## 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.

(x - x_{1})² + (y - y_{1})² = r² |
|z-d| = r | |

(x - x_{1})² + (y - y_{1})² < r² |
|z-d| < r | |

(x - x_{1})² + (y - y_{1})² > r² |
|z-d| > r | |

tan^{-1}(y/x) = θ |
arg(z)=θ |

z plane | w plane | |
---|---|---|

--> w=z² |
||

--> w=e |
||

--> w=1/z |

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:

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

e^{-x²} = √π

e^{-x²} cos(2 b x) dx = √π e^{b²}

e^{-x²} sin(2 b x) dx = 0

proof that:

^{}e^{-z²} dz - > 0 as a - > ∞

principles used:

- | f(z) dz | <= | f(z) dz |
- |f(z) dz | = | f(z)| |dz | <= ml

## Poles of a function

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

Laurent Series:

f(z) = a_{n}/(z-a)^{n} + a_{n-1}/(z-a)^{n-1} + … +a_{1}/(z - a) + b_{0} + b_{1}(z-a) + …

where:

- a = a pole of function
- a
_{1}= residue of pole

### Evaluation of residue at simple pole

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

i.e. a_{1} = lim_{z->a}(z - a) f(z)

### Evaluation of residue at pole of order n

a_{1} = 1/(n-1)! lim_{z->a} d^{n-1}/ d z^{n-1}{(z - a)^{n} f(z)}

### Application of poles: complex integration

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

### Evaluation of integrals between ±∞

- Find Poles
- Take suitable path
- Find residues of poles inside path
- Check if =0 (denominator of f(z) 2 more than numerator) (proof uses ≤ml)