Lecture 3 Applications of de Moivre’s theorem Curves … 3 • Applications of de Moivre’s...

Lecture 3

• Applications of de Moivre’s theorem

• Curves in the complex plane

• Complex functions as mappings

Uses of de Moivre and complex exponentials

(cos θ + i sin θ)n = cosnθ + i sinnθ

• We saw application to trigonometric identities,functional relations for trig. and hyperb. fctns.

• We next see examples of two more kindsof applications:

♦ solution of differential equations

♦ summation of series


Ex. 2 Solving differential equations

d2 y

d 2!

+ y " Im( d2 z

d 2!

+ z) = 0, v cos sinA B! != +

z = Cei!

where C = A+ ıB is a complex constant.

The solution to d2z

d2

!

+ z = 0 is simply given by

y = Im(z) = Im((A+ ıB)(cos! + ısin!))

= Asin! + Bcos!

z = x + i y

c.f.

0 1r< <20

(1 )sinsin(2 1) ,

1 2 cos 2

n

n

rr n

r r

!!

!

"

=

++ =

# +$


Ex. 3 Summing series

e.g. Show

n=0

!

" rn sin(2n +1)# = Im

n

" rn(eı(2n+1)# ) = Im(eı#

n

" (re2ı# )n )

= Im(eı!1

1" re2ı!)

= Im(eı! (1" re"2ı! )

(1" re2ı! )(1" re"2ı! ))

=sin! + r sin!

1" 2r cos2! + r2

Homework

(Question from Maths Collection Paper 2006)

♦ Find Re and Im of 1/(1− z), where z = reiθ, and hence

sum the series∑

∞

n=0rn cos(nθ) for r < 1.

Answ. :1

1− z=

1− z∗

|1− z|2=

1− r cos θ

1 + r2 − 2r cos θ︸︷︷︸

Re

+ ir sin θ

1 + r2 − 2r cos θ︸︷︷︸

Im

∞∑

n=0

rn cos(nθ) =∞∑

n=0

rnRe einθ = Re∞∑

n=0

(reiθ)n

= Re1

1− reiθ=

1− r cos θ

1 + r2 − 2r cos θ

♦ Show that 1 + cos θ2

+ cos 2θ4

+ cos 3θ8

+ . . . = 4−2 cos θ5−4 cos θ

.

Answ. :∞∑

n=0

cos(nθ)

2n=

∞∑

n=0

1

2nRe einθ = Re

∞∑

n=0

(eiθ

2

)n

= Re1

1− eiθ/2=

1− cos θ/2

1 + 1/4− cos θ=

4− 2 cos θ

5− 4 cos θ

CURVES IN THE COMPLEX PLANE

• locus of points satisfying constraint in complex variable z

Examples

Re z

Im z

Re z

z − z*

|z| = 1

arg z = π / 3

1

Im z

i/2= i

Im z

Re z

Curves in the complex plane

1z| |=

2 2

0 0 ( ) ( ) 1a a b b! + ! =

Circle centre z0, radius 1

1, any i

z re r!

!= " =

Ex 1

01z z| ! |=

Circle centre (0,0), radius 1

(a)2+ (b)2

= 1, z = a + ibor


Ex 2 | z! ız+ ı

|= 1

| | | |z i z i! = +

Distance from (0,1) = distance from (0,-1) Real axis

2 2 2 2 (a) ( 1)or (a) ( 1)b b+ ! = + +

0, arbitraryb a=

z = a + ib

Another way to define a curve:

γ : [a, b]→C

γ : t 7→ z = γ(t) = x(t) + iy(t)

Im z

Im z

Re z

R

Re z

Im z

Re z

x(t) = x

y(t) = y

0 < t < 2

0 + R cos t

0 + R sin t

π

x(t) = x 0 + R1 cos t

y(t) = y0 + R 2 sin t

y(t) = t sin t

x(t) = t cos t

i ti.e., z = t e

Homework(Maths Paper 2007)

Sketch the curve in the complex plane

z = ateibt , −6 ≤ t ≤ 6

where a and b are real constants given by a = 1.1, b = π/3.

− 6

Im z

Re z1

1.16

1.1

Ex 3 1 4arg( )z

z

!

+=


i.e. arg(z) ! arg(z +1) = "

4

2 21 1 12 2 2

( ) ( )a b+ + ! =

2( 1)b a a b= + +

tan(A! B) =tan A! tan B

1+ tan A tan B

1

1

11 .

b b

a a

b b

a a

+

+

!=

+2

( 1)

( 1)

b a ba

a a b

+ !=

+ +

Take tan of both sides :

z = a + ib

…but BEWARE…not all of circle satisfies equation…

1 4arg( )z

z

!

+=

z

z +1=

z

z +1.z *+1

z*+1

=z(z* +1)

z +12

=(a + ib)(a +1! ib)

z +12

=a(a +1) + b2 + ib

z +12

! b > 0 since argz

z +1

"#$

%&'

positive

b < 0 solution introduced by tangent ambiguity( )

(a + 1

2)2 + (b ! 1

2)2 = 1

2, b > 0

Alternative solution

Solution : portion of circle through (0,0) and (-1,0)

Circle centre (-1/2,1/2) and radius 1/ 2

1 4arg( )z

z

!

+=

i.e. arg(z) ! arg(z +1) = "

4

31 4

arg( )zz

!

+= "

The lower portion of the circle is given by :

Homework(Question from Maths Collection Paper 2009)

Sketch the locus of points in the complex plane for

(a) Im(z2 + i) = −1 , (b) arg(z2 + i) = π/2

Answ. :

2= =

=

Im z

1 Re z

Im z

Re z

y = − 1 / x

1 + 2 x y = − 1 x y = − 1

θ = arctg [(1 + 2 x y) / (x2

− y2)] = π / 2

x2

= y2 x = y

z = x + i y ; z2= x − y

2 2+ 2 i x y

x = − y , 1 − 2 x

(b)

(a)

> 0

Summary on Curves in the Complex Plane

• locus of points satisfying constraint in complex variable z

♠ Curves can be defined

⊲ either directly by constraint in complex variable z

ex. : |z| = R

⊲ or by parametric form γ(t) = x(t) + iy(t)

ex. : x(t) = R cos t

y(t) = R sin t , 0 ≤ t < 2π

♦ intersection of algebraic and geometric approaches

COMPLEX FUNCTIONS AS MAPPINGS

f : z 7→ w = f(z)

• Complex functions can be viewed as mapping sets of points in C (e.g. curves,

regions) into other sets of points in C.

Example:

w = f(z) = ez . Set z = x+ iy ; w = ρeiφ . Then ρ = ex ;φ = y .

lines x = af

−→ circles ρ = ea

lines y = bf

−→ rays φ = b

1

��

��

Im z

Re w

f

Im w

b

b

b2

b1

1

2

Re za a2

• For any subset A of C, the “image of A through f” is the set of points w

such that w = f(z) for some z belonging to A, and is denoted f(A).

• f maps A on to f(A)

Examples

• Let f(z) = 1/z. What does f do to the interior and exterior of theunit circle |z| = 1? What does f do to points on the unit circle?

Answ. : w = f(z) =1

z; z = reiθ ⇒ w =

1

re−iθ

1

Im z

Re z

f

Im w

Re w1

⊲ f maps the interior of the unit circle on to the exterior, and viceversa.

Unit circle is mapped on to itself.

• Let f(z) = (z − 1)/(z + 1). What is the image through f of theimaginary axis? [Answ.: unit circle]

And of the right and left half planes? [Answ.: interior and exterior of

unit circle]

• Let f(z) = ln z (principal branch). What is the image through fof the upper half plane? [Answ.: horizontal strip between 0 and iπ]

Lecture 3 Applications of de Moivre’s theorem Curves … 3 • Applications of de Moivre’s...

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Transcript of Lecture 3 Applications of de Moivre’s theorem Curves … 3 • Applications of de Moivre’s...