Lecture 3
• Applications of de Moivre’s theorem
• Curves in the complex plane
• Complex functions as mappings
Uses of de Moivre and complex exponentials
n
(cos θ + i sin θ) = cos nθ + i sin nθ
• We saw application to trigonometric identities,
functional relations for trig. and hyperb. fctns.
• We next see examples of two more kinds
of applications:
♦ solution of differential equations
♦ summation of series
Uses of de Moivre and complex exponentials
Ex. 2
Solving differential equations
The solution to
z = Ce i!
d2z
d! 2
+ z = 0 is simply given by
where C = A + ıB is a complex constant.
z = x + iy
y = Im(z) = Im(( A + ıB)(cos! + ısin ! ))
= Asin ! + Bcos!
c.f.
d2 y
d!
2
d2z
+ y " Im( d 2 + z) = 0, v = A cos ! + B sin !
!
Uses of de Moivre and complex exponentials
Ex. 3 Summing series
"
$r
e.g. Show
n
sin(2n + 1)! =
n =0
(1 + r ) sin !
,
1 # 2r cos 2! + r 2
0 < r <1
!
"r
n
sin(2n + 1)# = Im " r n (eı(2n+1)# ) = Im(eı# " (re 2ı# ) n )
n
n=0
= Im(eı!
n
1
)
2ı!
1 " re
eı! (1 " re "2ı! )
= Im(
)
(1 " re 2ı! )(1 " re "2ı! )
=
sin ! + r sin !
1 " 2r cos 2! + r 2
Homework
(Question from Maths Collection Paper 2006)
♦ Find Re and Im of 1/(1 − z), where z = reiθ , and hence
P∞
sum the series n=0 r n cos(nθ) for r < 1.
1
r sin θ
1 − z∗
1 − r cos θ
Answ. :
+
i
=
=
1−z
|1 − z|2
1 + r2 − 2r cos θ
1 + r2 − 2r cos θ
{z
}
{z
}
|
|
Re
∞
X
n
r cos(nθ) =
n=0
∞
X
r Re e
inθ
= Re
n=0
= Re
♦ Show that 1 +
n
Im
cos θ
2
+
∞
X
(reiθ )n
n=0
1
1 − r cos θ
=
1 − reiθ
1 + r2 − 2r cos θ
cos 2θ
4
+
cos 3θ
8
+ ... =
4−2 cos θ
5−4 cos θ
.
∞
∞ iθ n
∞
X
X
X
1
e
cos(nθ)
inθ
=
Re
e
=
Re
Answ. :
2n
2n
2
n=0
n=0
n=0
= Re
1 − cos θ/2
4 − 2 cos θ
1
=
=
1 − eiθ /2
1 + 1/4 − cos θ
5 − 4 cos θ
CURVES IN THE COMPLEX PLANE
• locus of points satisfying constraint in complex variable z
Examples
Im z
1
|z| = 1
Re z
Im z
z − z* = i
i/2
Re z
Im z
arg z = π / 3
Re z
Curves in the complex plane
Ex 1
| z |= 1
z = re
or
i!
" r = 1, any !
(a)2 + (b)2 = 1,
Circle centre (0,0), radius 1
| z ! z0 |= 1
Circle centre z0, radius 1
(a ! a0 ) 2 + (b ! b0 ) 2 = 1
z = a + ib
Curves in the complex plane
Ex 2
|
z !ı
z +ı
|= 1
| z ! i |=| z + i |
Distance from (0,1) = distance from (0,-1)
or (a) 2 + (b ! 1) 2 = (a) 2 + (b + 1) 2
b = 0, a arbitrary
Real axis
z = a + ib
Another way to define a curve:
γ : [a, b] → C
γ : t 7→ z = γ(t) = x(t) + iy(t)
R
Im z
x(t) = x
0 + R cos t
y(t) = y + R sin t
0
0<t<2 π
Re z
Im z
x(t) = x 0 + R cos t
1
y(t) = y0 + R sin t
2
Re z
Im z
x(t) = t cos t
y(t) = t sin t
i.e., z = t e i t
Re z
Homework
(Maths Paper 2007)
Sketch the curve in the complex plane
z = at eibt ,
−6 ≤ t ≤ 6
where a and b are real constants given by a = 1.1, b = π/3.
Im z
1.1
1
−6
1.1
6
Re z
Curves in the complex plane
Ex 3
arg( z +z 1 ) = !4
i.e. arg(z) ! arg(z + 1) =
Take tan of both sides :
tan( A ! B) =
tan A ! tan B
1 + tan Atan B
b(a + 1) ! ba
! ab+1
=1 =
b
b
a (a + 1) + b 2
1 + a . a +1
b
a
b = a (a + 1) + b 2
(a + 12 ) 2 + (b ! 12 ) 2 =
1
2
…but BEWARE…not all of circle satisfies equation…
z = a + ib
"
4
arg( z +z 1 ) = !4
z
z z * +1 z(z * + 1) (a + ib)(a + 1 ! ib) a(a + 1) + b2 + ib
=
. *
=
=
=
2
2
2
z +1 z +1 z +1
z +1
z +1
z +1
" z %
positive
! b > 0 since arg $
# z + 1 '&
(a + 12 )2 + (b ! 12 )2 = 12 , b > 0
(b < 0 solution introduced by tangent ambiguity )
Alternative solution
arg( z +z 1 ) = !4
i.e. arg(z) ! arg(z + 1) =
Solution : portion of circle through (0,0) and (-1,0)
Circle centre (-1/2,1/2) and radius 1/ 2
"
4
The lower portion of the circle is given by :
arg( z +z 1 ) = " 34!
Homework
(Question from Maths Collection Paper 2009)
Sketch the locus of points in the complex plane for
(a) Im(z 2 + i) = −1
(b) arg(z 2 + i) = π/2
,
Answ. :
2
2 2
z=x+iy;z =x −y +2ixy
1+2xy=−1=
(a)
Im z
1 Re z
xy=−1
y=−1/x
2
2
θ = arctg [(1 + 2 x y) / (x − y )] = π / 2
Im z
Re z
(b)
=
2
2
x =y
=
x=y
2
x=−y,1−2x >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 .
Im z
f
b2
f
b1
lines x = a −→ circles ρ = ea
lines y = b −→ rays φ = b
Im w
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a
1
a2
b
2
b
1
f
Re z
Re w
• 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 the
unit circle |z| = 1? What does f do to points on the unit circle?
Answ. :
w = f (z) =
Im z
1
;
z
z = reiθ ⇒ w =
Im w
1 −iθ
e
r
f
1
Re z
1
Re w
⊲ 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 the
imaginary 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 f
of the upper half plane?
[Answ.: horizontal strip between 0 and iπ]