Download NOTES ON COMPLEX VARIABLES 1. The complex Exponential

NOTES ON COMPLEX VARIABLES
1. The complex Exponential Function ez , where z = x + iy
The complex exponential function is defined by extending the Taylor series of ex from real values
of x to complex values:
ez = 1 + z + z 2 /2 + z 3 /3! + . . .
(1)
The partial sums are well defined since products and sums of complex numbers are well defined.
With this definition, ez agrees with ex when z is real. Furthermore, one can show that ez satisfies
z1
d
the properties of exponentials: ez1 +z2 = ez1 ez2 , ez1 −z2 = eez2 , (ez )n = enz , dz
[ez ] = ez .
We showed in class that the definition (1) implies Euler’s formula: if θ is a real number, then
eiθ = cos θ + i sin θ
(2)
which is a point on the unit circle that subtends an angle θ with the real axis.
Every complex number z in the complex plane is obtained by taking a point on the unit circle
times its magnitude r. Thus, Euler’s formula enables us to write any complex number z = x + iy
in the form
z = reiθ = r cos(θ) + ir sin(θ)
(3)
where r is the modulus and θ is the argument. We’ll refer to this form as the polar representation
of z in terms of the coordinates r, θ.
The relation between the polar representation z = reiθ and the Cartesian representation
z = x + iy can be seen from (3) to be given by
x = r cos θ ,
and
r=
y = r sin θ
p
x2 + y 2 , θ = atan(y/x) .
By writing complex numbers in polar form it is easy to take powers and roots of complex
numbers, as shown in the following examples.
Example 1: Plot the complex number 1 + i in the complex plane
√ and find its polar
√ representation.
Solution: By plotting the point we see that θ = π/4 and r = 2. Thus 1 + i = 2eiπ/4 .
Example 2: Find the real and imaginary parts of z =
+ i)10 .
√ (1iπ/4
and the properties of the exponential
Solution: We use the polar representation 1 + i = 2e
function to find that
√ π
5π
π
z = (1 + i)10 = ( 2ei 4 )10 = 25 ei 2 = 25 ei 2 = 25 i
so Re(z) = 0, Im(x) = 25 . That is, z = (1 + i)10 = 32i.
Notice that taking a complex number z = reiθ to a power
z n = rn einθ
1
simply takes the power of the modulus and multiplies the argument by n. For example squaring a
number squares the modulus and doubles the argument.
√
Exercise: Plot all integer powers of the complex number on the unit circle (1 + i)/ 2.
Euler’s formula also enables us to identify the real and imaginary parts of ez , where z = x + iy, as
follows
ez = ex+iy = ex eiy = ex (cos y + i sin y) = ex cos y + iex sin y .
(4)
so Re(z) = ex cos y and Im(z) = ex sin y.
Example 3: Find the real and imaginary parts of z = e1+i .
Solution: z = e1+i = e1 ei = e(cos 1 + i sin 1), so Re(z) = e cos 1 and Im(z) = e sin 1.
Note that the argument of a complex number is not uniquely defined since you can add any multiple
2nπ of 2π, where n is any positive or negative integer, to θ and get the same points. That is, if
z = reiθ then also
z = rei(θ+2nπ)
For example, the number 1 = e0i = e(0+2nπ)i = e2nπi . We’ll use this to find all complex roots of
equations. In particular, all roots of unity.
Example 4: Find all complex roots of z 3 = 1.
Note: The fundamental theorem of algebra states that a polynomial of degree n has precisely n
complex roots, possibly repeated. This implies that z 3 − 1 = 0 has precisely 3 complex roots.
Solution: By taking the modulus on both sides we find that |z 3 | = |z|3 = 1 so |z| = 1 so z must be
a point on the unit circle,
z = eiθ
By plugging this into the equation and writing 1 = e2nπi we get that
ei3θ = e2nπi
Now we can equate powers to get
for n = 0 : θ = 0
for n = 1 : θ = 2π/3
for n = 2 : θ = 4π/3
Any other values of n only give values of θ that agree with one of the above to within a multiple
of 2π. So the 3 distinct cubic roots of unity are
z = 1, e
2π
3 i
,e
4π
3 i
.
Alternatively, one can think of this problem as follows: we want to find a number z such that the
argument of z 3 is 2π. Since cubing it triples the argument of z we deduce that z0 = ei2π/3 is a
solution. Since z03 = 1 it follows that any power z02 , z03 , . . . , z0j , . . . , also solve the equation, since
they satisfy (z0j )3 = (z03 )j = 1j = 1. You can check that only 3 of those powers are distinct. z0 is
called the principal root of unity.
2
2. Deriving Trig identities using Euler’s Formula
Euler’s formula and the properties of exponentials make it is easy to derive several of the trigonometric identities we have been using.
Example 10: Use exponentials to derive formulas for sin 2x and cos 2x.
Solution: Start with the identity e2ix = (eix )2 . Using Euler’s formula we rewrite this identity as
cos 2x + i sin 2x = (cos x + i sin x)2 = cos2 x − sin2 x + 2i sin x cos x
By equating real and imaginary parts on the left and right hand side of this equation we
get
cos 2x = cos2 x − sin2 x ,
sin 2x = 2 sin x cos x
3. Integration using Complex Variables
Complex variables can also be used to determine integrals thatR are tedious to evaluate using conventional methods. For example, consider integrals of the form eat cos bt dt. Recall how to integrate
using integration by parts. Here it is shown that the use of complex variables can sometimes reduce
the complecity of the integration of such functions. You will need to use Euler’s equation and the
definition of equality for complex numbers – namely that if z1 = z2 then Re(z1 ) = Re(z2 ) and
Im(z1 ) = Im(z2 ).
We first define what the integral of a complex function means. Consider a complex valued
function w(t) = u(t) + iv(t), where u and v are real-valued function and t is a real variable. (That
is, u and v are the real and imaginary parts of w.) Then the integral of w is defined by
Z
b
w(t)dt =
a
Z
b
u(t)dt + i
a
Z
b
v(t)dt
(10)
a
R1
Example 11: Evaluate the integral 0 1 + 2it2 dt
1
R1
R1
R1
1
Solution: 0 1 + 2it2 dt = 0 1 dt + 2i 0 t2 dt = [t]0 + 2i t3 /3 0 = 1 + 2i/3
It follows that the standard rules of integration can be used by treating i as a constant.
R π/3
Example 12: Evaluate the integral 0 e2it dt
√
√
2it π/3
R π/3
1
1
Solution: 0 e2it dt = 2i
3/4 + 3i/4
e 0 = 2i
e2iπ/3 − e0 = −i
2 −3/2 + i 3/2 =
We now illustrate that some of the integrals we computed earlier this semester are evaluated
quite easily if we use complex variables instead.
R
Example 13: Evaluate the integral e(a+ib)t dt. What do you learn by equating real and imaginary
parts on the left and right hand sides of the resulting equation?
Z
1
Solution:
e(a+ib)t dt =
e(a+ib)t + C . Equating real and imaginary parts gives
a + ib
Z
eat cos bt dt =
eat
(a cos bt + b sin bt) + C1 ,
a 2 + b2
Z
eat sin bt dt =
eat
(a sin bt − b cos bt) + C2 ,
a 2 + b2
3
Notice that to compute these two integrals without complex variables, we would have to
use integration by parts twice. These integrals occur frequently in mechanical systems.
The functions eat cos bt, eat sin bt represent oscillating functions whose magnitude grows or
decays exponentially.
Problems: Part I
Powers of complex numbers:
√
1. Write√
the following numbers in the form reiθ and in the form a+ib: (a) (1+i)20 , (b) (1− 3i)5 ,
(c) (2 3 + 2i)5 , (d) (1 − i)8 .
2. Find all solutions to the given equation. Sketch the roots in the complex plane.
(a) z 8 = 1 (the eigth roots of unity)
(b) z 5 = 1 (the fifth roots of unity)
(c) z 5 = 32 (the fifth roots of 32)
Problems: Part II
Exponentials and Logarithms:
1. Show that:
(a) e2±3πi = −e2
(b) e
2+πi
4
=
pe
2 (1
(c) ez+πi = −ez
+ i)
2. Find the real and imaginary parts of e(2−3i)t (t is real)
Trigonometric Identities:
3. Derive the formulas for sin(a + b) and cos(a + b) using complex variables. (Hint: look at the
real and imaginary parts of the equation ei(a+b) = eia eib .)
Integration:
4. Evaluate the following integrals
R1
R π/4
(a) 0 (1 + it2 ) dt
(b) 0 eit dt
R
5. Find the indefinite integral e−st cos(at) dt, where s, a are positive constants
(a) using integration
parts twice
R by
−st iat
(b) by integrating e e dt, and then equating real and imaginary parts.
6. The Laplace transform of a function f (t) is defined to be
F (s) = L[f ](t) =
Z
∞
e−st f (t) dt
0
(a) Find the Laplace transform of f (t) = cos(at). You may use your result in problem 5
above. Use proper notation to evaluate the improper integral.
(b) Find the Laplace transform of
f (t) = uc (t) =
4
0 if t < c
1 if t ≥ c
,
c > 0.