msc math complex variables
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Mathematical PhysicsMSc courseDr. Ali Abdulateef Kareem
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In the early days of modern mathematics, people were puzzled by
equations like this one:
For this reason, mathematicians dubbed an imaginary number. We
abbreviate this by writing i in its place, that is:
Now have solutions in terms of complex numbers, i.e
2 1 0x
1
1i
x i
Which have no solution in the real number system
It is useful to note that
2
2 4 2
2
3 2
1 3
2 3
1, , 1
1 1 1, 1,
i i i i
ii i i i
ii
i
i i
i i
i
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Functions of a complex variable provide us some powerfuland widely useful tools in in theoretical physics.
Some important physical quantities are complex variables(the wave-function )
Evaluating definite integrals.
Integral transforms
, i kx t x t A e
representing a wave travelling in the positivexdirection,
recall that: cos sinix
e ix x
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The familiar numbers, such as 1, 1/3, 2, and which arerepresented by points on a line, will be referred to as real numbers.
A complex number is an expression of the form
z x iy
where xandyare real numbers. The number xis called the real part
of zand is writtenRex z
The numbery, despite the fact that it is also a real number, is called
the imaginary part of zand is written
Imy z
If x= 0, then z= iy is a pure imaginary number.
Two complex numbers are equal if and only if their real parts are
equal and their imaginary parts are equal.
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A complex number (The sum of real and an imaginary umber) can beplotted on a plane with two perpendicular coordinate axes
The horizontal x-axis, called the real axis
The vertical y-axis, called the imaginary axis.
Each complex number z = x + iy corresponds to the point P(x, y)in the xy-plane.
r is the absolute value or modulus of z, then, is just the distancefrom the point P(x, y) to the origin.
2 2z r x y
*
2 2
( )( )zz x iy x iy
x y
Note that :
x
yP(x, y)
z r
Re
Im
O
,z z
z x iy
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There is a further interest and significance if we make use the usualpolar coordinates in the xy-plane for representation the complex
numbers which gives
x
yP(x, y)
z r
Re
Im
O
cos , sinx r y r
siny r
cosx r
2 2
1tan , if 0, tan
r x y
y yx
x x
2 2 2 2sin , cos
y x
x y x y
z x iy cos sinz r i or r cis
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From the complex number we have2 4z i
4 24 16 20, sin , cos
20 20z r
2 4z i
2 4z i The complex conjugate of
orz z 2 4z i
Then 20, 20z z
Therefore, =z z
In other words in the diagram, the
complex conjugate is the mirror image
of z in the real axis
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1 2 1 2
1 2 1 2
1 1
2 2
2
. .
(a)
(b)
(c)
(d)
(e)
(f)
z z z z
z z z z
z z
z z
z z
z z
z z z
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Example:Find the polar representation 1z i
Solution:
cos sinz r i
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Addition and Subtraction of Complex Numbers
The complex numbers satisfy the commutative and associativelaws
(10 4i) (5 2i)
= (10 5) + [4 (2)]i
= 5 2i
(4 6i) + (3 + 7i)
= [4 + (3)] + [6 + 7]i
= 1 + i
1 2 1 1 2 2
1 2 1 2 1 2
z z x iy x iy
z z x x i y y
1 1 1 2 2 2If and are two complex numbersthen
z x iy z x iy
1 2 2 1
1 2 3 1 2 3
z z z z
z z z z z z
Examples
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Multiplication of Complex Numbers
The product of and is defined as
1 2 1 1 2 2
1 2 1 2 1 2 2 1
( )( )
= ( ) ( ).
z z x iy x iy
x x y y i x y x y
Division of Complex NumbersThe division of and is defined as1z
2 2 1 12 2 2
1 1 1 1 1 1 1
x iy x iyz x iy
z x iy x iy x iy
2
2 1 2 1 2 1 2 1
2 2 2
1 1
x x x iy iy x i y y
x i y
2 1 2 1 2 1 2 12 2
1 1
x x y y y x x y i
x y
2 1 2 1 2 1 2 1
2 2 2 2
1 1 1 1
x x y y y x x yi
x y x y
Denominatorconjugate
2z
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4(a)
2 3
i
i
4 2 3 5 14 5 14
2 3 2 3 13 13 13
i i ii
i i
1 2 2(b)
3 4 5
i i
i i
1 2 3 4 2 5
3 4 3 4 5 5
5 10 5 10 2
25 25 5
i i i i
i i i i
i i
H.W.30 193
2 1
i i
i
Express each of the following complex numbers in polar form.
( ) 2 2 3 , ( ) 5 5 , ( ) 6 2 , ( ) 3a i b i c i d i
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We know that
find
1 1 1 1 2 2 2 2andcos sin cos sinz r i z r i 1
1 2
2
andz
z z
z
1 1 11
2 2 2 2
cos sin
cos sin
r iz
z r i
2 2
2 2
cos sin
cos sin
i
i
1 2 1 2 1 2 1 212 2
2 2 2
cos cos sin sin sin cos cos sin
cos sin
ir
r
1 1 1 2 1 22 2
cos sinz r
iz r
1 2 1 1 1 2 2 2cos sin cos sinz z r i r i 1 2 1 2 1 2 1 2 1 2cos cos sin sin sin cos cos sinr r i
1 2 1 2 1 2 1 2cos sinz z r r i
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Find the product of4(cos50 sin 50 ) and 2(cos10 sin10 ).i i
4(cos50 isin50) 2(cos10 isin10 )
4 2 cos(50 10) isin(50 10)
8(cos60 isin60)
8 1
2 i
3
2
4 4i 3
Find the division of 16(cos70 sin 70 ) 16= cos(70 40 ) sin(70 40 )4(cos40 sin 40 ) 44cos30 sin30
3 14 2 3 2
2 2
i ii
i
i i
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We know 1 2 1 2 1 2 1 2cos sinz z r r i
If we have a set of n complex numbers, a generalization of the
above equation leads to:
1 2 1 2 1 2 1 2cos sinn n n nz z z r r r i
putting
1 2 1 2
1 and givesn n
r r r
cos sin cos sinnnz i n i n
1 1cos sin cos sin .nn n
z r i r n i n
DeMoivres
theorem
n
i nie e The results is equivalent to the statement
In general
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First, find trigonometric notation for (
1
i) using
Theorem 1 2 cos225 sin 225i i
5
5
5
1 2 cos 225 sin 225
using De Moiver's theorem we get
cos( ) sin( )
4 2 cos112
2 5 225 5 22
5 sin1125
2 24 2
5
2 2
4 4
i i
i
i
i
i
52 , 225
4r
cos sinz r i
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2 3 4
12! 3! 4!
x x x xe x Substitute ixforx.
2 3 4 5 61
2! 3! 4! 5! 6!
ix ix ix ix ix ixe ix
2 2 3 3 4 4 5 5 6 6
12! 3! 4! 5! 6!
ix i x i x i x i x i xe ix
2 3 4 5 6
12! 3! 4! 5! 6!
ix x ix x ix xe ix
2 4 6 3 5
12! 4! 6! 3! 5!
xi x x x x xe i x
Factor out the iterms.
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2 4 6 3 5
12! 4! 6! 3! 5!
ix x x x x xe i x
This is the series for
cosine.
This is the series
for sine.
cos sinz r i or r cis
cos sinix
e ix x Eulers Formula
Using EulerMs formula, the polar form of a complex number canbe rewritten as :
(cos sin )
i
z r i x y
z r e
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Show that
(a) (b)
cos 2
i ie e
sin
2
i ie e
i
Solution
we have cos sin and cos sini ie i e i
(a) Adding (1) and (2)
2cos or cos2
i ii i e ee e
(a) subtracting (2) and (1)
2 sin or sin2
i i
i i e ee e ii
H.W: Prove
(a) (b)33 1
sin sin sin 3 ,4 4
41 1 3
sin cos 4 cos 2 .8 2 8
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If , then to each value of wthere correspondsone value of z. Conversely, to a given there correspond
precisely n distinct values of w. Each of these values is called annth rootof z, and we write
( 1, 2, )nz w n 0z
nw z
Hence this symbol is multivalued, namely, n-valued. The n
values of can be obtained as follows.n z
andcos sin cos sinz r i w R i Then the equation becomes, by De Moivresformulanw z
cos sin nnw R i cos sin cos sinnR n i n z r i
We write zand win polar form
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i.e. cos sin cos sinnR n i n r i
Form the Equating of these equations, The modulus
n nR r R r
and the arguments
thu2
2 , s k
n kn n
where kis an integer. For we get ndistinct values
of w. Further integers of kwould give values already obtained. For
instance,
0, 1, 2, ..., 1k n
gives 2 / 2k n k n
Re
Im
4
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hence the wcorresponding to k = 0, etc. Consequently,
for , has the n distinct values
n z
0z
2 2cos sinn n
k kz r i
n n
where . These n values lie on a circle of radius
with center at the origin and constitute the vertices of
polygon of n sides.
0, 1, 2, ..., 1k n n r
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Find the square roots of
Trigonometric notation:
For k= 0, root is
For k= 1, root is
1 3i
1 3 2 cos60 sin 60i i
11
2260 360 60 360
2 cos60 sin 60 2 cos sin2 2 2 2
2 cos 30 180 sin 30 180
i k i k
k i k
2 cos30 isin30
2 cos210 isin210
2 2cos sinn n
k kz r i
n n
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Find all fourth roots of Write the roots inrectangular form.
Write in trigonometric form.
Here r= 16 and = 120. The fourth roots of this number haveabsolute value
8 8 3.i
8 8 3 16 cis 120i
4
16 2.120 360
30 904 4
kk
0 30 90 30
1 30 90 120
2 30 90 210
3 30 90
0
1
2
3 300
k
k
k
k
There are four fourth roots, let k= 0, 1, 2 and 3.
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Using these angles, the fourth roots are:
2 cis 30 , 2 cis 120 ,
2 cis 210 , 2 cis 300
written in rectangular form
3
1 3
3
1 3
i
i
i
i
The graphs of the roots areall on a circle that has center
at the origin and radius 2.
13
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Find each of the indicated roots and locate themgraphically.
1 41 3
1 , ( ) 2 3 2a i b i
Find the square roots of 15 8i
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When dealing with integrals that have complex numbers wetreat the imaginary number ias a constant
Example: The integral 1 i xe dx
Let 1
then 11
u i x
dudu i dx dx
i
Making the substitution yields
11
.1 1 1
i xuu e ee du
i i i
Removing the complex denominator by multiplying it by the conjugate
gives
111 1
1 1 2
i xi xe i i
ei i
the constant will be a complex number as well thus 1 2c c ic 1 1
1 2
1
2
i x i xi
e dx e c ic
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Some real-valued integrals can be solved using complex-valued
integrals.Integration such as and canbe solved using integration by parts.
Using the integral we can solve integration easily
cosxe x dx sinxe x dx 1 i x
e dx
rearrangement of the integral and its value 1 i xe dx 1 1 212
i xie c ic
cos sin
cos sin
x ix
x
x x
e e dx
e x i x dx
e x dx i e x dx
Firstly, rearranging and applying Euler's Formula gives 1 i xe dx
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Rearranging the value of the integral yields 1 1 21
2
i xie c ic
11 21
2 2i xi e c ic
1 2
1 2
1 2
1 2
1 2
1
1cos sin
2 2
1cos sin cos sin2 2
1 1cos sin cos sin
2 2
1 1 1 1cos sin cos sin
2 2 2 21 1
cos sin sin cos2 2
1 1cos sin
2
x
x x
x x
x x x x
x x
x
ie x i x c ic
ie x i x e x i x c ic
e x i x e i x x c ic
e x i e x i e x e x c ic
e x x c i e x x ic
e x x c i
2sin cos
2
xe x x c
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Since the integral and its value
rearranged, are equivalent, it means that the real parts areequal to each other, as well as the imaginary parts
cos sinx x
e x dx i e x dx
Therefore
1cos cos sin
21
sin sin cos2
x x
x x
e x dx e x x c
e x dx e x x c
cos sinx xe x dx i e x dx 1 2
1 1cos sin sin cos
2 2
x xe x x c i e x x c
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