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Student Solutions Manual to accompany Advanced Engineering Mathematics VOL 2

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Student Solutions Manual to accompany Advanced Engineering Mathematics VOL 2

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PART D
Complex
Analysis



Chap. 13 Complex Numbers and Functions.
Complex Differentiation

Complex numbers appeared in the textbook before in different topics. Solving linear homogeneous ODEs
led to characteristic equations, (3), p. 54 in Sec. 2.2, with complex numbers in Example 5, p. 57, and
Case III of the table on p. 58. Solving algebraic eigenvalue problems in Chap. 8 led to characteristic
equations of matrices whose roots, the eigenvalues, could also be complex as shown in Example 4, p. 328.
Whereas, in these type of problems, complex numbers appear almost naturally as complex roots of
polynomials (the simplest being x 2 C 1 D 0), it is much less immediate to consider complex analysis—the
systematic study of complex numbers, complex functions, and “complex” calculus. Indeed, complex
analysis will be the direction of study in Part D. The area has important engineering applications in
electrostatics, heat flow, and fluid flow. Further motivation for the study of complex analysis is given on
p. 607 of the textbook.
We start with the basics in Chap. 13 by reviewing complex numbers z D x C yi in Sec. 13.1 and
introducing complex integration in Sec.13.3. Those functions that are differentiable in the complex, on
some domain, are called analytic and will form the basis of complex analysis. Not all functions are
analytic. This leads to the most important topic of this chapter, the Cauchy–Riemann equations (1),
p. 625 in Sec. 13.4, which allow us to test whether a function is analytic. They are very short but you have
to remember them! The rest of the chapter (Secs. 13.5–13.7) is devoted to elementary complex functions
(exponential, trigonometric, hyperbolic, and logarithmic functions).
Your knowledge and understanding of real calculus will be useful. Concepts that you learned in real
calculus carry over to complex calculus; however, be aware that there are distinct differences between
real calculus and complex analysis that we clearly mark. For example, whereas the real equation e x D 1
has only one solution, its complex counterpart e z D 1 has infinitely many solutions.

,258 Complex Analysis Part D

Sec. 13.1 Complex Numbers and Their Geometric Representation
Much of the material may be familiar to you, but we start from scratch to assure everyone starts at the
same level. This section begins with the four basic algebraic operations of complex numbers (addition,
subtraction, multiplication, and division). Of these, the one that perhaps differs most from real numbers is
division (or forming a quotient). Thus make sure that you remember how to calculate the quotient of two
complex numbers as given in equation (7), Example 2, p. 610, and Prob. 3. In (7) we take the number z2
from the denominator and form its complex conjugate zN2 and a new quotient zN2 =Nz2 . We multiply the given
quotient by this new quotient zN2 =Nz2 (which is equal to 1 and thus allowed):
z1 z1 z1 zN2
zD D 1D  ;
z2 z2 z2 zN2

which we multiply out, recalling that i 2 D 1 [see (5), p. 609]. The final result is a complex number in a
form that allows us to separate its real (Re z/ and imaginary (Im z/ parts. Also remember that 1= i D i
(see Prob. 1), as it occurs frequently. We continue by defining the complex plane and use it to graph
complex numbers (note Fig. 318, p. 611, and Fig. 322, p. 612). We use equation (8), p. 612, to go from
complex to real.


Problem Set. 13.1. Page 612
1. Powers of i. We compute the various powers of i by the rules of addition, subtraction,
multiplication, and division given on pp. 609–610 of the textbook. We have formally that

i2 D ii
D .0; 1/.0; 1/ [by (1), p. 609]
D .0  0 1  1; 0  1 C 1  0/ [by (3), p. 609]
(I1)
D .0 1; 0 C 0/ (arithmetic)
D . 1; 0/
D 1 [by (1)],

where in (3), that is, multiplication of complex numbers, we used x1 D 0, x2 D 0, y1 D 1, y2 D 1.

(I2) i 3 D i 2 i D . 1/  i D i:

Here we used (I1) in the second equality. To get (I3), we apply (I2) twice:

(I3) i 4 D i 2 i 2 D . 1/  . 1/ D 1:

(I4) i 5 D i 4 i D 1  i D i;

and the pattern repeats itself as summarized in the table below.
We use (7), p. 610, in the following calculation:

1 1 iN 1 . i/ .1 C 0i/.0 i/ 10C01 00 11
(I5) D D D D Ci 2 D0 iD i:
i ii N i . i/ .0 C i/.0 i/ 2
0 C12 0 C 12

,Chap. 13 Complex Numbers and Functions. Complex Differentiation 259

By (I5) and (I1) we get

1 1 1
(I6) 2
D  D . i/. i/ D . 1/i  . 1/i D 1  i 2 D 1;
i i i
 2  
1 1 1
D D . 1/. i/ D i [from (I6) and (I5)];
i3 i i
 2  2
1 1 1
4
D D . 1/. 1/ D 1,
i i i

and the pattern repeats itself. Memorize that i 2 D 1 and 1=i D i as they will appear quite
frequently.

i8 i9 : .
i4 i5 i6 i7
Start ! i0 i i2 i3
1 i 1 i
1=i 4 1=i 3 1=i 2 1=i Start
1=i 8 1=i 7 1=i 6 1=i 5
. . 1=i 10 1=i 9

Sec. 13.1. Prob. 1. Table of powers of i

3. Division of complex numbers
a. The calculations of (7), p. 610, in detail are
z1 x1 C iy1
z D D (by definition of z1 and z2 )
z2 x2 C iy2
x1 C iy1 x2 iy2
D  (N.B. corresponds to multiplication by 1)
x2 C iy2 x2 iy2
.x1 C iy1 /.x2 iy2 /
D
.x2 C iy2 /.x2 iy2 /
x1 x2 x1 iy2 C iy1 x2 iy1 iy2
D (multiplying it out: (3) in notation (4), p. 609)
x2 x2 x2 iy2 C iy2 x2 iy2 iy2
x1 x2 ix1 y2 C ix2 y1 i 2 y1 y2
D (grouping terms, using commutativity)
x22 ix2 y2 C ix2 y2 i 2 y22
x1 x2 ix1 y2 C ix2 y1 C y1 y2
D (using i 2 D 1 and simplifying)
x22 C y22
x1 x2 C Cy1 y2 x2 y1 x1 y2
D 2 2
Ci (breaking into real part and imaginary part).
x2 C y2 x22 C y22
b. A practical example using (7) is
26 18i .26 18i/ .6 C 2i/ 26  6 C 26  2i 18  6i 18  2i 2
D D
6 2i .6 2i/ .6 C 2i/ 62 C 22
156 C 52i 108i C 36 192 56i
D D D 4:8 1:4i:
36 C 4 40

, 260 Complex Analysis Part D

5. Pure imaginary number a. If z D x C iy is pure imaginary, then zN D z:
Proof. Let z D x C iy be pure imaginary. Then x D 0, by definition on the bottom of p. 609.
Hence

(A) z D iy and (B) zN D iy (by definition. of complex conjugate, p. 612).

If we multipy both sides of (A) by 1, we get

zD iy;

N hence
which is equal to z,
z D z:
N
b. If zN D z then z D x C iy is pure imaginary.
Proof. Let z D x C iy so that zN D x iy. We are given that zN D z, so

zN D x iy D zD .x C iy/ D x iy:

By the definition of equality (p. 609) we know that the real parts must be equal and that the
imaginary parts must be equal. Thus

Re zN D Re. z/;
xD x;
2x D 0;
x D 0;

and

Im zN D Im. z/;
yD y;

which is true for any y. Thus
z D x C iy D iy:
But this means, by definition, that z is pure imaginary, as had to be shown.

11. Complex arithmetic

z1 z2 D . 2 C 11i/ .2 i/
D 2 C 11i 2Ci D. 2 2/ C .11 C 1/i D 4 C 12i
2
.z1 z2 / D . 4 C 12i/. 4 C 12i/ D 16 48i 48i 144 D 128 96i
.z1 z2 /2 128 96 8  16 25  3
D iD D 8 6i:
16 16 16 16 24
Next consider
z z2 2
1
:
4 4
We have
z1 1 2 11 z2 2 1
D . 2 C 11i/ D C i; D i:
4 4 4 4 4 4 4

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