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# Complex Numbers

This set of notes reviews complex numbers and their properties. We then examine
the use of complex numbers to solve second-order ordinary differential equations, using the
mass-spring system as our example.

Introduction

Complex numbers generalizes the idea of a real number to include an imaginary component.
Imaginary numbers contain a factor of i, where

(1)
i 1.
No real number follows this definition, so we invent the imaginary number. A complex
number, z, is a number of the form
z = x + iy,
(2)
where x and y are both real. So, for example
z = 2 + 2i
z = 4
z = 6i
are all complex numbers. In the first case z has both a real and imaginary part, while z is
purely real in the second case, and purely imaginary in the third. With this in mind, we can
define x in Eq. (2) to be the real part of the complex number z; in other words,
x Re [z] ,

(3)

## while y (not iy) is defined as the imaginary part,

y Im [z] .

(4)

So, in our earlier examples of complex numbers, when z = 2 + 2i, then Re [z] = 2, while
Im [z] = 2, and when z = 6i, then Re [z] = 0, and Im [z] = 6.
We can define another quantity, called the complex conjugate of z, denoted z , which
is obtained from z simply by changing the sign of the imaginary part; i.e., by letting i i.
Then, if z is given by Eq. (2), then
z x iy.

(5)

## We can multiply together z and z to get a purely real number,

zz = z z |z|2 = x2 + y 2 .

(6)

## For example, if z = 2 + 2i, then z = 2 2i, and zz = (2 + 2i) (2 2i) = 4 4i + 4i + 4 = 8.

These definitions will be extremely useful in what follows.
1

## Because a complex number is, in a sense, a

two-dimensional object, we can easily treat
it like a vector, as seen in the figure to
the right. We imagine graphing a complex
number z = x + iy, placing the real part of
z along the x-axis, and the imaginary part
of z along the y-axis, much like we would
graph a vector ~r = xi + yj; for example, if
we have z = 2 + 2i, then we would treat
the graph like ~r = 2i + 2j.

Im[z]
(x,y)
r
T
Re[z]

Looking at the graph, we see that the vector describing the complex number has a
magnitude, and also makes an angle with the real axis. Treating these values like ordinary
polar coordinates we see that we can define
Re [z] = x = r cos
Im [z] = y = r sin .

(7)

## Furthermore, we can see that

r=

p
x2 + y 2 = |z|2 = zz ,

## while the angle is

1

= tan

(8)

y

,
(9)
x
and so complex numbers really do mimic (two-dimensional) vectors. Now, based on the
graph we can write our complex number in another way. Using Eqs. (7) we can write
z = x + iy = r (cos + i sin ) .

(10)

## Notice that this implies that z = r (cos i sin ), and so zz = r2 , as expected.

Now, we can rewrite our complex number in a very interesting and unexpected way.
Let us expand Eq. (10) in a Taylor series, using
P
2
4
n 2n
cos = 1 2! + 4! + =
n=0 (1) (2n)!
P
3
5
n 2n+1
sin = 3! + 5! + = =
n=0 (1) (2n+1)! ,
then

"
X

2n
2n+1
X

r (cos + i sin ) = r
(1)n
+i
(1)n
.
(2n)!
(2n
+
1)!
n=0
n=0

## Writing out the first few terms of this series gives



2
i3
4
r (cos + i sin ) = r 1 + i 2! 3! + 4! +


2
(i)3
(i)4
= r 1 + i + (i)
+
+
+

,
2!
3!
4!
2

which we can recognize as the Taylor series for ei . Thus, we find the extremely useful Eulers
Law (or relation),
ei = cos + i sin ,
(11)
which we can use to rewrite our complex number as
z = x + iy = rei = r (cos + i sin ) .

(12)

## Notice that all our old results still hold,

z = rei

r2 = rei rei ,
and so on.
Using Eulers relation can give us some interesting and important results. For instance,
we can use it to determine a useful trigonometric identity. Taking r = 1, then z = ei =
cos + i sin . Squaring both sides gives
2
z 2 = ei = e2i = (cos + i sin )2 = cos2 sin2 + 2i sin cos .
But, e2i = cos (2) + i sin (2), and equating gives
cos (2) + i sin (2) = cos2 sin2 + 2i sin cos .
Since this equality should be true for any values of , we can equate the real and imaginary
parts to find that
cos (2) = cos2 sin2
sin (2) = 2 sin cos ,
which is exactly right. Notice further that the following identities are true, and easy to
prove:
 
Re ei  = cos
(13)
Im ei = sin ,
and

ei +ei
2
ei ei
.
2i

cos =
sin =

(14)

## In additionwe can perform other

useful arithmetic manipulations. We know, for instance, that i = 1, but what is i? From Eq. (12)
 
 
i = cos
+ i sin
= ei/2 ,
2
2
which means that

i=

 
  1 + i

## ei/2 = ei/4 = cos

+ i sin
= .
4
4
2

We can do similar calculations such as 11/4 , which has four roots! So, we see that complex
numbers can be extremely useful. We now turn to another application, which is just as
useful.
3

## We are constantly solving differential equations in physics - in fact, F = ma is a differential

equation we solve all the time. Complex numbers are essential for solving a particular type
of differential equation that shows up all the time in physics, namely a second-order linear
ordinary differential equation with constant coefficients. This type of equation describes
linear simple harmonic oscillator equations such as a mass on a spring, alternating current,
the effect of an electric field on an electron, and, in fact, the fundamental equations of
physics, themselves! Here we will see many examples, starting with the simplest case.

3.1

## Frictionless, unforced spring.

We begin with the simple case of an undamped, unforced mass on a spring. The mass obeys
Hookes law, which says that F = kx = m
x, and so the acceleration,
x =

k
x 02 x,
m

p
where we have defined the harmonic frequency, 0
k/m. Hookes law is a secondorder (two derivatives), ordinary (no partial derivatives), linear (no terms like x 2 , etc.),
differential equation with constant coefficients (there are no terms like x
x, and so on), and
is homogeneous (the differential equation is equal to zero). While its easy enough to guess
the solutions (sines and cosines) lets approach this equation from another direction, using
complex numbers. Suppose that we have the following differential equation,
x = 02 x,
subject to the initial conditions, x (0) = x0 and x (0) = 0; in other words, the system starts
from rest at x = x0 . Heres how we approach the problem (which always works for these
types of problems): we guess a solution of the form
x (t) = Aet ,
where A and are constants to be determined (from our two initial conditions). In order
to see if this is a solution we simply plug it in to our differential equation. Two derivatives
gives x = 2 Aet = 2 x, which is a solution only if 02 = 2 , meaning that = i0 .
Notice what we have done - weve changed a differential equation into an algebraic equation.
Since our differential equation is second-order, this means that we will have two solutions,
one involving +i0 , and the other i0 . Thus,
x (t) = Aei0 t + Bei0 t ,
where A and B are constants. Now, since x (0) = x0 then
x (0) = x0 = A + B,
4

## while x (0) = 0 means

x (0) = 0 = i0 (A B) A = B.
So, we have that A = B = 12 x0 , giving
x (t) =


x0 i0 t
e
+ ei0 t = x0 cos (0 t) ,
2

where we have used Eq. (14). Solving differential equations (of this form) always proceeds
in the same way, as well see in more complicated examples.

3.2

## Now we include a damping term, Fd = bv = bx,

where b is the damping coefficient. Then
m
x + bx + kx = 0,
or, dividing by m and defining b/m 2 (for convenience) we have
x + 2 x + 02 x = 0.
This solution is not so easy to guess as the first one, but we try the same trick, guessing
a solution of the form x (t) = Aet . Plugging it in gives (after canceling off the common
factors)
2 + 2 + 02 = 0.
Once again, our differential equation has been transformed into an algebraic (quadratic)
equation. We can easily solve for the terms,
q
= 2 02 ,
where

2 02 .

## So, our solution is



x (t) = et Aet + Bet .
Now, the behavior of our solution depends on how much damping we have. If 2 > 02 ,
then the term inside the square root (the discriminant) is positive, and we have a solution
exp [ ( + ) t], falling away quickly (wed likely have to set A = 0 in this solution to keep
the spring from blowing up!). This corresponds to lots of damping, leading to a overdamped
solution.
On the other hand, if 2 = 02 , then our solution is exp [t], which is still damped, but
not as much. This is a critically-damped solution. Finally, if 2 < 02 , then the discriminant
is negative, meaning that is imaginary, and we can redefine
q
= 02 2 ,
with a solution


x (t) = et Aeit + Beit .
5

Again, lets take the case that x (0) = x0 and x (0) = 0, giving
x (0) = x0 = A + B,
while
x (0) = 0 = ( i) A ( + i) B,
which is solved by
A =
B =
and so

x0
2i
x0
,
2i

x0
0
0
+ x
eit + x20 hx
eiti
2i i
2i
h2 it
it
it
it
x0 et e +e
+ x0 et e e
2
2i
x0 et cos (t) + x0 et sin (t) .

x (t) = et
=
=

x0
2
x0
2

This is our solution, but it can be put into another form, as follows: First, multiply
and divide by a factor of
r
1p 2
2
0
1+ 2 =
+ 2 = ,

to find



0
cos (t) + sin (t) =
cos (t) +
sin (t)

0
0

Z

,
0

cos =
sin =

## as seen in the figure to the right (notice

2
2
that cos2 +sin2 = +
= 1, as needed).
2

Z
Then we have


0

0
0
cos (t) +
sin (t) =
[cos (t) cos + sin (t) sin ] =
cos (t ) .
0
0

x (t) =

x0 0 t
e cos (t ) ,

3.3

## Forced, undamped spring.

Now we will include a forcing function, which we take to be F (t) = F0 cos (t), where F0 is
a constant, and is the driving frequency. We will take the damping term to be zero for
now, b 0. Then, our differential equation is (after dividing by m) gives
x + 02 x =

F0
cos (t) .
m

Again, this is not too tough a solution to guess, but well solve it using our complex methods.
Now, since the cosine is the real part of the complex exponential,
 
cos (t) = Re eit ,
we can imagine solving a different differential equation,
z + 02 z =

F0 it
e ,
m

and note that, since the cosine is the real part of our forcing function, then our solution
x (t) = Re [z (t)]. So, in other words, if we can figure out a solution to the differential
equation for z, well get a complex function z = x (t) + iy (t), then we can take the real part
of z (since we have a cosine forcing function) to get our solution for x (t). Notice that, if we
had a sine function, instead, then we would take the imaginary part of z.
So, we proceed as before, taking a solution z = Aet . Plugging in to the differential
equation gives
F0 it
2 Aet + 02 Aet =
e .
m
Comparing tells us that = i, meaning that the spring oscillates at the same frequency as
the forcing term. Canceling off the exponential provides the algebraic equation,

F0
,
2 + 02 A =
m
which tells us A,
A=
and so
z (t) =

F0 /m
,
02 2
F0 /m it
e .
02 2

Now, to get our solution to the x (t) equation, we just take the real part,
x (t) = Re [z (t)] =

F0 /m
cos (t) .
02 2

Notice that we can get resonance when = 0 , with this function blowing up, since we
didnt include any friction.

3.4

## Driven, damped harmonic oscillator.

Now we consider the most general case, solving it in the same way. The method is the
same, but the algebra is a little messier. Our full differential equation is (making the same
definitions as before)
F0
cos (t) ,
x + 2 x + 02 x =
m
which we transform into
F0 it
z + 2 z + 02 z =
e .
m
Once again, well solve this equation, and then take the real part at the end. Now, we
proceed as before - this time well take for our guess
z (t) = Aeit ,
where we just set = i, since we know we have to cancel off the driving function. Plugging
this into the differential equation, and canceling of the exponentials gives
2 A + 2iA + 02 A =

F0
.
m

## Solving for the constant

A=

02

F0 /m
.
2 + 2i

Now, to make life easier, lets rationalize this expression, multiplying the top and bottom by
02 2 2i,
 2

F0 /m
A= 2
0 2 2i .
2
(0 2 ) + 4 2 2
This means that the constant A has a real and imaginary part, which means that we have
to be a little careful. We can play the same games as before, defining our angle by
cos =

sin =

02 2
2
2
0 2 +4 2 2

(02 2 )

+4 2 2

such that
F0 /m
F0 /m
A= q
(cos i sin ) = q
ei
2
2
(02 2 ) + 4 2 2
(02 2 ) + 4 2 2
So, our solution for z is
F0 /m
z (t) = q
ei(t) .
2
(02 2 ) + 4 2 2

This is why we did all the algebra with A: to get a simple form for z (t). Now all we have
to do is to take the real part,
x (t) = q

F0 /m
(02

2
2)

cos (t ) .
4 2 2

Trying to solve this without the complex variables is doable, but its much more of a pain.
The complex variables method is much nicer, once you get the hang of it!
The text passage is a detailed set of notes on complex numbers and their application in solving differential equations, particularly in the context of a mass-spring system. It covers the definition of complex numbers, their properties, and how they can be used to solve various types of differential equations, including undamped, damped, and forced systems.
no_question

the passage does not contain any explicit mathematical questions or problems for the reader to solve. it is primarily expository, providing definitions, derivations, and examples.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
You are on page 1of 9

# Complex Numbers

This set of notes reviews complex numbers and their properties. We then examine
the use of complex numbers to solve second-order ordinary differential equations, using the
mass-spring system as our example.

Introduction

Complex numbers generalizes the idea of a real number to include an imaginary component.
Imaginary numbers contain a factor of i, where

(1)
i 1.
No real number follows this definition, so we invent the imaginary number. A complex
number, z, is a number of the form
z = x + iy,
(2)
where x and y are both real. So, for example
z = 2 + 2i
z = 4
z = 6i
are all complex numbers. In the first case z has both a real and imaginary part, while z is
purely real in the second case, and purely imaginary in the third. With this in mind, we can
define x in Eq. (2) to be the real part of the complex number z; in other words,
x Re [z] ,

(3)

## while y (not iy) is defined as the imaginary part,

y Im [z] .

(4)

So, in our earlier examples of complex numbers, when z = 2 + 2i, then Re [z] = 2, while
Im [z] = 2, and when z = 6i, then Re [z] = 0, and Im [z] = 6.
We can define another quantity, called the complex conjugate of z, denoted z , which
is obtained from z simply by changing the sign of the imaginary part; i.e., by letting i i.
Then, if z is given by Eq. (2), then
z x iy.

(5)

## We can multiply together z and z to get a purely real number,

zz = z z |z|2 = x2 + y 2 .

(6)

## For example, if z = 2 + 2i, then z = 2 2i, and zz = (2 + 2i) (2 2i) = 4 4i + 4i + 4 = 8.

These definitions will be extremely useful in what follows.
1

## Because a complex number is, in a sense, a

two-dimensional object, we can easily treat
it like a vector, as seen in the figure to
the right. We imagine graphing a complex
number z = x + iy, placing the real part of
z along the x-axis, and the imaginary part
of z along the y-axis, much like we would
graph a vector ~r = xi + yj; for example, if
we have z = 2 + 2i, then we would treat
the graph like ~r = 2i + 2j.

Im[z]
(x,y)
r
T
Re[z]

Looking at the graph, we see that the vector describing the complex number has a
magnitude, and also makes an angle with the real axis. Treating these values like ordinary
polar coordinates we see that we can define
Re [z] = x = r cos
Im [z] = y = r sin .

(7)

## Furthermore, we can see that

r=

p
x2 + y 2 = |z|2 = zz ,

## while the angle is

1

= tan

(8)

y

,
(9)
x
and so complex numbers really do mimic (two-dimensional) vectors. Now, based on the
graph we can write our complex number in another way. Using Eqs. (7) we can write
z = x + iy = r (cos + i sin ) .

(10)

## Notice that this implies that z = r (cos i sin ), and so zz = r2 , as expected.

Now, we can rewrite our complex number in a very interesting and unexpected way.
Let us expand Eq. (10) in a Taylor series, using
P
2
4
n 2n
cos = 1 2! + 4! + =
n=0 (1) (2n)!
P
3
5
n 2n+1
sin = 3! + 5! + = =
n=0 (1) (2n+1)! ,
then

"
X

2n
2n+1
X

r (cos + i sin ) = r
(1)n
+i
(1)n
.
(2n)!
(2n
+
1)!
n=0
n=0

## Writing out the first few terms of this series gives



2
i3
4
r (cos + i sin ) = r 1 + i 2! 3! + 4! +


2
(i)3
(i)4
= r 1 + i + (i)
+
+
+

,
2!
3!
4!
2

which we can recognize as the Taylor series for ei . Thus, we find the extremely useful Eulers
Law (or relation),
ei = cos + i sin ,
(11)
which we can use to rewrite our complex number as
z = x + iy = rei = r (cos + i sin ) .

(12)

## Notice that all our old results still hold,

z = rei

r2 = rei rei ,
and so on.
Using Eulers relation can give us some interesting and important results. For instance,
we can use it to determine a useful trigonometric identity. Taking r = 1, then z = ei =
cos + i sin . Squaring both sides gives
2
z 2 = ei = e2i = (cos + i sin )2 = cos2 sin2 + 2i sin cos .
But, e2i = cos (2) + i sin (2), and equating gives
cos (2) + i sin (2) = cos2 sin2 + 2i sin cos .
Since this equality should be true for any values of , we can equate the real and imaginary
parts to find that
cos (2) = cos2 sin2
sin (2) = 2 sin cos ,
which is exactly right. Notice further that the following identities are true, and easy to
prove:
 
Re ei  = cos
(13)
Im ei = sin ,
and

ei +ei
2
ei ei
.
2i

cos =
sin =

(14)

## In additionwe can perform other

useful arithmetic manipulations. We know, for instance, that i = 1, but what is i? From Eq. (12)
 
 
i = cos
+ i sin
= ei/2 ,
2
2
which means that

i=

 
  1 + i

## ei/2 = ei/4 = cos

+ i sin
= .
4
4
2

We can do similar calculations such as 11/4 , which has four roots! So, we see that complex
numbers can be extremely useful. We now turn to another application, which is just as
useful.
3

## We are constantly solving differential equations in physics - in fact, F = ma is a differential

equation we solve all the time. Complex numbers are essential for solving a particular type
of differential equation that shows up all the time in physics, namely a second-order linear
ordinary differential equation with constant coefficients. This type of equation describes
linear simple harmonic oscillator equations such as a mass on a spring, alternating current,
the effect of an electric field on an electron, and, in fact, the fundamental equations of
physics, themselves! Here we will see many examples, starting with the simplest case.

3.1

## Frictionless, unforced spring.

We begin with the simple case of an undamped, unforced mass on a spring. The mass obeys
Hookes law, which says that F = kx = m
x, and so the acceleration,
x =

k
x 02 x,
m

p
where we have defined the harmonic frequency, 0
k/m. Hookes law is a secondorder (two derivatives), ordinary (no partial derivatives), linear (no terms like x 2 , etc.),
differential equation with constant coefficients (there are no terms like x
x, and so on), and
is homogeneous (the differential equation is equal to zero). While its easy enough to guess
the solutions (sines and cosines) lets approach this equation from another direction, using
complex numbers. Suppose that we have the following differential equation,
x = 02 x,
subject to the initial conditions, x (0) = x0 and x (0) = 0; in other words, the system starts
from rest at x = x0 . Heres how we approach the problem (which always works for these
types of problems): we guess a solution of the form
x (t) = Aet ,
where A and are constants to be determined (from our two initial conditions). In order
to see if this is a solution we simply plug it in to our differential equation. Two derivatives
gives x = 2 Aet = 2 x, which is a solution only if 02 = 2 , meaning that = i0 .
Notice what we have done - weve changed a differential equation into an algebraic equation.
Since our differential equation is second-order, this means that we will have two solutions,
one involving +i0 , and the other i0 . Thus,
x (t) = Aei0 t + Bei0 t ,
where A and B are constants. Now, since x (0) = x0 then
x (0) = x0 = A + B,
4

## while x (0) = 0 means

x (0) = 0 = i0 (A B) A = B.
So, we have that A = B = 12 x0 , giving
x (t) =


x0 i0 t
e
+ ei0 t = x0 cos (0 t) ,
2

where we have used Eq. (14). Solving differential equations (of this form) always proceeds
in the same way, as well see in more complicated examples.

3.2

## Now we include a damping term, Fd = bv = bx,

where b is the damping coefficient. Then
m
x + bx + kx = 0,
or, dividing by m and defining b/m 2 (for convenience) we have
x + 2 x + 02 x = 0.
This solution is not so easy to guess as the first one, but we try the same trick, guessing
a solution of the form x (t) = Aet . Plugging it in gives (after canceling off the common
factors)
2 + 2 + 02 = 0.
Once again, our differential equation has been transformed into an algebraic (quadratic)
equation. We can easily solve for the terms,
q
= 2 02 ,
where

2 02 .

## So, our solution is



x (t) = et Aet + Bet .
Now, the behavior of our solution depends on how much damping we have. If 2 > 02 ,
then the term inside the square root (the discriminant) is positive, and we have a solution
exp [ ( + ) t], falling away quickly (wed likely have to set A = 0 in this solution to keep
the spring from blowing up!). This corresponds to lots of damping, leading to a overdamped
solution.
On the other hand, if 2 = 02 , then our solution is exp [t], which is still damped, but
not as much. This is a critically-damped solution. Finally, if 2 < 02 , then the discriminant
is negative, meaning that is imaginary, and we can redefine
q
= 02 2 ,
with a solution


x (t) = et Aeit + Beit .
5

Again, lets take the case that x (0) = x0 and x (0) = 0, giving
x (0) = x0 = A + B,
while
x (0) = 0 = ( i) A ( + i) B,
which is solved by
A =
B =
and so

x0
2i
x0
,
2i

x0
0
0
+ x
eit + x20 hx
eiti
2i i
2i
h2 it
it
it
it
x0 et e +e
+ x0 et e e
2
2i
x0 et cos (t) + x0 et sin (t) .

x (t) = et
=
=

x0
2
x0
2

This is our solution, but it can be put into another form, as follows: First, multiply
and divide by a factor of
r
1p 2
2
0
1+ 2 =
+ 2 = ,

to find



0
cos (t) + sin (t) =
cos (t) +
sin (t)

0
0

Z

,
0

cos =
sin =

## as seen in the figure to the right (notice

2
2
that cos2 +sin2 = +
= 1, as needed).
2

Z
Then we have


0

0
0
cos (t) +
sin (t) =
[cos (t) cos + sin (t) sin ] =
cos (t ) .
0
0

x (t) =

x0 0 t
e cos (t ) ,

3.3

## Forced, undamped spring.

Now we will include a forcing function, which we take to be F (t) = F0 cos (t), where F0 is
a constant, and is the driving frequency. We will take the damping term to be zero for
now, b 0. Then, our differential equation is (after dividing by m) gives
x + 02 x =

F0
cos (t) .
m

Again, this is not too tough a solution to guess, but well solve it using our complex methods.
Now, since the cosine is the real part of the complex exponential,
 
cos (t) = Re eit ,
we can imagine solving a different differential equation,
z + 02 z =

F0 it
e ,
m

and note that, since the cosine is the real part of our forcing function, then our solution
x (t) = Re [z (t)]. So, in other words, if we can figure out a solution to the differential
equation for z, well get a complex function z = x (t) + iy (t), then we can take the real part
of z (since we have a cosine forcing function) to get our solution for x (t). Notice that, if we
had a sine function, instead, then we would take the imaginary part of z.
So, we proceed as before, taking a solution z = Aet . Plugging in to the differential
equation gives
F0 it
2 Aet + 02 Aet =
e .
m
Comparing tells us that = i, meaning that the spring oscillates at the same frequency as
the forcing term. Canceling off the exponential provides the algebraic equation,

F0
,
2 + 02 A =
m
which tells us A,
A=
and so
z (t) =

F0 /m
,
02 2
F0 /m it
e .
02 2

Now, to get our solution to the x (t) equation, we just take the real part,
x (t) = Re [z (t)] =

F0 /m
cos (t) .
02 2

Notice that we can get resonance when = 0 , with this function blowing up, since we
didnt include any friction.

3.4

## Driven, damped harmonic oscillator.

Now we consider the most general case, solving it in the same way. The method is the
same, but the algebra is a little messier. Our full differential equation is (making the same
definitions as before)
F0
cos (t) ,
x + 2 x + 02 x =
m
which we transform into
F0 it
z + 2 z + 02 z =
e .
m
Once again, well solve this equation, and then take the real part at the end. Now, we
proceed as before - this time well take for our guess
z (t) = Aeit ,
where we just set = i, since we know we have to cancel off the driving function. Plugging
this into the differential equation, and canceling of the exponentials gives
2 A + 2iA + 02 A =

F0
.
m

## Solving for the constant

A=

02

F0 /m
.
2 + 2i

Now, to make life easier, lets rationalize this expression, multiplying the top and bottom by
02 2 2i,
 2

F0 /m
A= 2
0 2 2i .
2
(0 2 ) + 4 2 2
This means that the constant A has a real and imaginary part, which means that we have
to be a little careful. We can play the same games as before, defining our angle by
cos =

sin =

02 2
2
2
0 2 +4 2 2

(02 2 )

+4 2 2

such that
F0 /m
F0 /m
A= q
(cos i sin ) = q
ei
2
2
(02 2 ) + 4 2 2
(02 2 ) + 4 2 2
So, our solution for z is
F0 /m
z (t) = q
ei(t) .
2
(02 2 ) + 4 2 2

This is why we did all the algebra with A: to get a simple form for z (t). Now all we have
to do is to take the real part,
x (t) = q

F0 /m
(02

2
2)

cos (t ) .
4 2 2

Trying to solve this without the complex variables is doable, but its much more of a pain.
The complex variables method is much nicer, once you get the hang of it!
user
The text passage is a detailed set of notes on complex numbers and their application in solving differential equations, particularly in the context of a mass-spring system. It covers the definition of complex numbers, their properties, and how they can be used to solve various types of differential equations, including undamped, damped, and forced systems.

TYPE: NO_QUESTION

The passage does not contain any explicit mathematical questions or problems for the reader to solve. It is primarily expository, providing definitions, derivations, and examples.