UNSW — School of Mathematics and Statistics

MATH2521 Complex Analysis

4. THE EXPONENTIAL AND OTHER FUNCTIONS

Apart from rational functions and “special” 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.

Comments.

• 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.

Comments.

• 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

about.

• 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

(

3±

√

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

14

欢迎咨询51作业君