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Home , Conformal map

Cauchy-Riemann

Equations and  

Conformal Mapping 26.2 

Introduction In this Section we consider two important features of complex functions. The Cauchy Riemann equations introduced on page 2 provide a necessary and sufficient condition for a f(z) to be analytic in some region of the complex ; this allows us to find f (z)inthat region by the rules of the previous Section. A mapping between the z-plane and the w-plane is said to be conformal if the between two intersecting in the z-plane is equal to the angle between their mappings in the w-plane. Such a mapping has widespread uses in solving problems in fluid flow and electromagnetics, for example, where the given problem geometry is somewhat complicated.

 

Prerequisites ① understand the idea of a complex function and its Before starting this Section you should ...  

✓ use the Cauchy Riemann equations to ob- Learning Outcomes tain the derivative of complex functions

After completing this Section you should be ✓ appreciate the idea of a conformal map- able to ... ping 1. Cauchy-Riemann equations Remembering that z = x+iy and w = u+iv we note that there is a very useful test to determine whether a function w = f(z)isanalytic at a point. This is provided by the Cauchy-Riemann equations. These state that w = f(z)isdifferentiable at a point z = z0 if, and only if, ∂u ∂v ∂u ∂v = and = − at that point. ∂x ∂y ∂y ∂x When these equations hold then it can be shown that the complex derivative may be determined df ∂f df ∂f by using either = or = −i . dz ∂x dz ∂y (The use of ‘if, and only if,’ means that if the equations are valid, then the function is differen- tiable and vice versa). If we consider f(z)=z2 = x2 − y2 +2ixy then u = x2 − y2 and v =2xy so that ∂u ∂u ∂v ∂v =2x, = −2y, =2y, =2x. ∂x ∂y ∂x ∂y It should be clear that, for this example, the Cauchy-Riemann equations are always satisfied; therefore, the function is analytic everywhere. We find that df ∂f df ∂f = =2x +2iy =2z or, equivalently = −i = −i(−2y +2ix)=2z dz ∂x dz ∂y This is the result we would expect to get by simply differentiating f(z)asifitwas a real function. For analytic functions this will always be the case.

Example Show that the function f(z)=z3 is analytic everwhere and hence obtain its derivative.

Solution

w = f(z)=(x + iy)3 = x3 − 3xy2 +(3x2y − y3)i Hence u = x3 − 3xy2 and v =3x2y − y3. Then ∂u ∂u ∂v ∂v =3x2 − 3y2, = −6xy, =6xy, =3x2 − 3y2. ∂x ∂y ∂x ∂y The Cauchy-Riemann equations are identically true and f(z)isanalytic everywhere. Furthermore df ∂f = =3x2 − 3y2 +(6xy)i=3(x +iy)2 =3z2 as we would expect. dz ∂x

We can easily find functions which are not analytic anywhere and others which are only analytic in a restricted region of the . Consider again the function f(z)=¯z = x − iy. Here

HELM (VERSION 1: April 8, 2004): Workbook Level 2 2 26.2: Cauchy-Riemann Equations and Conformal Mapping ∂u ∂u ∂v ∂v u = x so that =1, and =0; v = −y so that =0, = −1. ∂x ∂y ∂x ∂y The Cauchy-Riemann equations are never satisfied so thatz ¯ is not differentiable anywhere and so is not analytic anywhere. 1 By contrast if we consider the function f(z)= we find that z x y u = ; v = . x2 + y2 x2 + y2 As can readily be shown, the Cauchy-Riemann equations are satisfied everywhere except for 1 x2 + y2 =0, i.e. x = y =0(or, equivalently, z = 0). At all other points f (z)=− . This z2 function is analytic everywhere except at the single point z =0. Analyticity is a very powerful property of a function of a complex variable. Such functions tend to behave like functions of a real variable.

Example Show that if f(z)=zz¯ then f (z) exists only at z =0.

Solution f(z)=x2 + y2 so that u = x2 + y2,v=0. ∂u ∂u ∂v ∂v =2x, =2y, =0, =0. ∂x ∂y ∂x ∂y Hence the Cauchy-Riemann equations are satisfied only where x =0and y =0,i.e. where z =0. Therefore this function is not analytic anywhere.

Analytic Functions and Harmonic Functions Using the Cauchy-Riemann equations in a region of the z-plane where f(z)isanalytic, ∂2u ∂ ∂u ∂ ∂v ∂2v = = − = − ∂x∂y ∂x ∂y ∂x ∂x ∂x2 and ∂2u ∂ ∂u ∂ ∂v ∂2v = = = . ∂y∂x ∂y ∂x ∂y ∂y ∂y2 ∂2u ∂2u If these differentiations are possible then = so that ∂x∂y ∂y∂x ∂2u ∂2u + =0 (Laplaces equation). ∂x2 ∂y2 In a similar way we find that ∂2v ∂2v + =0 (can you show this?) ∂x2 ∂y2 When f(z)isanalytic the functions u and v are called conjugate harmonic functions.

3 HELM (VERSION 1: April 8, 2004): Workbook Level 2 26.2: Cauchy-Riemann Equations and Conformal Mapping Suppose u = u(x, y)=xy then it is easy to verify that u satisfies Laplace’s equation (try this). We now try to find the conjugate v = v(x, y). First, using the Cauchy-Riemann equations: ∂v ∂u ∂v ∂u = = y and = − = −x. ∂y ∂x ∂x ∂y 1 Integrating the first equation gives v = y2+afunction of x.Integrating the second equation 2 1 gives v = − x2+afunction of y. Bearing in mind that an additive constant leaves no trace 2 after differentiation we pool the information above to obtain 1 v = (y2 − x2)+C where C is a constant 2 1 Note that f(z)=u + iv = xy + (y2 − x2)i + D where D is a constant (replacing Ci). 2 1 We can write f(z)=− iz2 + D (as you can verify). This function is analytic everywhere. 2

Show that the function u = x2 − x − y2 is harmonic.

Your solution

∂y ∂x

2 2 Hence sharmonic. is u and =0 +

∂ u u ∂

2 2

∂y ∂y ∂x ∂x

2 2

y, 2 = , 1 x =2 . 2 = , =2

− − −

∂ ∂ ∂u u u ∂u

2 2

Now find the conjugate harmonic function v.

Your solution

HELM (VERSION 1: April 8, 2004): Workbook Level 2 4 26.2: Cauchy-Riemann Equations and Conformal Mapping

62 acyReanEutosadCnomlMapping Conformal and Equations Cauchy-Riemann 26.2:

5 EM(ESO :Arl8 04:Wrbo ee 2 Level Workbook 2004): 8, April 1: (VERSION HELM

1. f(z)issingular at z = 2i. Elsewhere

(z − 2i).1 − z.1 −2i f (z)= = (z − 2i)2 (z − 2i)2

−2i −2i 2 f (−i) = = = i (−3i)2 −9 9 2. u = x2 + x − y2 and v =2xy + y ∂u ∂u ∂v ∂v =2x +1, = −2y, =2y, =2x +1 ∂x ∂y ∂x ∂y

Here the Cauchy-Riemann equations are identically true and f(z)isanalytic everywhere.

df ∂f = =2x +1+2yi=2z +1

dz ∂x

. z of terms in v +i u )= z ( f find hence and v

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 z Exercises

f(z)=u + iv = x2 − x − y2 +2xyi − iy + D, D constant

Now z2 = x2 − y2 +2ixy and z = x +iy thus f(z)=z2 − z + D.

u solution our Y

. z of terms n )i z ( f find Finally,

∂v ∂v Integrating =2x − 1 gives v =2xy − y+ function of x.Integrating =+2y gives ∂y ∂x v =2xy+ function of y. Ignoring the duplication, v =2xy − y + C, where C is a constant.

62 acyReanEutosadCnomlMapping Conformal and Equations Cauchy-Riemann 26.2:

6 EM(ESO :Arl8 04:Wrbo ee 2 Level Workbook 2004): 8, April 1: (VERSION HELM

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∂2u ∂2u 3. =2, = −2 therefore u is harmonic. ∂x2 ∂y2 ∂v ∂u = =2x therefore v =2xy+ function of y ∂y ∂x ∂v ∂u = − =2y +2 therefore v =2xy +2x+ function of x ∂x ∂y ∴ v =2xy +2x + constant f(z)=x2 +2ixy − y2 +2xi − 2y = z2 +2iz to each other; the dashed curves are the lines of force and the solid curves are the equipotentials.

Figure 1 In ideal fluid flow the curves v = constant are the streamlines of the flow. In these situations the function w = u +iv is the complex potential of the field.

Conformal Mapping A function w = f(z) can be regarded as a mapping, which ‘maps’ a point in the z-plane to a point in the w-plane. Curves in the z-plane will be mapped into curves in the w-plane. Consider aerodynamics. The idea is that we are interested in the fluid flow, in a complicated geometry (say flow past an aerofoil). We first find the flow in a simple geometry that can be mapped to the aerofoil shape (the complex plane with a circular hole works here). Most of the calculations necessary to find physical characteristics such as and drag on the aerofoil can be performed in the simple geometry - the resulting integrals being much easier to evaluate than in the complicated geometry. Consider the mapping w = z2. The point z = 2+imaps to w =(2+i)2 = 3+4i. The point z = 2+ilies on the intersection of the two lines x =2and y =1.Towhat curves do these map? To answer this question we note that a point on the line y =1can be written as z = x +i. Then w =(x +i)2 = x2 − 1+2xi As usual, let w = u +iv, then u = x2 − 1 and v =2x Eliminating x we obtain: 4u =4x2 − 4=v2 =4 or v2 = 4+4u.

7 HELM (VERSION 1: April 8, 2004): Workbook Level 2 26.2: Cauchy-Riemann Equations and Conformal Mapping Example To what curve does the line x =2map?

Solution Apoint on the line is z =2+yi. Then w =(2+yi)2 =4− y2 +4yi Hence u =4− y2 and v =4y so that, eliminating y we obtain 16u =64− v2 or v2 =64− 16u

In Figure 2(a) we sketch the lines x =2and y =1and in Figure 2(b) we sketch the curves into which they map. Note these curves intersect at the point (3,4).

v y v2 =64− 16u x =2 v2 =4+4u

(3, 4)

(2, 1) y =1

x u (a) (b)

Figure 2 The angle between the original lines was clearly 900; what is the angle between the curves at the point of intersection?

2 dv The curve v = 4+4u has a gradient du . Differentiating the equation implicitly we obtain dv dv 2 2v =4 or = du du v dv 1 At the point (3,4) = . du 2

dv Find for the curve v2 =64− 16u and evaluate it at the point (3,4) du

Your solution

HELM (VERSION 1: April 8, 2004): Workbook Level 2 8 26.2: Cauchy-Riemann Equations and Conformal Mapping

62 acyReanEutosadCnomlMapping Conformal and Equations Cauchy-Riemann 26.2:

9 EM(ESO :Arl8 04:Wrbo ee 2 Level Workbook 2004): 8, April 1: (VERSION HELM

-plane. w the in origin the through line a is which , mu = v that so −

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dv − ∴ dv − 8 2v du = 16 du = v dv − At v =4we obtain du = 2 Consider the line ax + by + c =0where c =0 . This represents a line in the z-plane which does not pass through the origin. To what sort of curve does it map in the w-plane?

Your solution

paewihpse hog h origin. the through passes which -plane w the in circle a of equation the is which

c c

u =0 v u + v +

−

2 2 b a

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− 2 2

v + u v + u

2 2 2 2

=0 c +

−

au bv h apdcreis curve mapped The

Similarly, it can be shown that a circle in the z-plane passing through the origin maps to a line in the w-plane which does not pass through the origin and a circle in the z-plane which does not pass through the origin maps to a circle in the w-plane which does not pass through the origin. The inversion mapping is an example of the bilinear transformation: az + b w = f(z)= where we demand that ad − bc =0 cz + d (If ad − bc =0the mapping reduces to f(z)=constant).

Find the bilinear transformations which map z =2tow =1.

Your solution

d + c 2

1= d + c =2 b + a 2 Hence .

b + a 2

HELM (VERSION 1: April 8, 2004): Workbook Level 2 10 26.2: Cauchy-Riemann Equations and Conformal Mapping If in addition, z = −1ismapped to w =3find the class of transformation which is possible.

Your solution

d + c

−

3= d +3 c 3 = b + a Hence .

− −

b + a −

− − b − If, further, z =0is mapped to w = 5 then 5= d so that b = 5d. Substituting this last relation into the first two obtained we obtain

2a − 2c − 6d =0 −a +3c − 8d =0

Solving these two in terms of d we find 2c =11d and 2a =17d. Hence the transformation is: 17z − 10 w = (note that the d’s cancel in the numerator and denominator). 11z +2

Some other mappings are shown in Figure 3.

z2

z3

π/3

z1/2

zα π/α

z-plane w-plane

Figure 3 As an engineering application we consider the transformation 2 w = z − where  is a constant. z

11 HELM (VERSION 1: April 8, 2004): Workbook Level 2 26.2: Cauchy-Riemann Equations and Conformal Mapping It is used to map circles which contain z =1as an interior point and which pass through z = −1 into shapes resembling aerofoils. Figure 4 shows an example

y v

x u

z-plane w-plane

Figure 4 This creates a cusp at which the associated fluid velocity can be infinite. This can be avoided by adjusting the fluid flow in the z-plane. Eventually, this can be used to find the lift generated by such an aerofoil in terms of physical characteristics such as aerofoil shape and air density and speed.t

Exercises

1. Find a bilinear transformation which maps z =0, −1, −iintow =i, 0, 1 respectively.

+1 z d + dz

− −

= w and d = c Hence =

−

i d + τ i d +i z i

d i+ c

−

ie = 1 gives =1 ,w i = z i d + d = d i+ a = d i+ c that so

− − −

b i+ a −

d

ie = i gives =i ,w =0 z i d = b that so

b

d + cz

= w 1.

b + az

HELM (VERSION 1: April 8, 2004): Workbook Level 2 12 26.2: Cauchy-Riemann Equations and Conformal Mapping