Math 370 - Introductory Complex Variables€¦ · Math 370 - Introductory Complex Variables G.Pugh...
Transcript of Math 370 - Introductory Complex Variables€¦ · Math 370 - Introductory Complex Variables G.Pugh...
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Math 370 - Introductory Complex Variables
G.Pugh
Sep 11 2014
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Recap of Last Day
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1.3 - Vector and Polar Forms
I For z = a + ib ∈ C :
θ = arg(z)
r = |z|
z
I Consequently,
z = a + ib = r [cos θ + i sin θ] = reiθ
zn = (a + ib)n = rn[cos (nθ) + i sin (nθ)] = rneinθ,n ∈ N
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1.4 - The Complex ExponentialI The representation z = reiθ is a consequence of the
definition
ez =∞∑
k=0
zk
k !
This series converges absolutely for every z ∈ C; more onthis later.
I Letting z = iθ in this definition we find:
eiθ =
( ∞∑k=0
θ2k
(2k)!
)+ i
( ∞∑k=0
θ2k+1
(2k + 1)!
)= cos θ + i sin θ ,
I Also can show that
ezew = ez+w ,
and so for p, q ∈ R
epeiq = ep+iq
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Euler’s Equation
I The equation eiθ = cos θ + i sin θ is called Euler’s equation
I Letting θ = π we find
eiπ = cosπ + i sinπ
from whicheiπ + 1 = 0
I Called "The most beautiful theorem in mathematics" bysome
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Euler’s Equation continued
I Recall that for θ ∈ R:
cos (−θ) = cos θ, sin (−θ) = − sin θ
I This giveseiθ = cos θ + i sin θ
e−iθ = cos θ − i sin θ
I Now add and divide by 2 to find cos θ =eiθ + e−iθ
2.
I Subtract and divide by 2i to find sin θ =eiθ − e−iθ
2i.
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Periodicity of the complex exponential
I For k ∈ Z,
ei(θ+2kπ) = eiθei2kπ = eiθ · 1 = eiθ
I Say that eiθ is periodic with period 2π
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Example
Find an identity which expresses sin (4θ) in terms of sin θ andcos θ.
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1.5 - Powers and Roots
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Powers
I For n ∈ N it is easy to define zn:
z = |z|eiθ
sozn = |z|neinθ
I In fact, true for n ∈ Z if z−n = 1/zn.
I Value of zn is the same regardless of branch of arg(z)used to define θ:
|z|nein(θ+2kπ) = |z|neinθein2kπ = |z|neinθ · 1
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Roots
I For roots of complex numbers there is more to consider.
I Definition: For m ∈ N, ζ is an mth root of z if ζm = z
I To find all mth roots of a complex number z = |z|eiθ 6= 0, letζ = ρeiφ where ρ > 0.
I Then we must have ρmeimφ = |z|eiθ
I So ρ = m√|z| and eimφ = eiθ
continued...
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Roots, continued
I So mφ = θ + 2kπ, where k ∈ Z
I So φ =θ
m+
2kπm
, where k ∈ Z
I So all possible mth roots of z are given by
ζ = m√|z|ei(θ+2kπ)/m, k ∈ Z
continued...
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Roots, continued
I Notice: for k = 0,1, . . . ,m − 1 we have 0 ≤ 2kπm
< 2π
I Soζ = m
√|z|ei(θ+2kπ)/m, k = 0,1, . . . ,m − 1
represents m distinct mth roots of z.
I Are these m roots the only ones? That is, what if k ≤ −1 ork ≥ m?
continued...
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Roots, continued
I By the Division Algorithm there are integers q and r suchthat k = qm + r where 0 ≤ r ≤ m − 1
I So
ζ = m√|z|ei(θ+2kπ)/m
= m√|z|ei(θ+2(qm+r)π)/m
= m√|z|ei(θ+2rπ)/mei2qmπ/m
= m√|z|ei(θ+2rπ)/m
which, since 0 ≤ r ≤ m − 1, is one of the roots we foundalready.
continued...
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Roots, Conclusion
I Theorem: Let m ≥ 1 be an integer and z = reiθ with r ,θ ∈ R, and where θ is given by any branch of arg(z). Themth roots of z are given by
z1/m = m√|z|ei(θ+2kπ)/m, k = 0,1, . . . ,m − 1
I Corollary: If m and n are positive integers with nocommon factors, then (z1/n)m = (zm)1/n and this commonnumber, denoted by zm/n is given by
zm/n = n√|z|meim(θ+2kπ)/n, k = 0,1, . . . ,n − 1
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Example
Find all 6th roots of z =2i
1 + i.
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mth roots of unity
I Consider the special case z = 1 = ei·0
I Here the mth roots are ζ = ei2kπ/m, k = 0,1, . . . ,m − 1
I This gives roots
(ei2π/m)0 = 1
(ei2π/m)1 = ωm
(ei2π/m)2 = ω2m
...
(ei2π/m)m−1 = ωm−1m
continued...
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mth roots of unity, continued
I So the mth roots of unity are 1, ωm, ω2m, . . . , ω
m−1m
I Here ωm = ei2π/m is called a primitive mth root of unitysince all the other mth roots of 1 can be found by raising ωmto positive integer powers.
I Definition: ω is called a primitive mth root of unity ifωm = 1, but ωq 6= 1 for 1 ≤ q ≤ m − 1.
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1.6 - Planar Sets
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Open DisksDefinition: Let z0 ∈ C and ρ > 0 be real. The set
{z : |z − z0| < ρ}
is called an open disk of radius ρ and centre z0. The set issometimes called a circular neighbourhood of z0.
C :
ρ
z0
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Interior PointDefinition: Let S ⊂ C be a set and z0 ∈ S. z0 is an interiorpoint of S if there is some circular neighbourhood of z0completely contained in S. That is, there is some ρ > 0 suchthat
{z : |z − z0| < ρ} ⊂ S
C :
S
z0
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Open Set
Definition: A set S ⊂ C if open if every point of S is an interiorpoint.
C :S
S is open
C :
S
S is not open
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Connected Set
Definition: An open set S ⊂ C if connected if any two points Scan be joined by a path consisting of a finite number of linesegments which lie entirely in S.
C :S
S is connected
C :
A
B
S = A ∪ B
A and B are each connected, but S = A ∪ B is not
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Domain, Boundary Point, Closed sets
Definition: A domain is an open connected set.
Definition: Let S ⊂ C. z0 is called a boundary point of S ifevery circular neighbourhood of z0 contains at least one point inS and one point not in S.
Definition: A set S ⊂ C is called closed if it contains all of itsboundary points.
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Bounded sets, Compact sets, Regions
Definition: A set S ⊂ C is bounded if there is R ∈ R such that|z| < R for every z ∈ S .
Definition: A S ⊂ C is compact if it is both closed andbounded.
Definition: A region is a domain together with some, none, orall of its boundary points.
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Example
Let S be the set of complex numbers which satisfy1 < (Im(z))2 < 4.
1. Is S open?
2. Is S connected?
3. Is S a domain?
4. Is S bounded?
5. Describe the boundary points.
6. Is S a region?
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