Download TMA4115 - Calculus 3 - Lecture 3, Jan 23

TMA4115 - Calculus 3
Lecture 3, Jan 23
Toke Meier Carlsen
Norwegian University of Science and Technology
Spring 2013
www.ntnu.no
TMA4115 - Calculus 3, Lecture 3, Jan 23
Review of last week
Last week we
introduced complex numbers, both in a geometric way
and in an algebraic way,
defined Re(z), Im(z), |z| and arg(z) for a complex
number z,
defined addition and multiplication of complex numbers,
defined complex conjugation,
introduced polar representation of complex numbers,
computed powers of complex numbers,
defined and computed roots of complex numbers.
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TMA4115 - Calculus 3, Lecture 3, Jan 23, page 2
Today’s lecture
Today we shall
use complex numbers to solve polynomial equations,
look at the fundamental theorem of algebra,
introduce the complex exponential function,
and study extensions of trigonometric functions to the
complex numbers.
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TMA4115 - Calculus 3, Lecture 3, Jan 23, page 3
Solutions to second degree
equations
If a, b, c are complex numbers and a 6= 0, then the
√ solutions
b2 − 4ac
−b
±
to the equation az 2 + bz + c = 0 are z =
.
2a
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TMA4115 - Calculus 3, Lecture 3, Jan 23, page 4
Example
Let us solve the equation z 2 + 4iz − 4 − 4i = 0.
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TMA4115 - Calculus 3, Lecture 3, Jan 23, page 5
Solution
p
(4i)2 − 4(−4 − 4i)
, so we
2
must compute the square roots of
(4i)2 − 4(−4 − 4i) = −16 + 16 + 16i = 16i.
Since |16i| = 16 and Arg(16i) = π2 , the two square roots of
16i are
1
1
w0 = 4 cos(π/4) + i sin(π/4)) = 4 √ + i √
2
2
√
√
= 2 2 + i2 2
−1
−1
w1 = 4 cos(5π/4) + i sin(5π/4)) = 4 √ + i √
2
2
√
√
= −2 2 − i2 2
The solutions are z =
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−4i ±
TMA4115 - Calculus 3, Lecture 3, Jan 23, page 6
Solution
So the two solutions of z 2 + 4iz − 4 − 4i = 0 are
p
√
−4i ± (4i)2 − 4(−4 − 4i)
−4i ± 16i
z=
=
2
2
(
√
√
√
√
2 + ( 2 − 2)i
−4i ± (2 2 + i2 2)
√
√
=
=
2
− 2 − ( 2 + 2)i.
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TMA4115 - Calculus 3, Lecture 3, Jan 23, page 7
Problem 1 from the exam from
June 2012
√
Solve w 2 = (−1 + i 3)/2. Find all solutions of the equation
z 4 + z 2 + 1 = 0 and draw them in the complex plane. Write
the solutions in the form x + iy .
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TMA4115 - Calculus 3, Lecture 3, Jan 23, page 8
Solution
√
p
√
|(−1 + i 3)/2| = (−1/2)2 + ( 3/2)2 = 1/4 + 3/4 = 1
√
, so the
and Arg((−1 + i 3)/2) = arccos(−1/2) = 2π
3
√
2
solutions of w = (−1 + i 3)/2 are
√
3
1
w0 = cos(π/3) + i sin(π/3) = + i
,
2
2 √
3
1
w1 = cos(4π/3) + i sin(4π/3) = − − i
.
2
2
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q
TMA4115 - Calculus 3, Lecture 3, Jan 23, page 9
Solution
√
√
z 4 + z 2 + 1 = 0 if and only if z 2 = −1±2 −3 = −1±i2 3 . It follows
√
2
only if
from the
first √part
that z = (−1 + i 3)/2 if and
√
1
3
2
z = ± 2 + i 2 . We have that z = (−1 − i 3)/2 if and
√
only if (z)2 = (−1 + i 3)/2, and it follows from
part
the first
√ √
that (z)2 = (−1 + i 3)/2 if and only if z = ± 12 + i 23 .
So thefour solutions
of the equation z 4 + z 2 + 1 = 0 are
√ z = ± 12 ± i 23 .
z1
z0
z2
z3
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TMA4115 - Calculus 3, Lecture 3, Jan 23, page 10
Roots of polynomials
If P(z) is a polynomial, then a solution to the equation
P(z) = 0 is called a root (or zero) of P(z).
z0 is a root of P(z) if and only if (z − z0 ) is a factor of P(z)
(i.e., if P(z) = (z − z0 )Q(z) for some polynomial Q(z)).
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TMA4115 - Calculus 3, Lecture 3, Jan 23, page 11
The fundamental theorem of
algebra
Every complex polynomial of degree 1 or higher has a least
one complex root.
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TMA4115 - Calculus 3, Lecture 3, Jan 23, page 12
Roots of real polynomials
If w is aProot of a real polynomial
root of nk=0 ak z k .
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Pn
k =0
ak z k , then w is also a
TMA4115 - Calculus 3, Lecture 3, Jan 23, page 13
Exercise 34 on page xxvii
√
Check that z1 = 1 − 3i is a zero of
P(z) = z 4 − 4z 3 + 12z 2 − 16z + 16, and find all the zeros of
P.
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TMA4115 - Calculus 3, Lecture 3, Jan 23, page 14
Solution
√
√
√
√
√
z12 = (1 − 3i)(1
−
3i)
=
1
−
3i
−
3i
−
3
=
−2
−
2
3i,
√
√
√
√
3
z1 = (−2 − 2 3i)(1
√+ 2 3i − 2 3i − 6 = −8,
√ − 3i) = −2
and z14 = −8(1 − 3i) = −8 + 8 3i, so
z14 − 4z13 + 12z12 − 16z1 + 16
√
√
√
= −8 + 8 3i + 32 − 24 − 24 3i − 16 + 16 3i + 16 = 0.
Since z1 is a zero of
√ P and P has real coefficients, it follows
that also z1 = 1 + 3i is a zero of P.
It follows that (z − z1 )(z − z1 ) = z 2 − (z1 + z1 ) + z1 z1 =
z 2 − 2 Re(z1 )z + |z1 |2 = z 2 − 2z + 4 is a factor of P(z).
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TMA4115 - Calculus 3, Lecture 3, Jan 23, page 15
Solution
By doing polynomial division, we see that
P(z) = z 4 −4z 3 +12z 2 −16z +16 = (z 2 −2z +4)(z 2 −2z +4),
so z1 and z1 are the only zeros of P.
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TMA4115 - Calculus 3, Lecture 3, Jan 23, page 16
Complex functions
A complex function f is a rule that assigns a unique complex
number f (z) to each number z in some set of complex
numbers (called the domain of f ).
Examples of complex functions
f (z) = Re(z)
g(z) = Im(z)
h(z) = |z|
j(z) = Arg(z)
k (z) = z
p(z) = z 2 − 4z + 6
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TMA4115 - Calculus 3, Lecture 3, Jan 23, page 17
Graphic representations of
complex functions
We cannot draw the graph of a complex function since
we would need 4 dimensions to do that.
Instead, we can graphically represent the behavior of a
complex function w = f (z) by drawing the z-plane and
the w-plane separately, and showing the image in the
w-plane of certain, appropriately chosen set of points in
the z-plane.
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TMA4115 - Calculus 3, Lecture 3, Jan 23, page 18
Example
Consider the function f (z) = −2iz.
Arg(−2i) = −π/2 and | − 2i| = 2, so f maps the region
1/2 ≤ |z| ≤ 1, 0 ≤ arg(z) ≤ π/2 to the region 1 ≤ |z| ≤ 2,
−π/2 ≤ arg(z) ≤ 0.
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TMA4115 - Calculus 3, Lecture 3, Jan 23, page 19
Example
Consider the function g(z) = z 2 .
g maps the region 0 ≤ |z| ≤ 1/2, π/2 ≤ arg(z) ≤ π to the
region 0 ≤ |z| ≤ 1/4, π ≤ arg(z) ≤ 2π.
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TMA4115 - Calculus 3, Lecture 3, Jan 23, page 20
Complex functions
Limits, continuity and differentiability of complex functions
can be defined just as for real functions.
Examples of complex functions
Every complex polynomial is differentiable, and hence
continuous.
The functions f (z) = Re(z), g(z) = Im(z), h(z) = |z| and
k (z) = z are continuous, but not differentiable.
The function j(z) = Arg(z) is not continuous.
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TMA4115 - Calculus 3, Lecture 3, Jan 23, page 21
The exponential function
One can show that the series
∞
X
zn
n=0
n!
converges absolutely for
all complex numbers z.
∞
X
zn
We denote the sum of
as the exponential function ez .
n!
n=0
z n
z
e is also the limit lim 1 +
.
n→∞
n
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TMA4115 - Calculus 3, Lecture 3, Jan 23, page 22
The exponential function and cos
and sin
If y is a real number, then
∞
∞
∞
X
(iy )n X (iy )2n X (iy )2n+1
iy
e =
=
=
+
n!
(2n)!
(2n + 1)!
n=0
n=0
n=0
∞
∞
X
X
(−1)n y 2n
(−1)n y 2n+1
= cos(y ) + i sin(y ).
+i
(2n)!
(2n + 1)!
n=0
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n=0
TMA4115 - Calculus 3, Lecture 3, Jan 23, page 23
The exponential function
One can show that ez1 +z2 = ez1 ez2 . It follows that
ez = ex eiy = ex (cos y + i sin y ) for z = x + iy .
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TMA4115 - Calculus 3, Lecture 3, Jan 23, page 24
The exponential function and polar
representation
If z 6= 0, then
z = |z|(cos(arg(z)) + i sin(arg(z)))
=eln(|z|) ei arg(z) = eln(|z|)+i arg(z) .
The exponential function is not injective (because
ex+iy = ex+i(y +2π) ), and does therefore not have an inverse.
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TMA4115 - Calculus 3, Lecture 3, Jan 23, page 25
Properties of the exponential
function
If z = x + iy , then
ez = ez
Re(ez ) = ex cos y
Im(ez ) = ex sin y
|ez | = ex
arg(ez ) = y
One can also show that ez is differentiable and that
d z
e = ez .
dz
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TMA4115 - Calculus 3, Lecture 3, Jan 23, page 26
Sine and cosine
One can show that the series
∞
X
(−1)n z 2n+1
n=0
(2n + 1)!
n=0
We denote the sum of
∞
X
(−1)n z 2n+1
n=0
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(2n)!
and
converge absolutely for all complex numbers
z.
of
∞
X
(−1)n z 2n
(2n + 1)!
∞
X
(−1)n z 2n
n=0
(2n)!
as cos(z), and the sum
as sin(z).
TMA4115 - Calculus 3, Lecture 3, Jan 23, page 27
Properties of sin and cos
If z is a complex number, then
cos z =
eiz + e−iz
eiz − e−iz
and sin z =
.
2
2i
sin and cos are periodic with period 2π.
d
sin and cos are differentiable and dz
sin z = cos z and
d
cos
z
=
−
sin
z.
dz
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TMA4115 - Calculus 3, Lecture 3, Jan 23, page 28
Hyperbolic sine and cosine
∞
∞
X
X
z 2n
z 2n+1
One can show that the series
and
(2n)!
(2n + 1)!
n=0
n=0
converge absolutely for all complex numbers z.
∞
X
z 2n
We denote the sum of
as cosh(z), and the sum of
(2n)!
n=0
∞
X
z 2n+1
as sinh(z).
(2n + 1)!
n=0
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TMA4115 - Calculus 3, Lecture 3, Jan 23, page 29
Properties of sinh and cosh
If z is a complex number, then
ez + e−z
ez − e−z
cosh z =
and sinh z =
.
2
2
sinh and cosh are periodic with period 2πi.
d
sinh and cosh are differentiable and dz
sinh z = cosh z
d
and dz cosh z = sinh z.
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TMA4115 - Calculus 3, Lecture 3, Jan 23, page 30
Problem 1 from the exam from
June 2010
Solve the equation z 2 − 2iz − 1 − 2i = 0. Write your answer
in the form x + iy .
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TMA4115 - Calculus 3, Lecture 3, Jan 23, page 31
Solution
p
(2i)2 − 4(−1 − 2i)
The solutions are z =
, so we must
2
compute the square roots of
(2i)2 − 4(−1 − 2i) = −4 + 4 + 8i = 8i.
Since |8i| = 8 and Arg(8i) = π2 , the two square roots of 8i are
√
√
1
1
w0 = 8 cos(π/4) + i sin(π/4)) = 8 √ + i √
2
2
= 2 + 2i
√
√
−1
−1
w1 = 8 cos(5π/4) + i sin(5π/4)) = 8 √ + i √
2
2
= −2 − 2i
2i ±
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TMA4115 - Calculus 3, Lecture 3, Jan 23, page 32
Solution
So the two solutions of z 2 − 2iz − 1 − 2i = 0 are
p
√
2i ± (2i)2 − 4(−1 − 2i)
2i ± 8i
z=
=
2 (
2
1 + 2i
2i ± (2 + 2i)
=
=
2
−1.
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TMA4115 - Calculus 3, Lecture 3, Jan 23, page 33
Problem 1 from the exam from
August 2011
Find all complex numbers z such that Im(−z + i) = (z + i)2 .
Draw the solutions on a diagram.
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TMA4115 - Calculus 3, Lecture 3, Jan 23, page 34
Solution
Let z = x + iy where x and y are real numbers. Then
Im(−z + i) = Im(−x − iy + i) = 1 − y , and
(z + i)2 = (x + iy + i)2 = x 2 − (y + 1)2 + 2x(y + 1)i.
So Im(−z + i) = (z + i)2 if and only if 1 − y = x 2 − (y + 1)2
and x(y + 1) = 0. If x(y + 1) = 0, then either x = 0 or
y = −1.
Assume x = 0. Then x 2 − (y + 1)2 = −(y + 1)2 . So
1 − y = x 2 − (y + 1)2 if and only if y 2 + y + 2 = 0, but
y 2 + y + 2 = (y + 1/2)2 + 7/4 > 0 for all real numbers y . So
1 − y cannot be equal to x 2 − (y + 1)2 if x = 0.
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TMA4115 - Calculus 3, Lecture 3, Jan 23, page 35
Solution
Assume then that y = −1. Then 1 − y = 2 and
x 2 − (y√+ 1)2 = x 2 . So 1 − y = x 2 − (y + 1)2 if and only if
x = ± 2.
√
It follows that Im(−z + i) = (z + i)2 if and only if z = ± 2 − i.
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TMA4115 - Calculus 3, Lecture 3, Jan 23, page 36
Plan for tomorrow
Tomorrow we shall
study second-order linear differential equations,
introduce the Wronskian,
completely solve second-order homogeneous linear
differential equations with constant coefficients.
Section 4.1 and 4.3 in “Second-Order Equations” (pages
xxxv-xlv and xlix-lv).
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TMA4115 - Calculus 3, Lecture 3, Jan 23, page 37
Second-order linear differential
equations
A second-order linear differential equation is a differential
equation with can be written on the form
y 00 + p(t)y 0 + q(t)y = g(t).
Such an equation is homogeneous if g(t) = 0.
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TMA4115 - Calculus 3, Lecture 3, Jan 23, page 38