© Diola Bagayoko (2012)

APPLICATIONS OF COMPLEX VARIABLES

The integrals evaluated below, together, establish that the (1) Cos (nx), n=1 - ∞ are orthogonal. Similarly for sin(nx), n =0 - ∞ . These properties of sine and cosine functions are utilized in the Fourier expansion of a function.

I. Examples of the Evaluation of Integrals

Evaluate the integral ∫=π2

03 .)cos()cos( dxmxnxI

( ) ( ) dxeeeedxmxnxIimximxinxinx

⎥⎦

⎤⎢⎣

⎡ +⎥⎦

⎤⎢⎣

⎡ +==

−−

∫ ∫ 22)cos()cos(

2

0

2

03

π π(1)

[ ]dxeeeeI xmnixnmixmnixmni∫ −−−−+ +++=π2

0

)()()()(3 4

1(2)

For ( ) ,or 0)( and 0 mnn-mmn ≠≠≠+π2

0

)()()()(

3 )()()()(41

⎥⎦

⎤⎢⎣

⎡−−

+−

+−

++

=−−−−+

mnie

nmie

mnie

mnieI

xmnixnmixmnixmni

(3)

a) For ,mn ≠ using ,102 == eeik π one has 03 =I from the line above.

b) For ,0≠= mn clearly one must use line (2) as line (3) becomes meaningless. So

[ ]dxeeI nxinxi∫ −++=π2

0

223 2

41

or ππ

=⇒⎥⎦

⎤⎢⎣

⎡−

++=−

3

2

0

22

3 222

41 I

nie

niexI

nxinxi

c) for ,0== mn use line (2), as line (3) is meaningless in that situation:

[ ] [ ] πππ24

414

41 2

0

2

03 === ∫ xdxI

Evaluate the integral ∫=π2

04 .)cos()sin( dxmxnxI

( ) ( ) dxeeieedxmxnxI

imximxinxinx

∫ ∫ ⎥⎦

⎤⎢⎣

⎡ +−==

−−π π2

0

2

04 22)cos()sin( (1)

[ ]dxeeeei

I xmnixnmixmnixmni∫ −−−−+ −−+=π2

0

)()()()(4 4

1(2)

For mn ≠ [which also means (n+m)≠ 0 and (-n-m) ≠ 0]

( )

π2

0

)()()()(

4 )()()(41

⎥⎦

⎤⎢⎣

⎡−−

−−

−−

++

=−−−−+

mnie

nmie

mnie

mnie

iI

xmnixnmixmnixmni

(3)

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© Diola Bagayoko (2012)

a) So, for mn ≠ the above line gives I4 = 0. b) For 0≠= mn , from line (2) we have

I4 = 0. c) for n=m=0, sin(nx)=0 and the integrand is identically zero and so is the integral.

Evaluate the integral ∫=π2

05 .)sin()sin( dxmxnxI

( ) ( ) dxiee

ieedxmxnxI

imximxinxinx

∫ ∫ ⎥⎦

⎤⎢⎣

⎡ −−==

−−π π2

0

2

05 22)sin()sin( (1)

[ ] dxeeeeI xmnixnmixmnixmni∫ −−−−+ +−−−=π2

0

)()()()(5 4

1(2)

For mn ≠ (which also means n+m and -n-m are different from zero),

π2

0

)()()()(

5 )()()()(41

⎥⎦

⎤⎢⎣

⎡−−

+−

−−

−+

−=−−−−+

mnie

nmie

mnie

mnieI

xmnixnmixmnixmni

(3)

a) So, for ,mn ≠ the above line gives I5 = 0

b) For ,0≠= mn we must go back to line (2) and get:

[ ]π

π2

0

2

0

)2(2)2(2

5 )2(22

412

41∫ ⎥

⎦

⎤⎢⎣

⎡−

++−−=+−−=−

−

nie

niexdxeeI

xninxixninxi

.0for ,5 ≠== mnI π

c) For n = m = 0, from line (1) or line (2) shows that the integrand is identically zero

(i.e.. is zero for any value of x), then so is the integral and we have I5 = 0

NOTE: The above results hold for sine and cosine functions for integration from a to (a +

2 ),π i.e., ∫+ π2a

a, even though in the above example we have a = 0. These results, for instance,

are those one obtains for ∫−π

π. Recall that the period of sine and cosine functions is π2=T .

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© Diola Bagayoko (2012)

So, the above results hold for any integration from a to ..,.),( ∫+

+Ta

aeiTa These properties

are the basis for Fourier expansions of periodic functions satisfying the Dirichlet conditions. .

Evaluate the integral dxbxeI ax )cos(1 ∫= and ∫= .)sin(2 dxbxeI ax

[ ] [ ]221

)sin()cos(Re)cos(ba

bxbbxaedxeedxbxeIax

ibxaxax

++

=== ∫∫

[ ] [ ]∫ ∫ +

−=== 222

)cos()sin(Im)sin(ba

bxbbxedxeedxbxeIax

ibxaxax

With the detailed examples for Integrals 3 through 5 above, the reader should quickly carry out

the needed operations to obtain the results shown above.

II. Illustrative Properties of Elementary Functions of Complex Variables

Show that ,sinhcoscoshsinsin yxiyxz += where .iyxz +=

,2

)sin(sinieeiyxz

iziz −−=+= where yixiz −= .

So, i

xixexixeiee

zyyyixyix

2)sin(cos)sin(cos

2sin

−−+=

−=

−+−− and

yxiyxxiieex

ieez

yyyy

sinhcoscoshsinsin2

cos2

sin +=+

+−

=−−

.

Prove that .cossin2)2sin( zzz = (This relation is the same for real angles.)

( ) ( ) [ ]ziziiziziziz

eei

eeieezz 22 11

21

222cossin2 −

−−

−−+=+−

=

or ( ) zieezz

zizi

2sin2

cossin222

=−

=−

by definition.

Prove that zzz 22 sincos2cos −=

3

© Diola Bagayoko (2012)

42

2cos

2222

ziziiziz eeeez−− ++

=⎟⎟⎠

⎞⎜⎜⎝

⎛ +=

42

42

2sin

222222

ziziziziiziz eeeeieez

−−− −+−=

+−=⎟⎟

⎠

⎞⎜⎜⎝

⎛ −=

,2cos2

sincos22

22 zeezzzizi

=+

=−−

by definition.

Prove that .sinsinhcoscoshcosh iyxzwithyxiyxz +=+=

22

)cosh(coshiyxiyxzz eeeeeeiyxz

−−− +=

+=+=

[ ] ( ) ( )[ ]yiyeyiye xx

−+−++=−

sincos2

sincos2

yeeiyee xxxx

sin2

cos2

−− −+

+=

or yxiyxz sin)sinh(cos)cosh(cosh += .

For 32 −== yandx , we get

[ ][ ] )3sin()2sinh()32cosh(Im

)3cos()2cosh()32cosh(Re−=−

=−ii

and ( ) ( ) ( ) ( ) ( )3sin2sinh3cos2cosh32cosh 2222 +=− i

NOTE: We used, without comments, the following trivial identities:

( ) ( ) ( ) ( ) .1,sinsin,coscos ii

xxxx −=−=−−=

III. Many More Applications

You are urged to consult the textbook (Mathematical Physics for Scientists and Engineers byMary L. Boas) and other sources to gain further appreciation of the extensive applications ofcomplex numbers and of functions of a complex variable in Mathematics, Physics, Engineering,and related fields. The Cauchy-Goursat theorem, the residue theorem and its applications to theevaluations of integrals, and the utilization of ordinary expressions in conformal mapping are just

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© Diola Bagayoko (2012)

a few of the applications to be covered in Physics 411, Mathematical Physics II (or AdvancedMathematical Physics). Let me reiterate, one more time, that a total mastery of the elementaryfunctions and properties introduced in Physics 311 is necessary for an understanding of mostapplications of complex numbers and of functions of a complex variables.

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