# 程序代写案例-MATH2521

UNSW — School of Mathematics and Statistics
MATH2521 Complex Analysis
4. THE EXPONENTIAL AND OTHER FUNCTIONS
Apart from rational functions and “spec
ial” complex functions such
as modulus and conjugate, we don’t really know of any complex func-
tions yet. We’ll remedy this in the present chapter by defining various
important complex expressions. Mostly we shall try to create functions
f(z) which have properties corresponding to standard real functions,
and which will actually equal these real functions in the case z = x.
The terminology often used is that we seek to “extend functions from
the real line to the complex plane”. We shall find that most of the signif-
icant properties of the real functions remain true for our new functions,
though there may also be additional properties which are specific to the
complex case.
First, let’s consider the exponential function. The real exponential
function f(x) = ex is uniquely determined by the facts that it is defined
and differentiable for all real x, that it is equal to its own derivative and
that f(0) = 1. So –
Problem. Can we determine an entire complex function f such that
f(0) = 1 and f ′(z) = f(z) for all z ∈ C?
Solution. Let f(z) = u+ iv where u and v are real. If f is to be entire
then the Cauchy–Riemann equations
∂u
∂x
=
∂v
∂y
,
∂u
∂y
= −∂v
∂x
must hold for all real x, y. From the theorem “differentiability and the
Cauchy–Riemann equations” f ′(z) = ux + ivx, so if f
′(z) = f(z) then
∂u
∂x
= u ,
∂v
∂x
= v (∗)
for all x, y. We also have the conditions
u(0, 0) = 1 and v(0, 0) = 0 .
1
Equations (∗) involve differentiation with respect to x, keeping y con-
stant, so they are, in effect, just ordinary differential equations in one
variable. You will recall from first year that the general solution of
dF
dx
= F
is F (x) = Cex, where C is a constant. In the present case (compare
chapter 3, page 6) we have taken y as constant, so C could be a function
of y. Hence equations (∗), together with the initial conditions, yield
u = C(y)ex , v = D(y)ex , C(0) = 1 , D(0) = 0 .
Now substitute into the Cauchy–Riemann equations: we have
C(y)ex = D′(y)ex , C ′(y)ex = −D(y)ex .
Cancelling ex (which is not zero) and eliminating D from the equations,
we end up with
C ′′(y) = −C(y) , C(0) = 1 , C ′(0) = 0 .
Once again this type of initial value problem should be familiar from
first year. The solution is C(y) = cos y; hence D(y) = −C ′(y) = sin y;
substituting back and simplifying leads to
f(z) = ex cos y + iex sin y .
Definition. The (complex) exponential function is
f : C→ C where f(z) = f(x+ iy) = ex(cos y + i sin y) .
We shall write this function as f(z) = ez or f(z) = exp z.
• Taking x = 0 gives eiy = cos y + i sin y, which you will recall from
first year. Thus our definition of ez is consistent with our earlier
usage of eiθ.
2
• Taking y = 0 gives f(z) = f(x) = ex. So when z is real, the complex
exponential function is the same as the real exponential function.
In fact, the complex exponential function is the only possible entire
function which has the values ex on the real axis – proving this is an
easy application of the identity theorem for holomorphic functions
(chapter 3, page 12).
Lemma. Properties of the exponential function.
1. The function ez is entire, and d
dz
(ez) = ez for all z.
2. If z = x + iy with x, y real then |ez| = ex and arg(ez) = y + 2npi
with n ∈ Z.
3. For all z1, z2 in C we have e
z1+z2 = ez1ez2 .
4. The exponential function is periodic with period 2pii; and ez1 = ez2
if and only if z1 = z2 + 2npii for some integer n.
5. For all z ∈ C we have ez 6= 0 and (ez)−1 = e−z.
Proof. We give brief hints – please write these up as detailed proofs.
1. Use the Cauchy–Riemann equations.
2. Use the definition of modulus and argument.
3. Write z1 = x1 + iy1, z2 = x2 + iy2 and do the algebra.
4. For the second part, use property 2; the first part follows.
5. Use part 3 to show that eze−z = 1.
Examples.
• Solve ez = −1+ i. Solution. Let z = x+ iy and take the modulus
and argument of both sides: we have
ex =

2 and y = 34pi + 2npi , n ∈ Z .
The solution of the first equation is x = 12 ln 2, and hence
z = 1
2
ln 2 + i
(
3
4
pi + 2npi
)
, n ∈ Z .
• Solve ez = e−1+i. Solution. We could compute e−1+i and then
use the method of the previous problem; but it is easier to note that
3
z = −1+ i is an obvious solution, and then use property 4 from the
lemma to get the complete solution
z = −1 + i+ 2npii , n ∈ Z .
• For which z is it the case that ez = 1? Solution. It is easy to
see from the definition that z = 0 is one possibility; therefore the
second part of property 4 shows that
ez = 1 if and only if z = 2npii , n ∈ Z .
Similarly, we can check that eipi = −1, and hence
ez = −1 if and only if z = (2n + 1)pii , n ∈ Z .
Trigonometric functions. There are various ways to approach the
problem of defining complex trigonometric functions, though all “sensi-
ble” ways give the same final outcome. Here are two.
• Find an entire complex function f such that its derivative f ′ is also
entire, and f ′′(z) = −f(z) for all z, and f(0) = 1 and f ′(0) = 0.
Call this function the (complex) cosine function.
• Find two entire complex functions f and g such that f ′(z) = −g(z)
and g′(z) = f(z) and f(0) = 1 and g(0) = 0: call them cosine and
sine respectively.
Both of these methods involve similar working to that on page 2, though
the details are notably more intricate. Instead, we shall take the ap-
proach of directly extending the real cosine and sine functions. We
know the identities
cos θ =
eiθ + e−iθ
2
and sin θ =
eiθ − e−iθ
2i
for real θ, and we therefore say –
Definition. The complex cosine and sine functions are defined by
cos z =
eiz + e−iz
2
and sin z =
eiz − e−iz
2i
for all z ∈ C.
4
Lemma. Properties of the complex cosine and sine functions.
1. The cosine and sine functions are entire, and
d
dz
(cos z) = − sin z , d
dz
(sin z) = cos z
for all z.
2. The cosine function is even and the sine function is odd.
3. The cosine and sine functions have period 2pi.
4. For all real y we have cos(iy) = cosh y and sin(iy) = i sinh y.
5. We have cos z = 0 if and only if z = (n+ 1
2
)pi, and sin z = 0 if and
only if z = npi, with in each case n ∈ Z.
6. For all complex z and w we have
cos2 z + sin2 z = 1 ,
cos(z + w) = cos z cosw − sin z sinw ,
sin(z + w) = sin z cosw + cos z sinw .
7. If z = x+ iy then
cos z = cos x cosh y − i sin x sinh y ,
sin z = sinx cosh y + i cos x sinh y .
Proof (sketch). For part 1: iz and −iz are polynomials so they are en-
tire. Then eiz and e−iz are compositions of entire functions so they are
entire (chapter 2, page 15). So cos z is a constant times the sum of two
entire functions, and is entire; similarly for sin z. Use the known deriva-
tive of the exponential function, together with standard differentiation
rules, to verify the formulae for the derivatives.
For property 2 substitute −z for z in the definitions; for property 3,
substitute z + 2pi; for property 4, substitute z = iy and recall the defi-
nitions of cosh and sinh (MATH1131 calculus).
5
We prove property 5 for cosine, and leave sine as an exercise. Using
one of the examples from page 4,
cos z = 0 ⇔ eiz + e−iz = 0
⇔ eiz = −e−iz
⇔ e2iz = −1
⇔ 2iz = (2n + 1)pii , n ∈ Z
⇔ z = (n + 12 )pi , n ∈ Z .
For part 6, make appropriate substitutions and do the algebra. An
alternative, much simpler, proof for the first part: define
f(z) = cos2 z + sin2 z and g(z) = 1 .
These are both entire functions, and they coincide on the real line; so
by the identity theorem, they are equal for all z in C. To prove part 7,
use part 6 and part 4.
• We have shown that the complex cosine and sine functions are en-
tire; they coincide with the real cosine and sine functions when z
is real; and it follows from the identity theorem that they are the
only entire functions with this property.
• Other trigonometric functions are defined exactly as in the real case:
tan z =
sin z
cos z
and sec z =
1
cos z
provided cos z 6= 0;
and
cot z =
cos z
sin z
and csc z =
1
sin z
provided sin z 6= 0.
• For a complex function f , “even” and “odd” are defined alge-
braically as for real functions:
f(−z) = f(z) and f(−z) = −f(z)
6
respectively. However these concepts do not have the graphical
interpretations with which you are familiar in the real case.
• Part 5 of the lemma shows that the equations cos z = 0 and sin z = 0
have no solutions other than the real solutions we already know
• Other real trigonometric identities (double angle formulae, sums–
to–products and products–to–sums formulae, and so on) extend to
the complex case, and many can be proved by imitating our alter-
native proof for property 6.
• Notice that important inequalities for real trigonometric functions
are no longer true in the complex case. For example it is not true
for all complex z that sin z ≤ 1 (obviously – why?). Nor is it true
that | sin z| ≤ 1, because, for example, | sin(iy)| = | sinh y|, and this
tends to infinity as y →∞.
• It follows easily from the definition that for all complex z we have
eiz = cos z + i sin z .
But note that cos z and sin z are not always real, so this does not
mean that cos z and sin z are the real and imaginary parts of eiz.
Example. The equation cos z = 3 obviously has no real solutions. We
give two methods of finding its complex solutions.
• Let z = x + iy. Using part 7 of the lemma, we equate real and
imaginary parts to obtain
cos x cosh y = 3 and sinx sinh y = 0 .
From the second equation there are two possibilities: sinx = 0 or
sinh y = 0. However in the latter case we have y = 0 and the first
equation becomes cos x = 3 (with x real), so this must be rejected.
For sinx = 0 we have x = npi and hence
cosnpi cosh y = 3 ⇔ cosh y = 3cosnpi .
If n is odd this gives cosh y = −3, which is impossible; therefore n
is even, say n = 2m; hence, cosh y = 3 and y = ± cosh−1 3. So we
have the solutions
cos z = 3 if and only if z = 2mpi ± i cosh−1 3 , m ∈ Z .
7
• Alternatively, work directly from the definition:
cos z = 3 ⇔ e
iz + e−iz
2
= 3
⇔ eiz + e−iz = 6
⇔ eiz − 6 + e−iz = 0
⇔ (eiz)2 − 6eiz + 1 = 0 .
This is a quadratic in eiz, and solving by any method gives
eiz = 3±

8 . (∗)
Now let z = x + iy. Then eiz = e−y+ix; equating modulus and
argument in (∗) yields
e−y = 3±

8 and x = arg
(

8
)
= 2npi .
Solving the first equation and putting everything back together, we
find that
cos z = 3 if and only if z = 2npi − i ln(3±√8) , n ∈ Z .
Comment. This does not look the same as our previous answer!
However, if you recall from MATH1131 Calculus how to write in-
verse hyperbolic functions in terms of logarithms, you will be able
to reconcile the two solutions.
Note that there are sometimes traps in the “exponential” method, and
you will need to be careful. For instance, let’s try to solve sin z = −3i
in the same way. We have
sin z = −3i ⇔ eiz − 6− e−iz = 0 ⇔ eiz = 3±

10
and hence
sin z = −3i ⇔ z = 2npi − i ln(3±√10) , n ∈ Z .
8
Exercise. By carefully filling in the details we have omitted, explain
why this is wrong! Fix the errors and show that the correct solution is
sin z = −3i ⇔ z = mpi − (−1)mi ln(3 +√10) , m ∈ Z .
Also, give an alternative solution by starting with
sinx cosh y = 0 , cos x sinh y = −3 .
Example. Solve tan z = 3 + 2i. Solution. We have
tan z = 3 + 2i ⇔ sin z
cos z
= 3 + 2i
⇔ e
iz − e−iz
2i
= (3 + 2i)
eiz + e−iz
2
⇔ eiz − e−iz = (−2 + 3i)(eiz + e−iz)
⇔ (3− 3i)eiz = (−1 + 3i)e−iz
⇔ e2iz = −1 + 3i
3− 3i =
−2 + i
3
.
Writing z = x+ iy, this becomes
e−2y+2ix =
−2 + i
3
;
hence
e−2y =
∣∣∣−2 + i
3
∣∣∣ =

5
3
and
2x = arg
(−2 + i
3
)
= − tan−1
(1
2
)
+ (2n+ 1)pi , n ∈ Z .
Hence
z = −1
2
tan−1
(1
2
)
+
(
n+
1
2
)
pi − 1
2
i ln
(√5
3
)
, n ∈ Z .
9
Example. Solve cos z = cos 3i.
Solution. An obvious solution is z = 3i; by periodicity, z = 3i + 2npi
is a solution for every n ∈ Z. However, cos z is an even function, so
z = −3i is also a solution. The complete solution is
z = 2npi ± 3i , n ∈ Z .
Exercise. Solve sin z = sin 5i. (Note that sin is not an even function!)
Solution. As in the previous problem, z = 5i is a solution; the identity
sin(pi− z) = sin z shows that z = pi− 5i is also a solution. Extending by
periodicity, the complete solution is
z = 5i+ 2npi , z = −5i+ (2n+ 1)pi , n ∈ Z ,
which can be written more compactly as
z = mpi + (−1)m5i , m ∈ Z .
The complex hyperbolic functions are defined in terms of complex expo-
nentials in exactly the same way as real hyperbolics are defined in terms
of real exponentials.
Definition. The complex hyperbolic functions are defined by
cosh z =
ez + e−z
2
, sinh z =
ez − e−z
2
and
tanh z =
sinh z
cosh z
, sech z =
1
cosh z
provided cosh z 6= 0, and
coth z =
cosh z
sinh z
, csch z =
1
sinh z
provided sinh z 6= 0.
These definitions are very similar to those of cos and sin; so we should ex-
pect close connections between hyperbolic and trigonometric functions.
10
Lemma. Properties of complex hyperbolic cosine and sine functions.
1. For all z ∈ C we have
cosh z = cos(iz) and i sinh z = sin(iz) .
2. The functions cosh and sinh are entire, and
d
dz
(cosh z) = sinh z ,
d
dz
(sinh z) = cosh z
for all z.
3. We have cosh z = 0 if and only if z = (n + 1
2
)pii with n ∈ Z, and
sinh z = 0 if and only if z = npii with n ∈ Z.
Comment. We leave the proof of this lemma as an exercise. In fact,
once you have proved part 1, the rest follows easily from correspond-
ing facts about trigonometric functions, and most results on hyperbolic
functions can be proved in the same way. Two examples:
cosh(x+ iy) = cos(−y + ix)
= cos(−y) cos(ix) − sin(−y) sin(ix)
= cosh x cos y + i sinhx sin y (∗)
and
sech z =
1
cosh z
=
1
cos(iz)
= sec(iz) ,
i tanh z =
i sinh z
cosh z
=
sin(iz)
cos(iz)
= tan(iz) ,
d
dz
(sech z) =
d
dz
(sec(iz)) = i sec(iz) tan(iz) = − sech z tanh z .
Example. We know that the equation cosh z = −4 has no real solutions.
Find all its complex solutions.
Solution. Using (∗) we have
coshx cos y = −4 and sinhx sin y = 0 .
11
There is no solution with sinhx = 0 (why?), so sin y = 0 and hence
y = npi. Therefore cos y = ±1; since x is real, coshx must be positive,
cos y is negative and n is odd. So coshx = 4 and hence x = ± cosh−1 4.
The solution is
z = ± cosh−1 4 + (2m+ 1)pii , m ∈ Z .
Note. If you are not thoroughly familiar with the graphs of the real
hyperbolic cosine and sine functions, please revise!
Example. Solve cosh z = 1
2
.
Solution. We can follow the previous example (exercise), or turn the
question into a trigonometric problem. We have
cosh z =
1
2
⇔ cos(iz) = 1
2
⇔ iz = ±pi
3
+ 2npi
⇔ z = 1
i
(
±pi
3
+ 2npi
)
⇔ z = i
(
±pi
3
+ 2mpi
)
, m ∈ Z .
Note: make sure you understand the last step!
Complex transformations involving exponential, trigonometric
and hyperbolic functions. We have seen in chapter 1 various com-
plex transformations; exponential, trigonometric and hyperbolic func-
tions provide further important examples.
Example. Find the image of the rectangle
R = {x+ iy ∈ C | 0 ≤ x ≤ pi
3
, 0 ≤ y ≤ 1 }
under the transformation f(z) = eiz.
Solution. We have
w = ei(x+iy) = e−y+ix ⇒ |w| = e−y , argw = x ,
12
and so the image of R is given by
e−1 ≤ |w| ≤ 1 , 0 ≤ argw ≤ pi3 .
x
y
0 pi
3
1
u
v
0 e−1 1
Example. For the function f : C → C where f(z) = sin z, find the
image of
• the lines Re(z) = 0 and Re(z) = pi2 and Re(z) = c, with 0 < c < pi2 ;
• the region
A = {x+ iy | 0 < x ≤ pi2 and y > 0 } .
Solution. We write
w = u+ iv = f(z) = sin z = sinx cosh y + i cos x sinh y .
• For x = 0 we have u = 0, v = sinh y; since the range of the real
sinh function is R, any real value is possible for v. So the image is
the v–axis.
For x = pi2 we have u = cosh y, v = 0. For real y, the values of
u = cosh y are u ≥ 1. So the image is the interval [ 1,∞ ) on the
u–axis.
For x = c we have u = sin c cosh y, v = cos c sinh y and so
( u
sin c
)2

( v
cos c
)2
= cosh2 y − sinh2 y = 1 .
This is the equation of a hyperbola; however we also have the con-
dition u = sin c cosh y > 0, and so the image is only the right half
of this hyperbola.
13
Here are the three lines and their images.
x
y
0 pi2 u
v
0 1
• The region A is the union of all the half–lines x = c, y > 0 for
0 < c ≤ pi2 . The image of each half–line is the upper half of the
corresponding hyperbola in the above problem (the upper half be-
cause we have v = cos c sinh y > 0); except for the case c = pi2
when the image is an interval on the u–axis, as above (omitting
the endpoint u = 1 because we have y 6= 0). The image of A is
made up of hyperbolas and an interval, as shown. The hyperbolas
approach arbitrarily close to the v–axis; and arbitrarily close to the
u–axis for u > 1; so the image of A is the entire first quadrant,
including the interval ( 1,∞ ) on the u–axis, but excluding the rest
of the boundary.
u
v
0 1
Here are A and its image under f(z) = sin z.
x
y
0 pi2 u
v
0 1
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