TMA4110 - Calculus 3Lecture 3
Toke Meier CarlsenNorwegian University of Science and Technology
Fall 2012
www.ntnu.no TMA4110 - Calculus 3, Lecture 3
Review of last week
Last week weintroduced complex numbers, both in a geometric way and inan 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,solved complex, second degree equations.
www.ntnu.no TMA4110 - Calculus 3, Lecture 3, page 2
Solutions to second degree equations
If a,b, c are complex numbers and a2 6= 0, then the solutions to the
equation az2 + bz + c = 0 are z =−b ±
√b2 − 4ac
2a.
www.ntnu.no TMA4110 - Calculus 3, Lecture 3, page 3
Problem 1 from the exam from June 2012
Solve w2 = (−1 + i√
3)/2. Find all solutions of the equationz4 + z2 + 1 = 0 and draw them in the complex plane. Write thesolutions in the form x + iy .
www.ntnu.no TMA4110 - Calculus 3, Lecture 3, page 4
Problem 1 from the exam from June 2012
Solve w2 = (−1 + i√
3)/2. Find all solutions of the equationz4 + z2 + 1 = 0 and draw them in the complex plane. Write thesolutions in the form x + iy .
www.ntnu.no TMA4110 - Calculus 3, Lecture 3, page 5
The fundamental theorem of algebra
Every complex polynomial of degree 1 or higher has a least onecomplex root.
www.ntnu.no TMA4110 - Calculus 3, Lecture 3, page 6
Roots of real polynomials
If w is a root of a real polynomial∑n
k=0 akzk , then w is also a rootof∑n
k=0 akzk .
www.ntnu.no TMA4110 - Calculus 3, Lecture 3, page 7
Exercise 34 on page xxvii
Check that z1 = 1−√
3i is azero of P(z) = z4−4z3+12z2−16z+16, and find all the zeros of P.
www.ntnu.no TMA4110 - Calculus 3, Lecture 3, page 8
Exercise 34 on page xxvii
Check that z1 = 1−√
3i is azero of P(z) = z4−4z3+12z2−16z+16, and find all the zeros of P.
www.ntnu.no TMA4110 - Calculus 3, Lecture 3, page 9
Exercise 34 on page xxvii
Check that z1 = 1−√
3i is azero of P(z) = z4−4z3+12z2−16z+16, and find all the zeros of P.
www.ntnu.no TMA4110 - Calculus 3, Lecture 3, page 10
Complex functions
A complex function f is a rule that assigns a unique complexnumber f (z) to each number z in some set of complex numbers(called the domain of f ).
Examples of complex functionsf (z) = Re(z)g(z) = Im(z)h(z) = |z|j(z) = Arg(z)k(z) = zp(z) = z2 − 4z + 6
www.ntnu.no TMA4110 - Calculus 3, Lecture 3, page 11
Graphic representations of complexfunctions
We cannot draw the graph of a complex function since wewould need 4 dimensions to do that.Instead, we can graphically represent the behavior of acomplex function w = f (z) by drawing the z-plane and thew-plane separately, and showing the image in the w-plane ofcertain, appropriately chosen set of points in the z-plane.
www.ntnu.no TMA4110 - Calculus 3, Lecture 3, page 12
Example
Consider the function f (z) = −2iz.Arg(−2i) = −π/2 and | − 2i | = 2, so f maps the region1/2 ≤ |z| ≤ 1, 0 ≤ arg(z) ≤ π/2 to the region 1 ≤ |z| ≤ 2,−π/2 ≤ arg(z) ≤ 0.
www.ntnu.no TMA4110 - Calculus 3, Lecture 3, page 13
Example
Consider the function g(z) = z2.g maps the region 0 ≤ |z| ≤ 1/2, π/2 ≤ arg(z) ≤ π to the region0 ≤ |z| ≤ 1/4, π ≤ arg(z) ≤ 2π.
www.ntnu.no TMA4110 - Calculus 3, Lecture 3, page 14
Complex functions
Limits, continuity and differentiability of complex functions can bedefined just as for real functions.
Examples of complex functionsEvery complex polynomial is differentiable, and hencecontinuous.The functions f (z) = Re(z), g(z) = Im(z), h(z) = |z| andk(z) = z are continuous, but not differentiable.The function j(z) = Arg(z) is not continuous.
www.ntnu.no TMA4110 - Calculus 3, Lecture 3, page 15
The exponential function
One can show that the series∞∑
n=0
zn
n!converges absolutely for all
complex numbers z.
We denote the sum of∞∑
n=0
zn
n!as the exponential function ez .
ez is also the limit limn→∞
(1 +
zn
)n.
www.ntnu.no TMA4110 - Calculus 3, Lecture 3, page 16
The exponential function and cos and sin
If y is a real number, then
eiy =∞∑
n=0
(iy)n
n!=∞∑
n=0
(iy)2n
(2n)!+∞∑
n=0
(iy)2n+1
(2n + 1)!=
∞∑n=0
(−1)ny2n
(2n)!+ i
∞∑n=0
(−1)ny2n+1
(2n + 1)!= cos(y) + i sin(y).
www.ntnu.no TMA4110 - Calculus 3, Lecture 3, page 17
The exponential function
1
y
eiy = cos(y) + i sin(y)
www.ntnu.no TMA4110 - Calculus 3, Lecture 3, page 18
The exponential function
One can show that ez1+z2 = ez1ez2 . It follows thatez = exeiy = ex(cos y + i sin y) for z = x + iy .
www.ntnu.no TMA4110 - Calculus 3, Lecture 3, page 19
The exponential function
ex
ez = ex(cos(y) + i sin(y))
yex
www.ntnu.no TMA4110 - Calculus 3, Lecture 3, page 20
Example
ez maps the region 0 ≤ Re(z) ≤ 1/2, 0 ≤ Im(z) ≤ π/4 to theregion 0 ≤ |z| ≤ e1/2, 0 ≤ arg(z) ≤ π/4.
www.ntnu.no TMA4110 - Calculus 3, Lecture 3, page 21
The exponential function and polarrepresentation
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 (becauseex+iy = ex+i(y+2π)), and does therefore not have an inverse.
www.ntnu.no TMA4110 - Calculus 3, Lecture 3, page 22
Properties of the exponential function
If z = x + iy , thenez = ez
Re(ez) = ex cos yIm(ez) = ex sin y|ez | = ex
arg(ez) = yOne can also show that ez is differentiable and that d
dz ez = ez .
www.ntnu.no TMA4110 - Calculus 3, Lecture 3, page 23
Sine and cosine
One can show that the series∞∑
n=0
(−1)nz2n
(2n)!and
∞∑n=0
(−1)nz2n+1
(2n + 1)!converge absolutely for all complex numbers z.
We denote the sum of∞∑
n=0
(−1)nz2n
(2n)!as cos(z), and the sum of
∞∑n=0
(−1)nz2n+1
(2n + 1)!as sin(z).
www.ntnu.no TMA4110 - Calculus 3, Lecture 3, page 24
Properties of sin and cos
If z is a complex number, then
cos z =eiz + e−iz
2and sin z =
eiz − e−iz
2i.
sin and cos are periodic with period 2π.sin and cos are differentiable and d
dz sin z = cos z andddz cos z = − sin z.
www.ntnu.no TMA4110 - Calculus 3, Lecture 3, page 25
Hyperbolic sine and cosine
One can show that the series∞∑
n=0
z2n
(2n)!and
∞∑n=0
z2n+1
(2n + 1)!converge
absolutely for all complex numbers z.
We denote the sum of∞∑
n=0
z2n
(2n)!as cosh(z), and the sum of
∞∑n=0
z2n+1
(2n + 1)!as sinh(z).
www.ntnu.no TMA4110 - Calculus 3, Lecture 3, page 26
Properties of sinh and cosh
If z is a complex number, then
cosh z =ez + e−z
2and sinh z =
ez − e−z
2.
sinh and cosh are periodic with period 2πi .sinh and cosh are differentiable and d
dz sinh z = cosh z andddz cosh z = sinh z.
www.ntnu.no TMA4110 - Calculus 3, Lecture 3, page 27
Problem 1 from the exam from August2011
Find all complex numbers z such that Im(−z + i) = (z + i)2. Drawthe solutions on a diagram.
www.ntnu.no TMA4110 - Calculus 3, Lecture 3, page 28
Problem 1 from the exam from August2012
Write all of the solutions of z3 = 1 in the form z = x + iy . Write thesolutions of z3 = −3+i√
2(2+i)in the form z = x + iy and draw the
solutions in the complex plane.
www.ntnu.no TMA4110 - Calculus 3, Lecture 3, page 29