(Print[\*""Re[\!\(z\_1\)] = ""\ , \ 
        Re[z\_1]];\)\ \)

(b) Compute Re[(Subscript[z, 2])].

Clear[V, z]; 
(Subscript[z, 2])  =  9  +  4  I; 
(Subscript[V, 1])  =  
        Graphics[{Red, Arrow[{0,  0}, {Re[(Subscript[z, 2])], Im[(Subscript[z, 2])]}]}]; 
(Subscript[V, 2])  =  
        Graphics[{Blue, Arrow[{0,  0}, {Re[(Subscript[z, 2])], 0}]}]; 
Show[(Subscript[V, 1]), (Subscript[V, 2]), 
        PlotRange -> {{0, 9}, {0, 
              4}}, 
Ticks -> {Range[0, 9, 1], 
            Range[0, 4, 1]}, 
AspectRatio -> (4/9), 
        Axes -> True, 
        AxesLabel -> {"x", "y"}]; 
\(\(Print[\*""    \!\(z\_2\) = ""\ , \ z\_2];\)\ \)
\(\(Print[\*""Re[\!\(z\_2\)] = ""\ , \ 
        Re[z\_2]];\)\ \)

Solution 1.4 (b).

(a) Compute Im[(Subscript[z, 1])].

Clear[V, z]; 
(Subscript[z, 1])  =  -3  +  7  I; 
(Subscript[V, 1])  =  
        Graphics[{Red, Arrow[{0,  0}, {Re[(Subscript[z, 1])], Im[(Subscript[z, 1])]}]}]; 
(Subscript[V, 2])  =  
        Graphics[{Green, Arrow[{0,  0}, {0, Im[(Subscript[z, 1])]}]}]; 
Show[(Subscript[V, 1]), (Subscript[V, 2]), 
        PlotRange -> {{-3, 0.1}, {0, 
              7}}, 
Ticks -> {Range[-3, 0, 1], 
            Range[0, 7, 1]}, 
AspectRatio -> (7/3), 
        Axes -> True, 
        AxesLabel -> {"x", "y"}]; 
\(\(Print[\*""    \!\(z\_1\) = ""\ , \ z\_1];\)\ \)
\(\(Print[\*""Im[\!\(z\_1\)] = ""\ , \ 
        Im[z\_1]];\)\ \)

(b) Compute Im[(Subscript[z, 2])].

Clear[V, z]; 
(Subscript[z, 2])  =  9  +  4  I; 
(Subscript[V, 1])  =  
        Graphics[{Red, Arrow[{0,  0}, {Re[(Subscript[z, 2])], Im[(Subscript[z, 2])]}]}]; 
(Subscript[V, 2])  =  
        Graphics[{Green, Arrow[{0,  0}, {0, Im[(Subscript[z, 2])]}]}]; 
Show[(Subscript[V, 1]), (Subscript[V, 2]), 
        PlotRange -> {{0, 9}, {0, 
              4}}, 
Ticks -> {Range[0, 9, 1], 
            Range[0, 4, 1]}, 
AspectRatio -> (4/9), 
        Axes -> True, 
        AxesLabel -> {"x", "y"}]; 
\(\(Print[\*""    \!\(z\_2\) = ""\ , \ z\_2];\)\ \)
\(\(Print[\*""Im[\!\(z\_2\)] = ""\ , \ 
        Im[z\_2]];\)\ \)

Solution 1.4 (c).

(a) Compute  \(\(\(\(z\_1\)\&_\)\(\ \)\)\).

Clear[V, z]; 
(Subscript[z, 1])  =  -3  +  7  I; 
(Subscript[V, 1])  =  
        Graphics[{Red, Arrow[{0,  0}, {Re[(Subscript[z, 1])], Im[(Subscript[z, 1])]}]}]; 
(Subscript[V, 2])  =  
        Graphics[{Blue, 
            Arrow[{0,  0}, {Re[(Subscript[z, 1])], -Im[(Subscript[z, 1])]}]}]; 
Show[(Subscript[V, 1]), (Subscript[V, 2]), 
        PlotRange -> {{-3, 0}, {-7, 
              7}}, 
Ticks -> {Range[-3, 0, 1], 
            Range[-7, 7, 1]}, 
AspectRatio -> 
          (7/3), Axes -> True, 
        AxesLabel -> {"x", "y"}]; 
\(\(Print[\*""\!\(z\_1\) = ""\ , \ z\_1];\)\ \)
\(\(Print[\*""\!\(\(z\_1\)\&_\) = ""\ , 
        Conjugate[z\_1]\ ];\)\ \)

(b) Compute  \(\(z\_2\)\&_\).

Clear[V, z]; 
(Subscript[z, 2])  =  9  +  4  I; 
(Subscript[V, 1])  =  
        Graphics[{Red, Arrow[{0,  0}, {Re[(Subscript[z, 2])], Im[(Subscript[z, 2])]}]}]; 
(Subscript[V, 2])  =  
        Graphics[{Blue, 
            Arrow[{0,  0}, {Re[(Subscript[z, 2])], -Im[(Subscript[z, 2])]}]}]; 
Show[(Subscript[V, 1]), (Subscript[V, 2]), 
        PlotRange -> {{0, 9}, {-4, 
              4}}, 
Ticks -> {Range[0, 8, 1], 
            Range[-4, 4, 1]}, 
AspectRatio -> 
          (8/9), Axes -> True, 
        AxesLabel -> {"x", "y"}]; 
\(\(Print[\*""\!\(z\_2\) = ""\ , \ z\_2];\)\ \)
\(\(Print[\*""\!\(\(z\_2\)\&_\) = ""\ , \ 
        Conjugate[z\_2]];\)\ \)

Theorem 1.1,  Page 12.  Suppose (Subscript[z, 1]),(Subscript[z, 2]) and (Subscript[z, 3]) are arbitrary complex numbers. Then if (Subscript[z, 2]) ≠ 0 

		({{(1.10)     , \(\(\(\(\(\(z\)\(\ \)\)\&_\)\(\ \)\)\&_\)\(=\)\(z\)\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \)\)}, {(1.11)     , \(\(z\_1 + z\_2\)\&_\  = \(\(\(z\_1\)\&_\)\(+\)\(\(z\_2\)\&_\)\(\ \ \ \ \)\)\)}, {(1.12)     , \(\(\(z\_1\) z\_2\)\&_\  = \(\(\(z\_1\)\&_\)\(\ \ \)\(\(z\_2\)\&_\)\(\ \ \ \ \ \ \ \ \ \ \)\)\)}, {(1.13)     , \(\(z\_1\/z\_2\)\&_\) =\(\(z\_1\)\&_\/\(\(\ \)\(\(\(z\_2\)\(\ \)\)\&_\)\(\ \)\)\)if (Subscript[z, 2])≠0}, {(1.14)     , \(\(Re[
                z]\)\(=\)\(\(\(\ \)\(z\  + \ \ \(\(z\)\(\ \)\)\&_\)\)\/2\)\(\ \ \ \ \ \ \ \)\)}, {(1.15)     , \(\(Im[
                z]\)\(=\)\(\(\(\ \)\(z\  - \ \ \(\(z\)\(\ \)\)\&_\)\)\/\(2  i\)\)\(\ \ \ \ \ \ \ \ \)\)}, {(1.16)     , Re[i z] = - 
                Im[z]}, {(1.17)     , Im[i z]=Re[
                z]   }})

Because of what it erroneously connotes, it is a shame that the term imaginary is used in definition the definition of complex numbers. It was coined by the brilliant mathematician and philosopher René Descartes (1596-1650) during an era when quantities such as (Sqrt[-1]) were thought to be just that. Gauss, who was successful in getting mathematicians to adopt the phrase complex number rather than imaginary number, also suggested that we use lateral part of z in place of imaginary part of z. Unfortunately, this suggestion never caught on, and it appears we are stuck with what history has handed down to us.

Library Research Experience for Undergraduates

Project I. Write a report about complex numbers.

1. Apostol, Tom M., Gulbank D. Chakerian, Geraldine C. Darden and John D. Neff (Editors), (1977), Selected Papers on Precalculus: Chapter 2(c) Complex Numbers, The Math. Asso. of Am.

2. Ballantine, J. P. ''Equiangular Complex Numbers (in Classroom Notes),'' Am. Math. M., Vol. 67, No. 7. (Aug. - Sep., 1960), pp. 674-678, Jstor.   

3. Baker, John A. and Willem Sluis, (1997), ''Arg,'' Math. Mag., Vol. 70, No. 4, p. 291.

4. Blake, Robert G. (1962), ''Plane Geometry and Complex Numbers'', Math. Mag., Vol. 35 , pp. 239-242.

5. Bosch, W. and Krajkiewicz, P., (1970), ''A Categorical System of Axioms for the Complex Numbers,'' Math. Mag., Vol. 43, pp. 67-70.

6. Fuchs, L. and T. Szele, ''Introduction of Complex Numbers as Vectors of the Plane (in Classroom Notes),'' Am. Math. M., Vol. 59, No. 9. (Nov., 1952), pp. 628-631, Jstor.   

7. Holden, Herbert L., (1981), ''Applying Complex Arithmetic,'' T. Y. C. Math. J., V. 12, No. 3, pp. 190-194.

8. Jackson, Margaret, (1987), ''Complex Numbers and Pythagorean Triples,'' The Math. Gazette, V. 71, No. 456, p. 127

9. Kimberling, Clark, (1987), ''Power of Complex Numbers,'' The Math. Teach., V. 80, No. 1, pp. 63-67.

10. Kittappa, R., ''Complex Variables and Line-Coordinates, (1968), '' Math. Mag., Vol.. 41, pp. 269-272.

11. McGaughey, A. W., ''The Imaginary Number Problem (in Classroom Notes),'' Am. Math. M., Vol. 64, No. 3. (Mar., 1957), pp. 193-194., Jstor.   

12. Phillips, George M., ''Archimedes and the Complex Plane,'' Am. Math. M., Vol. 91, No. 2. (Feb., 1984), pp. 108-114, Jstor.   

13. Pineda, Anton and David C. Arney, (1991), ''Extending the Complex Number System,'' Math. and Comp. Ed., V. 25, No. 1, pp. 10-16.

14. Potts, D. H., ''Axioms for Complex Numbers (in Math. Notes) Am. Math. M., Vol. 70, No. 9. (Nov., 1963), pp. 974-975, Jstor.   

15. Richmond, D. E., ''Complex Numbers and Vector Algebra (in Classroom Notes),'' Am. Math. M., Vol. 58, No. 9. (Nov., 1951), pp. 622-628, Jstor.   

16. Richmond, D. E., ''Complex Numbers and Trigonometry,'' Am. Math. M., Vol. 64, No. 7. (Aug. - Sep., 1957), pp. 478-485, Jstor.   

17. Schelkunoff, S. A., ''A Note on Geometrical Applications of Complex Numbers (in Questions and Discussions),'' Am. Math. M., Vol. 37, No. 6. (Jun. - Jul., 1930), pp. 301-303, Jstor.   

18. Small, Lloyd L., ''Some Geometrical Applications of Complex Numbers,'' Am. Math. M., Vol. 36, No. 10. (Dec., 1929), pp. 504-511., Jstor.   

19. Subramaniam, K. B., (1979), ''On Euler's Formula (e^(i Θ)) = 
        cos Θ  +  i sin Θ,'' Int. J. of Math. Ed. in Sci. and Tech., V. 10, No. 2, pp. 279.

20. Webb, H. E., ''Discussion: Complex Numbers in Advanced Algebra (in Questions and Discussions),'' Am. Math. M., Vol. 27, No. 11. (Nov., 1920), pp. 411-413, Jstor.   

21. Weimer, Richard C., (1976), ''Can the Complex Numbers Be Ordered?,'' College Math. J., Vol. 7, No. 4, pp.10-12

22. Willson, William Wynne, (1970), ''An Approach to Complex Numbers,'' The Math. Gazette, V. 54, No. 390, pp. 342-346.

Section 1.2  Exercises for The Algebra of Complex Numbers.
See textbook page 13.

Section 1.3	The Geometry of Complex Numbers

Section 1.4	The Geometry of Complex Numbers, Continued

Section 1.5	The Algebra of Complex Numbers, Revisited

Section 1.6	The Topology of Complex Numbers

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