TEST BANK FOR Advanced Engineering Mathematics [Volume 2] By Herbert Kreyszig and Erwin Kreyszig

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 x2 C1 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 ex D 1
has only one solution, its complex counterpart ez 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 Nz2 and a new quotient Nz2=Nz2. We multiply the given
quotient by this new quotient Nz2=Nz2 (which is equal to 1 and thus allowed):
z D
z1
z2
D
z1
z2
1 D
z1
z2

Nz2
Nz2
;
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
i 2 D i i
D .0; 1/.0; 1/ [by (1), p. 609]
D .0 0 􀀀 1 1; 0 1 C 1 0/ [by (3), p. 609]
D .0 􀀀 1; 0 C 0/ (arithmetic)
D .􀀀1; 0/
D 􀀀1 [by (1)],
(I1)
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 2i 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 2i 2 D .􀀀1/ .􀀀1/ D 1:
(I4) i 5 D i 4i 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:
(I5)
1
i
D
1
i
Ni
Ni
D
1
i
.􀀀i /
.􀀀i /
D
.1 C 0i/.0 􀀀 i /
.0 C i/.0 􀀀 i /
D
1 0 C 0 1
02 C 12
C i
0 0 􀀀 1 1
02 C 12
D 0 􀀀 i D 􀀀i: