tma4110 - calculus 3 - lecture 3
TRANSCRIPT
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TMA4110 - Calculus 3Lecture 3
Toke Meier CarlsenNorwegian University of Science and Technology
Fall 2012
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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.
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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.
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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 .
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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
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The fundamental theorem of algebra
Every complex polynomial of degree 1 or higher has a least onecomplex root.
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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 .
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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.
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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.
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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.
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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
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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.
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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.
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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.
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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.
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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.
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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).
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The exponential function
1
y
eiy = cos(y) + i sin(y)
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The exponential function
One can show that ez1+z2 = ez1ez2 . It follows thatez = exeiy = ex(cos y + i sin y) for z = x + iy .
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The exponential function
ex
ez = ex(cos(y) + i sin(y))
yex
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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.
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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.
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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 .
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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).
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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.
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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).
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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.
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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.
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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.
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