Complex Differentiation The complex f is said to be differentiable at z0 if z z0 f z f z0 z z0 exists, by which we mean that f zn f z0 zn z0 converges to the same value for every sequence zn that converges to z0 . In this case we naturally call the limit z f z0 To illustrate the concept of 'for every' mentioned above, we utilize the following picture. We assume the point z0 is differentiable, which means that any conceivable sequence is going to converge to z0. We outline three sequences in the picture: real numbers, imaginary numbers, and a spiral pattern of both.

Sequences Approaching A Point In The Complex Plane The green is real, the blue is imaginary, and the red is the spiral.

Examples The derivative of z 2 is 2 z . z z0 z 2 z0 2 z z0 z z0 z z0 z z0 z z0 2 z0 The exponential is its own derivative. 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 The real part of z is not a differentiable function of z. We show that the limit depends on the angle of approach. First, when zn z0 on a line parallel to the real axis, e.g., zn x0 1 n y0 , we find n x0 1 n x0 x0 1 n y0 x0 y0 1 while if zn z0 in the imaginary direction, e.g., zn x0 y0 1 n , then n x0 x0 x0 y0 1 n x0 y0 0

Conclusion This last example suggests that when f is differentiable a simple relationship must bind its partial derivatives in x and y. Partial Derivative Relationship If f is differentiable at z0 then z f z0 x f z0 y f z0 With z x y0 , z f z0 z z0 f z f z0 z z0 x x0 f x y0 f x0 y0 x x0 x f z0 Alternatively, when z x0 y then z f z0 z z0 f z f z0 z z0 y y0 f x0 y f x0 y0 y y0 y f z0

Cauchy-Reimann Equations In terms of the real and imaginary parts of f this result brings the Cauchy-Riemann equations. x u y v and x v y u Regarding the converse proposition we note that when f has continuous partial derivatives in region obeying the Cauchy-Reimann equations then f 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 and q 1 , 2 in the expression 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 1 2 and so arrive at z 2 q 1 , 1 z 1 q 1 , 2 On setting z -1 this gives q 1 , 2 1 . With q 1 , 2 computed, takes the simple form z 1 q 1 , 1 z 1 and so q 1 , 2 1 as well. Hence, 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 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 3. This leaves us with z 2 2 q 1 , 1 z 1 2 q 1 , 2 z 1 q 1 , 3 Setting z -1 again produces the last coefficient, here q 1 , 3 1 . We are left however with one equation in two unknowns. Well, not really one equation, for is to hold for all z. We exploit this by taking two derivatives, with respect to z, of . This produces 2 z 2 2 q 1 , 1 z 1 q 1 , 2 and 2 q 1 , 1 The latter of course needs no comment. We derive q 1 , 2 from the former by setting z -1 . This example will permit us to derive a simple expression for the partial fraction expansion of the general proper rational function q f g where g has h distinct roots λ1 … λh of respective degrees d1 … dh . We write q z j 1 h k 1 dj q j , k z λj k and note, as above, that q j , k is the coefficient of z dj dj k in the rational function rj z q z z λj dj Hence, q j , k may be computed by setting z λj in the ratio of the dj k th derivative of rj to dj k . That is, q j , k z λj 1 dj k d dj k d z dj k z λj dj q z As a second example, let us take B 1 0 0 1 3 0 0 1 1 and compute the Φ j , k matrices in the expansion 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 element. We write 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 that q 1 , 1 z 1 z 3 1 14 and q 1 , 2 1 z 3 1 14 and q 2 , 1 1 z 3 2 3 14 It now follows that 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 In closing, let us remark that the method of partial fraction expansions has been implemented in Matlab. In fact, , , and 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 f z 1 while the second argument corresponds to the denominator g z z 1 2 z 3 z 3 5 z 2 7 z 3 .