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Holomorphic Functions and the Wirtinger Derivative

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Holomorphic Functions and the Wirtinger Derivative Ioakim I. Ioannis [email protected] Holomorphic functions and the Wirtinger derivative Definition 0.1 (Holomorphic function). If f is (complex) differentiable at every point z0 of U⊆C open, we say that f is holomorphic on U. If f is (complex) differentiable on some (open) neigh- bourhood of z0, then we say that f is holomorphic at z0. The word ’holomorphic’ was introduced by two of Cauchy’s students, Briot and Bouquet and derives from the Greek wordsoλoς ´ (holos) meaning ’entire’ or ’whole’, and µo̺ϕη´ (morph¯e) meaning ’form’ or ’shape’. We shall see later on where this terminology stems from. Remarks 1. It follows immediately that if f is differentiable at z0, then it is continuous at z0. 2. Complex differentiability (or holomorphicity) if often referred to as ’analyticity’. This is due to the fact, as we shall see later, that a holomorphic function can be expanded locally as a (convergent) series, hence the term ’analyticity’. The main tool for proving such a powerful result is Cauchy’s Integral formula. Nevertheless, there exists a proof that holomorphicity implies C∞ without the help of Cauchy’s Integral formula. It is based on a topological result by G. T. Whyburn (namely, if f is differentiable and non-constant in a region D of the complex plane, then f is an open map). The proof is presented in [9]. Since in what follows we will study holomorphicity under the scope of differentiability, it is useful that we keep this fact in mind. Definition 0.2 (Entire function). A function that is holomorphic on the entire complex plane is called an entire function. We shall denote O(U) the set of all holomorphic functions on U : O(U) = {f : U → C : f isholomorphic}. Examples 1. The function f(z) = z, z ∈ C is nowhere differentiable. Lets first proof that f ′(0) does not exist. We have that f(h) − f(0) h ∆x − i∆y lim = lim = lim (1.1) h→0 h h→0 h ∆x+i∆y→0 ∆x + i∆y Ioakim Ioannis | [email protected] | [email protected] 1 Complex analysis (where we’ve written h = ∆x + i∆y). If we restrict ourselves in the case were h → 0 along the x−axis, i.e. ∆x → 0, then ∆y = 0 and the above limit then becomes ∆x lim =1 ∆x→0 ∆x and similarly, if we restrict ourselves in the case were h → 0 along the y−axis, i.e. ∆y → 0, then ∆x = 0 and the above limit then becomes −i∆y lim = −1. ∆y→0 i∆y Thus, since we approached 0 in two different directions and found the corresponding limits different, we conclude that the limit (1.1) does not exist. For z 6=0, we could follow the same iarg(z0) steps as above, but it would perhaps be easier to write z in its polar form: z0 = |z0|e . Then, f(z + h) − f(z ) z + h − z z + h − z lim 0 0 = lim 0 0 = lim 0 0 h→0 h h→0 h h→0 h −iarg(z0) h |z0|e −2iarg(z0) = lim = lim iarg(z ) = lim e (1.2) h→0 h h→0 |z0|e 0 h→0 If arg(z0)=0or= π, i.e. h → 0 along the x−axis, then 2iarg(z0) = 1 and thus (1.2) equals π 3π to 1. Similarly, if arg(z0)= 2 or = 2 , i.e. h → 0 along the y−axis, then 2iarg(z0)= −1 and thus (1.2) equals to -1. The result follows. Note that since there cannot be an open set around (0, 0) such that f is (complex) differentiable, then f is not holomorphicat (0, 0). 2. Let 2 z z 6=0 f(z)= z (1.3) (0 z =0 2 C z2 |z| Obviously, f is continuous in \{0}. Since for z 6=0, |f(z)| = z = |z| = |z|, we have that f(z) → 0= f(0) as z → 0, thus f is continuous at the origin also. f is (complex) differentiable at the origin: f(h) − f(0) |h|2 lim = lim = lim h =0. h→0 h h→0 h h→0 Thus, f ′(0) = 0. Viewing the above limit in x, y coordinates, we have |h|2 lim = lim (x − iy)=0. h=(x,y)→(0,0) h (x,y)→(0,0) Now, for z 6=0, we have f(z + h) − f(z) |z + h|2 − |z|2 zh + zh lim = lim = lim h→0 h h→0 h h→0 h 2 Ioakim Ioannis | [email protected] | [email protected] Complex analysis h which limit does not exist, since the limh→0 h does not exist. Otherwise, using x, y coordinates: • Let z → 0 along the x− axis (y = 0): x3 f(x +0i) − f(0) 2 lim = lim x =1 (x,0)→(0,0) (x +0i) − 0 (x,0)→(0,0) x • Let z → 0 along the diagonal (y = x): the diagonal is parameterized by x = t, y = x = t and thus f(t + ti) − f(0) −t(t + i) lim = lim = −1 (t,t)→(0,0) (t + ti) − 0 t→0 t(t + i) Therefore, f is not differentiable at z 6= 0 in the complex sense. 1.1. The Cauchy-Riemann equations We shall now establish the relationship between the (complex) differentiability of a function f = u+iv and the partial derivatives of u and v. Let f the function defined in ((1.3)). Viewing f as a function from R2 to R, f is written as f(x, y)= x2 + y2 and since the partial derivatives of f are continuous functions all over R2, we conclude that f is differentiable (in R2). But as we saw earlier, f is (complex) differentiable only at the origin. Thus, the differentiability of u and v does not imply the (complex) differentiability of f = u + iv. Also consider the function f(z)=z, ¯ z ∈ C. Write f(x + iy) = x − iy. Then the partial derivatives of all orders of u(x, y) = x and v(x, y) = −y exist but, as we already proved, f is nowhere differentiable. The following Theorem states that the opposite is true: differentiability of f = u + iv implies (partial) differentiability of u and v. Recall Definition 1.1 (Smooth Functions). Let U ⊆ R2 open and f : U → R a continuous function. We say that f is of class Ck (or f is k− continuously differentiable, or simply f is Ck) if the partial derivatives of f up to order k and including order k exist and are continuous (functions) on U, (where k ∈ N ∪{0}). In this case, we write f ∈ Ck(U). The space C0(U) is just the space of all continuous functions on U, the space C1(U) is the ∂f ∂f 2 space of all differentiable functions on U such that ∂x and ∂y are smooth, C (U) is the space of ∂f ∂f ∂f 2 ∂f 2 all differentiable functions on U such that ∂x , ∂y ∂x∂y and ∂y∂x are smooth and so on. The space C∞(U) is the space of all functions infinitely differentiable on U. At this point, lets recall some basic facts about multivariate calculus, restricting ourselves mainly in the bivariate case. 1. Separately continuous 6⇒ continuity. It is possible that a function f of two real variables x, y is separately continuous in x, y, at some point (x0,y0) of its domain, but not continuous at the point. Consider for example the function xy , (x, y) 6= (0, 0) f(x, y)= x2+y2 (0, (x, y)=(0, 0) We can easily verify that f is continuous as well as separately continuous on R2 \{(0, 0)}. Since limx→0 f(x, y) = limy→0 f(x, y)=0= f(x, y), we conclude that f is separately continuous at Ioakim Ioannis | [email protected] | [email protected] 3 Complex analysis the point (0, 0) also. But f is not continuous at (0, 0), for it where, the lim(x,y)→(0,0) f(x, y) would equal to f(0, 0) = 0 on every direction. This is not the case on the diagonal {y = x} : t2 1 lim f(x, y) = lim = . (t,t)→(0,0) t→0 t2 + t2 2 Here is a plot for the surface f(x, y) = 0 : ·10−2 1 0 2 −1 0 −2 0 2 −2 y x 2. Existence of partial derivatives 6⇒ differentiability. It is possible that both partial derivatives of a function f of two real variables x, y exist at some point of its domain, but it is not differentiable there. See the previous example. 3. Existence of partial derivatives, function is continuous 6⇒ differentiability. Example of a function of two real variables x, y where the partial derivatives exist at some point of its domain, is also continuous at that point, but it is not differentiable there: 1 f(x, y) = (xy) 3 . We immediately see that for (x, y) 6= (0, 0) 1 1 ∂f(x, y) y 3 ∂f(x, y) x 3 = 2 and (by symmetry) = 2 ∂x 3x 3 ∂y 3y 3 ∂f(x,0) ∂f(x,0) Since f(x, 0)=0, for x 6=0, ∂x =0, (for x 6= 0) but ∂y does not exist (verify this). By symmetry, the similar results hold if we interchange x and y. Now, since the restriction of f ∂f(0,0) ∂f(0,0) along the x− axis (or the y axis) is identically zero, the partial derivatives ∂x and ∂y exist and are both equal to 0.
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