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686 CHAPTER 19 Complex Numbers and Functions

a
Since e = 0, then sin(b) = 0, so b = kπ for some integer k. But then
a
k a
a
e cos(b) = e cos(kπ) = (−1) e = 1.
a
k
This requires that (−1) be positive, so k =2n for some integer n. But then e =1, which for real
a means that a = 0. Therefore z = a + bi = 2nπi for some integer n.
z
p
For conclusion (2), suppose that e z+p = e for all z. Then e = 1, so p = 2nπi for some
integer n,by(1).
For conclusion (3), let n be a nonzero integer. By conclusion (1),
z
z 2nπi
e z+2nπi = e e = e ,
z
for all z,so e has period 2nπi. For the rest of conclusion (3), we need to show that any period of
z
z
e is an integer multiple of 2π. Thus, suppose that e z+p = e for all z and for some number p = 0.
Then by conclusion (2), p = 2nπi for some nonzero integer n.
z
Every period of e is therefore pure imaginary, explaining why no period is evident for the
real exponential function.
z
Another difference between the real and complex exponential functions is that e can be
z
negative. For example, it is easy to check that e =−1 exactly when z = (2n + 1)π for some
integer n.
EXAMPLE 19.12
Solve the equation
z
e = 1 + 2i.
We want all z such that
z
x
x
e = e cos(y) + ie sin(y) = 1 + 2i.
This requires that
x
x
e cos(y) = 1 and e sin(y) = 2.
If we square these equations and add, we obtain
2
2x
2x
2
e (cos (y) + sin (y)) = e = 5,
which implies that x = ln(5)/2. Next,
x
e sin(y)
= tan(y) = 2,
x
e cos(y)
z
so y = arctan(2). All solutions of e are
1
z = ln(5) + i arctan(2).
2
The complex sine and cosine functions are defined by
1 iz −iz 1 −z −iz
cos(z) = e + e and sin(z) = e − e .
2 2i
A routine calculation gives us
cos(z) = cos(x)cosh(y) − i sin(x)sinh(y)

and
sin(z) = sin(x)cosh(y) + i cos(x)sinh(y),

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October 15, 2010 18:5 THM/NEIL Page-686 27410_19_ch19_p667-694