Physics 264L: Assignment 7
Made available:
Due:
Friday, October 16, 2015
Wednesday, October 21, 2015, by 7 pm
Problems
1. The following tutorial problem involves your working through an example
in which
you use the complex
£
¤
exponential representation of a wave u(x, t) = a cos(kx − ωt) = a Re ei(kx−ωt) to derive the total wave
intensity I created by N equally-spaced equally-strong coherent (in phase) point wave sources. This
could correspond to the light or sound pattern produced by a plane wave (the wave you get a far
distance from a point source) striking a diffraction grating consisting of N equally spaced slits in an
opaque wall, or of N equally-spaced equally bright light bulbs or equally loud sound speakers placed
on a line, or of N equally-spaced identical atoms on the surface of some substance that scatters a beam
of electrons, like the Davisson-Germer experiment.
All of the details in this tutorial are important for your being able to carry out various manipulations
and simplifications related to solving the Schrodinger equation so please make sure you become fluent
with these ideas. These techniques are used in many other areas of science, engineering, and applied
mathematics.
I have also attached at the end of this assignment an optional guided tutorial of complex numbers and
you should work through that first if you are not familiar with complex numbers or need some review.
To begin, consider N identical point wave sources (light, sound, water, it doesn’t matter) of equal
amplitude a, of equal wavenumber k = 2π/λ, and of equal angular frequency ω = 2π/T where λ and T
are respectively the common wavelength and period of the waves. These sources are equally spaced
with separation d on some horizontal line like this:
P
r
θ
N
N-1
2
d sinθ
1
d
where each black dot corresponds to a wave source.
Now consider a point P in space where we would like to calculate the effect of all the wave sources.
For general points P , there is no simple formula to obtain and one would need to add up the effects
numerically using a computer. But if P is far enough away from all the sources (the distance r of P is
large compared to the length N d of the sources itself), it is possible to get a concise useful analytical
expression for the effects of all the waves at P . The reason is that, when P is sufficiently far away, all
the line segments connecting the sources to P become approximately parallel, and then the distance
of one source to P differs by the same constant amount
δ = d sin(θ),
1
(1)
from the distance of a nearest neighbor, where θ is the angle that P makes from the perpendicular
bisector of the line segment of sources, as shown.
If the wave produced by a given wave source is described by a simple sinusoidal form
u(x, t) = a cos(kx − ωt),
(2)
then you should be able to see that the sum of all the waves at the point P is given by
utotal (t, P ) = u1 (t, P ) + u2 (t, P ) + u3 (t, P ) + · · · + uN (t, P )
= a cos[kr − ωt] + a cos[k(r + δ) − ωt] + a cos[k(r + 2δ) − ωt]
+ · · · + a cos[k(r + (N − 1)δ) − ωt]
= a cos(φ) + a cos(φ + ∆φ) + a cos(φ + 2∆φ) + · · · + a cos(φ + (N − 1)∆φ),
(3)
(4)
(5)
(6)
where we have introduced the abbreviations
φ = kr − ωt
and
∆φ = kδ = kd sin(θ),
(7)
to reduce the number of symbols to write (these are dimensionless phase angles).
So adding up all the waves from equally-spaced identical in-phase sources at some sufficiently remote
point P corresponds to the mathematical problem Eq. (6) of adding up N cosine functions that each
differ by a constant phase ∆φ, and this is the problem we are going to solve using complex exponentials.
(The main goal here is to give you practice with complex exponentials, rather than to derive the
intensity pattern of a diffraction grating.) I will list the key steps and your homework answer is then
for you to explain how to go from one step to the next, as described below.
Starting from Eq. (6), we can consider each cosine function as the real part of a corresponding complex
exponential to get:
utotal (t, P ) = a cos(φ) + a cos(φ + ∆φ) + a cos(φ + 2∆φ) + · · · + a cos(φ + (N − 1)∆φ)
h
i
h
i
h
i
£ ¤
= a Re eiφ + a Re ei(φ+∆φ) + a Re ei(φ+2∆φ) + · · · + a Re ei(φ+(N −1)∆φ)
i
h
= a Re eiφ + ei(φ+∆φ) + ei(φ+2∆φ) + · · · + ei(φ+(N −1)∆φ)
´i
h ³
= a Re eiφ 1 + ei∆φ + ei2∆φ + · · · + ei(N −1)∆φ
h ³
¡
¢2
¡
¢N −1 ´i
= a Re eiφ 1 + ei∆φ + ei∆φ + · · · + ei∆φ
¶¸
· µ
1 − eiN ∆φ
= a Re eiφ
1 − ei∆φ
¸
·
iN ∆φ/2
e−iN ∆φ/2 − eiN ∆φ/2
iφ e
= a Re e · i∆φ/2 ·
e
e−i∆φ/2 − ei∆φ/2
"
¡N
¢#
i(φ+((N −1)/2)∆φ) sin ¡ 2 ∆φ¢
= a Re e
sin 21 ∆φ
¢ h
¡
i
sin N2 ∆φ
¡1
¢ Re ei(φ+((N −1)/2)∆φ)
=a
sin 2 ∆φ
³
´

¸
·
sin N
∆φ
N −1
2


³
´
∆φ .
= a
cos φ +
2
sin 12 ∆φ
(8)
(9)
(10)
(11)
(12)
(13)
(14)
(15)
(16)
(17)
In your homework, please explain clearly how each line Eq. (9)-Eq. (17) was
obtained from the previous line. You can refer to equations (34)-(40) and
equations (43)-(50) below when justifying some of the steps.
2
In various areas of science, engineering, and mathematics, when one is working with equally-spaced
coherent sources of waves like a diffraction grating or X-ray crystallography, Eq. (17) is the key result
of interest, e.g., it justifies why diffraction gratings are greatly superior to prisms for spectroscopy
because, as the number of slits N in a grating becomes larger (N ≈ 104 is typical for a diffraction
grating), the spectral lines become much brighter and much more narrow than can be achieved with
a prism, improving one’s ability to see and to resolve distinct spectral lines. But I will let you read
about this further in your intro physics book.
I would like to mention just one brief interesting insight of Eq. (17), which is that it helps to explain how
one can combine waves of infinite spatial extent to obtain functions that are highly spatially localized,
a somewhat paradoxical result that suggests the possibility that perhaps particles like electrons could
be made up of waves in the first place, in which case it would not be surprising for a particle to show
wave-like properties. If we ignore the original context of adding up waves and just consider Eq. (17)
to be some mathematical identity, we can set φ = 0, a = 1, and relabel ∆φ to be some symbol x to get
a concise analytical expression1 for the sum of the first N cosine functions cos(nx) for 0 ≤ n ≤ N − 1:
³
´
¸
·
sin N2x
N −1
³ ´ cos
x .
(18)
1 + cos(x) + cos(2x) + · · · + cos((N − 1)x) =
2
sin x
2
Now each function cos(nx) is an oscillatory function that extends from x = −∞ to x = +∞ and so
one would expect the sum of all these functions to be a highly wiggly thing that also has structure
spread out over the entire real line. But a plot of the amplitude sin(N x/2)/ sin(x/2) for increasing
values of N :
Manipulate[
Plot[
Sin[N x / 2]
{ x, -Pi, Pi
PlotRange ->
] ,
{ N, 1, 100, 5
]
/ Sin[x / 2] ,
} ,
All
} (* make plots for N varying from 1 to 100 in steps of 5 *)
shows that, as the number N of cosine functions increases, the sum of all of these cosines becomes an
extremely tall, extremely narrow peak of finite area that periodically repeats every 2π spatial units.
(How to use the Manipulate command is explained as part of the next problem.) This highly-localized
object is an example of something called a “Dirac delta function” that we will use many times in our
discussions of quantum mechanics, as a convenient mathematical way to model point particle or model
a highly localized potential energy, such as the strong and weak interactions of a nucleus, that decay
to zero rapidly just outside of a nucleus.
1 The coefficient sin(N x/2)/ sin(x/2) can be rewritten in terms of the cardinal sine or sinc function sinc(x) = sin(x)/x like
this: [sin(N x/2)/(N x/2)][N (x/2)/ sin(x/2) = N sinc(N x/2)/sinc(x/2). The sinc function has a characteristic peak of value 1
at x = 0 and decays in an oscillatory way as 1/|x| for large x. The sinc function appears in many places in physics and also in
signal analysis.
3
2. Problem 2-2 on page 96 of French and Taylor, but solve this problem only by expressing the
three sine functions as the imaginary part of a complex exponential, sin(x) = Im[ eix ]. The first two
steps look like this:
y0 + y1 + y2 = A sin(kx − ωt)
¶
µ
¶ ¸
·µ
∆ω
∆k
A
x− ω−
t
+ sin k −
2
2
2
¶
µ
¶ ¸
·µ
A
∆ω
∆k
x− ω+
t
+ sin k +
2
2
2
i
h
= A Im ei(kx−ωt)
i
h
A
+ Im ei[(k−∆k/2)x−(ω−∆ω/2)t]
2
i
h
A
+ Im ei[(k+∆k/2)x−(ω+∆ω/2)t]
2·
¸
1 i[(k−∆k/2)x−(ω−∆ω/2)t] 1 i[(k+∆k/2)x−(ω+∆ω/2)t]
i(kx−ωt)
= A Im e
+ e
,
+ e
2
2
(19)
(20)
(21)
(22)
(23)
(24)
(25)
where I obtained this last line by using the fact (which you should take the time to prove) that
Im(z1 ) + Im(z2 ) = Im(z1 + z2 ),
(26)
for two complex numbers z1 = x1 + iy1 and z2 = x2 + iy2 . Eq. (26) says that “taking the imaginary
part” of a complex number z distributes over the addition of complex numbers.
(a) Starting from Eq. (25) and showing all intermediate steps, derive this answer to Problem 2-2 of
French and Taylor:
µ
·
¶¸
∆k
∆ω
2
y1 + y2 + y3 = 2 cos
(27)
x−
t × A sin(kx − ωt),
4
4
using facts such as ea+b = ea eb and eix +e−ix = 2 cos(x). This answer is not only satisfying because
it is concise, it is physically insightful since Eq. (27) is the product of a slowly varying wave u1 (x, t)
(the quantity in brackets) that multiplies and so slowly modulates the wave u2 = A sin(kx − ωt)
that looks like the original wave.
(b) What is the wavelength λ1 , period T1 , speed v1 , and direction (+x or −x) of the wave u1 that
slowly modulates the magnitude of the basic wave u2 ?
(c) To get some intuition about what the wave y1 + y2 + y3 looks like, download and execute the
Mathematica notebook
www.phy.duke.edu/~hsg/264L/files/homeworks/assignment-7-notebook.nb
by selecting all the cells simultaneously via the keystroke “control-a” and then executing all the
cells together by typing Shift-Return.
The results of the Manipulate command, which is used to make movies of some plot as some
variable like time t is varied, is a panel that looks like this:
4
You will notice a horizontal slider in the upper left corner of the Manipulate box, labeled by the
variable (here time t) that varies during the movie. You can move the little box in the slider back
and forth with your mouse cursor to see images at different times, or you can click on the + sign
at the end of the slider box to display a small video control panel that looks like this:
If you click on the black right-pointing triangle in this control panel, the triangle changes to a
stop icon || and a movie starts. You can control the speed of the movie by clicking multiple times
on the double up arrow (speed up) or multiple times on the double down arrow (slow down). You
can stop the movie at any time by clicking on the stop icon. You can also make the movie bigger
by selecting the movie (click on the plot with your left mouse button), and then dragging a corner
of the bounding box to make the image bigger.
The key parts of the code involve the Mathematica Manipulate command and look like this:
k
= 2 Pi ;
w
= 2 Pi ;
amp = 1. ;
(* set wavelength lambda of wave to be 1 *)
(* set period T of wave to be one or omega = 2pi *)
(* set amplitude of wave A = 1 *)
dk = 0.3 k ;
dw = 0.3 w ;
(* perturbation of wavenumber k, small compared to k *)
(* perturbation of ang freq omega, small compared to omega *)
Manipulate[ (* make movie of modulated wave *)
Plot[
( 2 Cos[ (dk/4) x - (dw/4) t]^2 ) * ( amp Sin[ k x - w t ] ) ,
5
{ x, 0, 10 Pi} ,
PlotPoints -> 400 (* use more plotting points to make smooth graph *)
] ,
{ t, 0, 100 } (* repeat plot for times varying from t=0 to t=100 *)
]
The Manipulate command is quite useful in physics and mathematics for getting intuition about
how one or more variables affects some mathematical expression, and I encourage you to use it
heavily in this and other courses. You can read about the format of the Manipulate command by
starting Mathematica, clicking on the menu option Help/Wolfram Documentation, then typing
“Manipulate” in the search window. For example, the Mathematica command (which you should
paste and execute in a Mathematica notebook)
t = 0. ;
Manipulate[ (* make movie of modulated wave as function of dk *)
Plot[
( 2 Cos[ (dk/4) x - (dw/4) t]^2 ) * ( amp Sin[ k x - w t ] ) ,
{ x, 0, 10 Pi } ,
PlotPoints -> 400
] ,
{ dk, -0.3 k, 0.3 k }
]
will make a movie of how the spatial structure of the modulated wave at a particular time t = 0
depends on small successive changes of the parameter ∆k (labeled dk in the Mathematica code)
over some range of negative and positive values. You can see directly from the Manipulate output
that, as you make the perturbation ∆k smaller in magnitude (while holding dω at some fixed
value), the modulation of the wave has a longer and longer wavelength.
Here is what you need to do: Run the above Mathematica notebook for the pairs of parameters
(dk, dω) = (αk, αω) for the three choices α = 0.4, 0.2, and 0.1 and summarize briefly and qualitatively what you learn as these perturbations become smaller. Also study briefly and summarize
what happens in the movie of the modulated wave when the group velocity vg = ∆ω/∆k has a
negative value, e.g., try the values (dk, dω) = (−0.3k, 0.3ω) and describe the resulting modulated
wave.
These four YouTube videos
https://www.youtube.com/watch?v=tlM9vq-bepA
https://www.youtube.com/watch?v=v9DPzMoWpc0
https://www.youtube.com/watch?v=v9DPzMoWpc0
https://www.youtube.com/watch?v=iVJNcANWmI0
v_g positive
v_g zero
v_g negative
v_phi zero
complement this problem by showing more clearly the phase velocity vφ = ω/k and group velocity vg = dω/dk of a propagating modulated wave, by adding a magenta dot that moves with a
local maximum of the slow modulating envelope, and by adding a green dot that moves with a
local minimum of the fast basic wave. From these movies, you can see how the group velocity can
be positive, zero, or negative while the phase velocity always remains positive, and that it is even
possible to have a zero phase velocity with a non-zero group velocity.
3. Problem 2-3 on page 97 of French and Taylor.
4. Problem 2-7 on pages 97-98 but skip part (b), since this is discussed in most intro physics books and
there is no need to repeat the derivation here .
6
5. To the nearest integer, please give the time in hours that it took you to complete this assignment,
including reading in the text.
Also, if you got help with this assignment, please give the names of the people from whom you got
help (classmates, the TA, myself, etc).
6. Optional Challenge Problem
Challenge problem 7.1: To appreciate the power of complex numbers over using trigonometry, solve
this problem using complex numbers:
Given a regular N -gon on the unit circle for N ≥ 2, prove that the product of the lengths of
the N − 1 chords from a given vertex to the other N − 1 vertices is N .
Hint: put one vertex of the N -gon at the point (x, y) = (1, 0) and represent the locations of the other
vertices by successive powers of the N th root of 1, z = e2πi/N .
7. Optional Bootcamp on Complex Numbers
The following tutorial problem is completely optional and I provide it as a convenient way for you to
review and practice working with complex numbers and complex exponentials at the level needed for
Physics 264L. The parts of the tutorial that require you to carry out some calculation are written in
underlined bold font like this, so look below for this format to see what you are actually supposed to
do.
A complex number z is defined to be an expression of the form
z = x + iy,
(28)
where x and y are real numbers and where2 i is a square root of −1 so that i2 = −1. (Don’t
be intimidated,
Eq. (28) is no more complicated or scary than writing an expression of the form
√
1 + 2 3.) The number x is called the real part of the complex number, the number y is called the
imaginary part of the complex number. A complex number for which y = 0 is called “real” and
a complex number for which x = 0 is called “imaginary” or “purely imaginary”. (This is a terrible
and unfortunate historical naming convention, there is nothing imaginary or pretend about imaginary
numbers.) Please note that, by definition, the imaginary part of a complex number x + iy is the real
number y, not the imaginary number iy.
Two complex numbers z1 = x1 + iy1 and z2 = x2 + iy2 are defined to be equal, z1 = z2 , if and only if
their real and imaginary parts are equal, x1 = x2 and y1 = y2 . The sum z1 +z2 of two complex numbers
is what you would expect based on the usual rules of algebra (treating i like any typical number) and
corresponds to adding their real and imaginary parts like this:
z1 + z2 = (x1 + iy1 ) + (x2 + iy2 ) = (x1 + x2 ) + i (y1 + y2 ) .
(29)
Similarly, the product z1 z2 of two complex numbers is what you expect based on the usual algebraic
manipulations provided that you remember that i2 = −1:
z1 z2 = (x1 + iy1 ) (x2 + iy2 )
= x1 (x2 ) + x1 (iy2 ) + (iy1 )x2 + (iy1 )(iy2 )
= x1 x2 + ix1 y2 + iy1 x2 + i2 y1 y2
= (x1 x2 − y1 y2 ) + i (x1 y2 + x2 y1 ) .
(30)
(31)
(32)
(33)
2 Some students have asked me what is the difference between physicists and engineers. There is no simple answer, there are
physicists who do engineering-like projects and engineers who study physics-like questions, but one defining difference is that
the engineering community uses the symbol j rather than i to denote a square root of -1.
7
The real-valued function Re(z) of the complex number z (pronounced “real of z”) is defined to be the
real part of a complex number z = x + iy, namely the number x, while the real-valued function Im(z)
(pronounced “imaginary of x”), is defined to be the imaginary part of z, namely the real number y.
You should make sure that you understand why:
z = Re(z) + i Im(z).
(34)
Re[ Im(z) ] = Im(z).
Im[ Re(z) ] = 0.
(35)
(36)
Re (z1 + z2 ) = Re (z1 ) + Re (z2 ) .
Im (z1 + z2 ) = Im (z1 ) + Im (z2 ) .
Re(cz) = c Re(z), for c real.
Im(cz) = c Im(z),
for c real.
(37)
(38)
(39)
(40)
Can you see why, generally, Re(z1 z2 ) 6= Re(z1 )Re(z2 ) and Im(z1 z2 ) 6= Im(z1 )Im(z2 )?
The complex conjugate z ∗ (usually pronounced as “z star” but some times as “z conjugate”) of a
complex number z is defined to be the complex number
z ∗ = x − iy,
(41)
obtained by reversing the sign of the imaginary part. The real-valued non-negative magnitude of a
complex number, denoted by |z|, is defined to be
p
|z| = x2 + y 2 .
(42)
The magnitude is also called the absolute value, size, modulus, or length of a complex number in
different contexts and by different communities (e.g., mathematicians vs physicists).
(a) Show that:
z is real
z = z∗.
⇔
z is imaginary
∗
⇔
z = −z .
(z1 + z2 ) = z1∗ + z2∗ .
∗
(z1 z2 ) =
2
∗ 2
(43)
∗
z1∗ z2∗ .
∗
(44)
(45)
(46)
∗
|z| = |z | = zz = z z.
(47)
|z1 z2 | = |z1 | |z2 |.
(48)
1
(49)
Re(z) = (z + z ∗ ) .
2
1
Im(z) =
(z − z ∗ ) .
(50)
2i
Note that there are two ways top
demonstrate the often used relation Eq. (48). One way is to use
directly the definition |x + iy| = x2 + y 2 , but a more efficient and clever ( and so recommended)
way is to prove that the square of Eq. (48) holds, |z1 z2 |2 = |z1 |2 |z2 |2 and then use Eq. (47) to
write |z1 z2 |2 = (z1 z2 ) · (z1 z2 )∗ and you should be able to finish the argument from here by using
Eq. (46) and the associative property of multiplication of complex numbers. In your homework,
please include both ways so that you can see why the latter way is faster and more general than
the former way.
The above results explain why it is often useful to work with the complex conjugate of complex
expressions. For example, the result |z|2 = zz ∗ , Eq. (47), provides a quick way to compute the
reciprocal of a complex number by solving for 1/z like this:
z∗
1
= 2,
z
|z|
8
(51)
or in terms of the real and imaginary parts:
a
a − ib
1
= 2
+i
= 2
2
a + ib
a +b
a + b2
µ
−b
2
a + b2
¶
.
(52)
Since z1 /z2 = z1 × (1/z2 ), Eq. (48) provides a quick and practical way to compute the magnitude
of the ratio of two complex numbers as the ratio of their magnitudes:
¯ ¯
¯ z1 ¯ |z1 |
¯ ¯=
(53)
¯ z2 ¯ |z2 | ,
since one avoids calculating explicitly the complex number z1 /z2 in the form x + iy. For example,
√
¯
¯
¶1/2
µ 2
¯ a + ib ¯ |a + ib|
a2 + b2
a + b2
¯
¯=
√
=
.
=
¯ c + id ¯ |c + id|
c2 + d2
c2 + d2
(54)
(b) If p(x) is a polynomial of degree N with real coefficients, so has the form
p(x) = c0 + c1 x + c2 x2 + · · · + cN −1 xN −1 + cN xN
=
N
X
ci xi ,
(55)
(56)
i=0
where all the coefficients ci are real numbers, show that if z = a + ib is a root of this
polynomial so that p(z) = 0, then z ∗ is also a root, so that p(z ∗ ) = 0.
Note: if you are not comfortable working with general expressions like Eq. (55), it will be fine
for you to show the truth of this statement for the case N = 2, which corresponds to a quadratic
equation
p(x) = a + bx + cx2 ,
(57)
where the coefficients a, b, and c are real numbers. Thus you want to show again that if p(z) = 0
for Eq. (57), then p(z ∗ ) = 0.
To get started, assume that z = a + ib is a complex number such that p(z) = 0 and then take the
complex conjugate of both sides to get [p(x)]∗ = 0∗ = 0. Now use the above relations Eqs. (43)–
(46) multiple times to express [p(z)]∗ in terms of z ∗ = a − ib.
This important result says that, for real polynomials and also for real-valued functions, complexvalued roots always come in pairs, z± = a ± ib. (The proof for some real function f (x) can be
obtained by considering a Taylor series of f (x) about x = 0, which looks like a polynomial of
infinite degree and your same argument works even in this infinite case, provided the Taylor series
is convergent.) This is an obvious result for a quadratic polynomial via the quadratic equation,
but your argument using complex conjugates is much more general and illustrates how scientists
and mathematicians work with complex numbers.
(c) Show that:
| cos(θ) + i sin(θ)| = 1
(58)
eix = cos(x) + i sin(x),
(59)
for any real angle θ. Thus Euler’s extremely useful and wonderful formula for x a real number3
implies that |eix | = 1 for any real x.
3 You can obtain Euler’s formula informally by substituting ix for x in the Taylor series expansion for the exponen2 + · · · about x = 0, by collecting real and imaginary parts of the series
tial function
ex = 1 + (1/1!)x +´(1/2!)x
`
`
´
eix = 1 − (1/2!)x2 + (1/4!)x4 − · · · + i x − (1/3!)x3 + (1/5!)x5 + · · · and by then recognizing the real and imaginary parts
are the Taylor series for cos(x) and sin(x) respectively.
9
(d) Further show that:
|ez | = eRe(z)
(60)
for a complex number z = a + ib.
(e) Use Euler’s formula Eq. (59) to show that
eix − e−ix
,
2i
eix + e−ix
cos(x) =
,
2
sin(x) =
(61)
(62)
and then use these relations to determine the values of sin(i) and cos(i).
Equations (61) and (62) provide a way to extend the sine and cosine functions to complex-valued
arguments.
(f) Use Euler’s formula Eq. (59) to derive de Moivre’s formula:
¡
¢α
cos(x) + i sin(x) = cos(αx) + i sin(αx),
(63)
for any real number α. Then use de Moivre’s formula Eq. (63) for the two cases α = 2 and α = 3 to
derive the trigonometric identities that express cos(2x) and sin(2x) in terms of cos(x) and sin(x),
and similarly for cos(3x) and sin(3x), by expanding the left side algebraically and equating real
and imaginary parts of both sides.
(g) A point (x, y) in the xy-coordinate plane can be thought of as corresponding to the complex
number z = x + iy. The same point can be described in polar coordinates (r, θ) where you should
know from a previous math course how to convert from polar to Cartesian coordinates like this
x = r cos(θ),
(64)
y = r sin(θ),
(65)
p
x2 + y 2 ,
³y´
θ = arctan
.
x
(66)
and from Cartesian to polar like this:
r=
(67)
With this information, use Euler’s formula Eq. (59) to show that an arbitrary complex
number z = a + ib can be written in the so-called polar form
z = reiθ ,
(68)
√
where r = |z| = a2 + b2 and θ = arctan(b/a) is always in radians (never in degrees!). The
angle θ in this polar form is also often called the phase or argument of the complex number z,
and there is a commonly used function arg(z), pronounced “arg of z” (for “argument of z”) that
has as its value the phase θ of a complex number z.
Note that while a given polar form reiθ corresponds to a unique complex number z = r cos(θ) +
ir sin(θ), there are infinitely many different polar forms that correspond to a given complex number
in the Cartesian form z = x + iy. The reason is that reiθ has the same value for different phases
θ + 2πn that all differ from θ by an integer multiple of 2π (since ei2πn = 1 for any integer n, which
you should verify). One common way to avoid this ambiguity is to choose the unique phase θ that
lies in the interval −π < θ ≤ π.
The polar form Eq. (68) is useful when one wants to emphasize the magnitude and direction of a
complex number, for example when one wants to interpret a complex number as a two-dimensional
vector in the xy plane. The polar form also helps one to understand the effect of multiplying one
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complex number by another. If one multiplies an arbitrary complex number x + iy by another
complex number of the form reiθ , the effect is to stretch the size of x + iy by the factor r, and to
rotate the vector (x, y) corresponding to x + iy by the angle θ counterclockwise around the origin.
In particular, since i = eiπ/2 , multiplying any nonzero complex number by i rotates the complex
number 90 degrees counterclockwise.
(h) Use Eq. (59) to find three of the infinitely many distinct values of ii (i raised to the i power),
and give their numerical values to three significant digits.
(i) Finally, graduate from bootcamp by showing me that you can express the
complex number
µ
¶1+2i
5 + 6i
,
3 − 4i
(69)
numerically in standard form a + ib and in polar form reiθ , i.e., give the values of a, b, r,
and θ (with θ in radians and separately in degrees) numerically to three significant digits. (To get
a unique polar form, choose all phases to be in the range −π < θ ≤ π.) To check your thinking,
you should get the values a = −0.038 and b = 0.018 to two significant digits
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