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# Complex Differentiation

Module by: Doug Daniels, Steven J. Cox. E-mail the authors

Summary: This module discusses complex differentiation and shows its similarity and differences to real differentiation.

## Complex Differentiation

The complex ff is said to be differentiable at z0z0 if limit  zz0fzfz0zz0 z z0 f z f z0 z z0 exists, by which we mean that fznfz0znz0 f zn f z0 zn z0 converges to the same value for every sequence zn zn that converges to z0 z0. In this case we naturally call the limit ddzfz0 z f z0

To illustrate the concept of 'for every' mentioned above, we utilize the following picture. We assume the point z0z0 is differentiable, which means that any conceivable sequence is going to converge to z0z0. We outline three sequences in the picture: real numbers, imaginary numbers, and a spiral pattern of both.

## Examples

### Example 1

The derivative of z2 z 2 is 2z 2 z .

limit  z z0 z2z02zz0=limit  zz0(zz0)(z+z0)zz0=2z0 z z0 z 2 z0 2 z z0 z z0 z z0 z z0 z z0 2 z0
(1)

### Example 2

The exponential is its own derivative.

limit  zz0ezez0zz0=ez0limit  zz0ezz01zz0=ez0limit  zz0n=0zz0n(n+1)!=ez0 z z0 z z0 z z0 z0 z z0 z z0 1 z z0 z0 z z0 n 0 z z0 n n 1 z0
(2)

### Example 3

The real part of zz is not a differentiable function of zz.

We show that the limit depends on the angle of approach. First, when znz0 zn z0 on a line parallel to the real axis, e.g., zn=x0+1n+iy0 zn x0 1 n y0 , we find

limit  nx0+1nx0x0+1n+iy0x0+iy0=1 n x0 1 n x0 x0 1 n y0 x0 y0 1
(3)
while if znz0 zn z0 in the imaginary direction, e.g., zn=x0+i(y0+1n) zn x0 y0 1 n , then
limit  nx0x0x0+i(y0+1n)x0+iy0=0 n x0 x0 x0 y0 1 n x0 y0 0
(4)

## Conclusion

This last example suggests that when ff is differentiable a simple relationship must bind its partial derivatives in xx and yy.

### Proposition 1: Partial Derivative Relationship

If ff is differentiable at z0z0 then ddzfz0=fz0x=(ifz0y) z f z0 x f z0 y f z0

#### Proof

With z=x+iy0 z x y0 ,

ddzfz0=limit  zz0fzfz0zz0=limit  xx0fx+iy0fx0+iy0xx0=fz0x z f z0 z z0 f z f z0 z z0 x x0 f x y0 f x0 y0 x x0 x f z0
(5)

Alternatively, when z=x0+iy z x0 y then

ddzfz0=limit  zz0fzfz0zz0=limit  yy0fx0+iyfx0+iy0i(yy0)=(ifz0y) z f z0 z z0 f z f z0 z z0 y y0 f x0 y f x0 y0 y y0 y f z0
(6)

## Cauchy-Reimann Equations

In terms of the real and imaginary parts of ff this result brings the Cauchy-Riemann equations.

ux=vy x u y v
(7)
and
vx=uy x v y u
(8)
Regarding the converse proposition we note that when ff has continuous partial derivatives in region obeying the Cauchy-Reimann equations then ff is in fact differentiable in the region.

We remark that with no more energy than that expended on their real cousins one may uncover the rules for differentiating complex sums, products, quotients, and compositions.

As one important application of the derivative let us attempt to expand in partial fractions a rational function whose denominator has a root with degree larger than one. As a warm-up let us try to find q 1 , 1 q 1 , 1 and q 1 , 2 q 1 , 2 in the expression z+2z+12= q 1 , 1 z+1+ q 1 , 2 z+12 z 2 z 1 2 q 1 , 1 z 1 q 1 , 2 z 1 2 Arguing as above, it seems wise to multiply through by z+12 z 1 2 and so arrive at

z+2= q 1 , 1 (z+1)+ q 1 , 2 z 2 q 1 , 1 z 1 q 1 , 2
(9)
On setting z=-1 z -1 this gives q 1 , 2 =1 q 1 , 2 1 . With q 1 , 2 q 1 , 2 computed, Equation 9 takes the simple form z+1= q 1 , 1 (z+1) z 1 q 1 , 1 z 1 and so q 1 , 2 =1 q 1 , 2 1 as well. Hence, z+2z+12=1z+11z+12 z 2 z 1 2 1 z 1 1 z 1 2 This latter step grows more cumbersome for roots of higher degrees. Let us consider z+22z+13= q 1 , 1 z+1+ q 1 , 2 z+12+ q 1 , 3 z+13 z 2 2 z 1 3 q 1 , 1 z 1 q 1 , 2 z 1 2 q 1 , 3 z 1 3 The first step is still correct: multiply through by the factor at its highest degree, here 33. This leaves us with
z+22= q 1 , 1 z+12+ q 1 , 2 (z+1)+ q 1 , 3 z 2 2 q 1 , 1 z 1 2 q 1 , 2 z 1 q 1 , 3
(10)
Setting z=-1 z -1 again produces the last coefficient, here q 1 , 3 =1 q 1 , 3 1 . We are left however with one equation in two unknowns. Well, not really one equation, for Equation 10 is to hold for all zz. We exploit this by taking two derivatives, with respect to zz, of Equation 10. This produces 2(z+2)=2 q 1 , 1 (z+1)+ q 1 , 2 2 z 2 2 q 1 , 1 z 1 q 1 , 2 and 2= q 1 , 1 2 q 1 , 1 The latter of course needs no comment. We derive q 1 , 2 q 1 , 2 from the former by setting z=-1 z -1 . This example will permit us to derive a simple expression for the partial fraction expansion of the general proper rational function q=fg q f g where gg has hh distinct roots λ1λh λ1 λh of respective degrees d1dh d1 dh . We write
qz=j=1hk=1dj q j , k zλjk q z j 1 h k 1 dj q j , k z λj k
(11)
and note, as above, that q j , k q j , k is the coefficient of zdjdjk z dj dj k in the rational function rjzqzzλjdj rj z q z z λj dj Hence, q j , k q j , k may be computed by setting z=λj z λj in the ratio of the djk dj k th derivative of rjrj to (djk)! dj k . That is,
q j , k =limit  zλj1(djk)!ddjkdzdjkzλjdjqz q j , k z λj 1 dj k d dj k d z dj k z λj dj q z
(12)
As a second example, let us take
B=( 100 130 011 ) B 1 0 0 1 3 0 0 1 1
(13)
and compute the Φ j , k Φ j , k matrices in the expansion zIB-1=( 1z100 1(z1)(z3)1z30 1z12(z3)1(z1)(z3)1z1 )=1z1 Φ 1 , 1 +1z12 Φ 1 , 2 +1z3 Φ 2 , 1 z I B -1 1 z 1 0 0 1 z 1 z 3 1 z 3 0 1 z 1 2 z 3 1 z 1 z 3 1 z 1 1 z 1 Φ 1 , 1 1 z 1 2 Φ 1 , 2 1 z 3 Φ 2 , 1 The only challenging term is the 3 1 3 1 element. We write 1z12(z3)= q 1 , 1 z1+ q 1 , 2 z12+ q 2 , 1 z3 1 z 1 2 z 3 q 1 , 1 z 1 q 1 , 2 z 1 2 q 2 , 1 z 3 It follows from Equation 12 that
q 1 , 1 =ddz1z31=1/4 q 1 , 1 z 1 z 3 1 14
(14)
and
q 1 , 2 =1z31=1/4 q 1 , 2 1 z 3 1 14
(15)
and
q 2 , 1 =1z323=1/4 q 2 , 1 1 z 3 2 3 14
(16)
It now follows that
zIB-1=1z1( 100 -1/200 -1/4-1/21 )+1z12( 000 000 -1/200 )+1z3( 000 1/210 1/41/20 ) z I B -1 1 z 1 1 0 0 -12 0 0 -14 -12 1 1 z 1 2 0 0 0 0 0 0 -12 0 0 1 z 3 0 0 0 12 1 0 14 12 0
(17)
In closing, let us remark that the method of partial fraction expansions has been implemented in Matlab. In fact, Equation 14, Equation 15, and Equation 16 all follow from the single command: [r,p,k]=residue([0 0 0 1],[1 -5 7 -3]). The first input argument is Matlab-speak for the polynomial fz=1 f z 1 while the second argument corresponds to the denominator gz=z12(z3)=z35z2+7z3 g z z 1 2 z 3 z 3 5 z 2 7 z 3 .

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