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Math 370 - Introductory Complex Variables G.Pugh Sep 11 2014 1 / 26
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### Transcript of Math 370 - Introductory Complex Variables€¦ · Math 370 - Introductory Complex Variables G.Pugh... Math 370 - Introductory Complex Variables

G.Pugh

Sep 11 2014

1 / 26 Recap of Last Day

2 / 26 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

3 / 26 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

4 / 26 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

5 / 26 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.

6 / 26 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

7 / 26 Example

Find an identity which expresses sin (4θ) in terms of sin θ andcos θ.

8 / 26 1.5 - Powers and Roots

9 / 26 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

10 / 26 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...

11 / 26 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...

12 / 26 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...

13 / 26 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...

14 / 26 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

15 / 26 Example

Find all 6th roots of z =2i

1 + i.

16 / 26 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...

17 / 26 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.

18 / 26 1.6 - Planar Sets

19 / 26 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

20 / 26 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

21 / 26 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

22 / 26 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

23 / 26 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.

24 / 26 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.

25 / 26 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?

26 / 26