SOLUTION MANUAL
All Chapters Included
ADVANCE
ENGINEERIN
MATHEMATI
VOLUME
,Chap. 13 Complex Numbers and Functions. Complex Differentiation 259
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
C D
naturally as complex roots of polynomials (the simplest being x2 1 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 numbersD C z x 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.
,260 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 zN 2 and a new quotient zN 2 =zN 2 . We multiply the given quotient by
this new quotient zN 2 =zN 2 (which is equal to 1 and thus allowed):
z1 z1 z1 zN2
zD D · 1D · ;
z2 z2 z2 zN 2
which we multiply out, recalling that D -i 2 1
D -
[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 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 2i D .-1/ · i D -i:
, Chap. 13
Here we used (I1)Complex Numbers
in the second and Functions.
equality. Complex
To get (I3), Differentiation
we apply (I2) 259
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:
1 1 iN 1 .-i/ .1 C 0i/.0 - i/ 1·0C0·1 0 ·0 -1·1
(I5) D D D D Ci D 0 - i D -i:
i i iN i .-i/ .0 C i/.0 - i/ 0 C1
2 2 02 C 12