Maths - Functions of a Complex Variable

Conformal Transformations

Transformations where the angle between two intersecting curves in Z plane equals the angle between corresponding curves in W plane.

A transformation is conformal where dw/dz ≠ 0 and the function is regular.

Differentiation

Is defined if dw/ dz is independent of angle.

In this case: dw/dz = dw/idy

therefore dv/dy - j du/dy = du/dx + j dv/dx

This gives the Cauchy Riemann equations:

dv/dy = du/dx

dv/dx = - du/dy

Harmonic Functions

∂u/∂x = ∂v/∂y —> ∂²u/∂x² = ∂v²/∂y∂x

∂u/∂y = -∂v/∂x —> ∂²u/∂y² = -∂v²/∂y∂x

therefore

∂²u/∂x² = -∂²u/∂y²

∂²u/∂x² + ∂²u/∂y² = 0

Therefore if a function obeys the Cauchy riemann equations then the real part is a harmonic function.

cv1 (x - x1)² + (y - y1)² = r² |z-d| = r
cv2 (x - x1)² + (y - y1)² < r² |z-d| < r
cv3 (x - x1)² + (y - y1)² > r² |z-d| > r
cv4 tan-1(y/x) = θ arg(z)=θ

 

z plane   w plane
z plane

-->

w=z²

w squared
z plane

-->

w=ez

w exponent
z plane

-->

w=1/z

w invert

Specific information about these functions is shown in the complex number part of the site (functions of a complex variable).

Cauchy's Theorem

If f(z) is analytic in a region R (no poles in R) then:

circular integral f(z) dz = 0

Extension of Cauchys Theorem

∫ round same singularities are equal.

∫ round several singularities = sum of ∫ round each singularity.

Use of Cauchys Theorem to evaluate Unknown Integral

integral between plus and minus infinity e-x² = √π

integral between plus and minus infinity e-x² cos(2 b x) dx = √π e

integral between plus and minus infinity e-x² sin(2 b x) dx = 0

proof that:

line integral e-z² dz - > 0 as a - > ∞

principles used:

  1. |line integral f(z) dz | <= line integral| f(z) dz |
  2. line integral |f(z) dz | = line integral| f(z)| |dz | <= ml

Poles of a function

A function can have several laurent series each corresponding to a pole:

Laurent Series:

f(z) = an/(z-a)n + an-1/(z-a)n-1 + … +a1/(z - a) + b0 + b1(z-a) + …

where:

Evaluation of residue at simple pole

  1. multiply by (z-pole)
  2. take limit as z -> a

i.e. a1 = limz->a(z - a) f(z)

Evaluation of residue at pole of order n

a1 = 1/(n-1)! limz->a dn-1/ d zn-1{(z - a)n f(z)}

Application of poles: complex integration

line integral f(z) d z = 2 π i (sum of enclosed resides)

Evaluation of integrals between ±∞

  1. Find Poles
  2. Take suitable pathintegral path
  3. Find residues of poles inside path
  4. Check if circular integral =0 (denominator of f(z) 2 more than numerator) (proof usescircular integral ≤ml)

 


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see also:

 

Correspondence about this page

Book Shop - Further reading.

Where I can, I have put links to Amazon for books that are relevant to the subject, click on the appropriate country flag to get more details of the book or to buy it from them.

flag flag flag flag flag flag Visual Complex Analysis - If you already know the basics of complex numbers but want to get an in depth understanding using an geometric and intuitive approach then this is a very good book. The book explains how to represent complex transformations such as the Möbius transformations. It also shows how complex functions can be differentiated and integrated.

 

Terminology and Notation

Specific to this page here:

 

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