2 Analytic functions - Massachusetts Institute of Technology

1 Topic 2 NotesJeremy Orloff2 Analytic IntroductionThe main goal of this topic is to define and give some of the important properties ofcomplex Analytic functions . A functionf(z) is Analytic if it has a complex derivativef (z).In general, the rules for computing derivatives will be familiar to you from single variablecalculus. However, a much richer set of conclusions can be drawn about a complex analyticfunction than is generally true about real differentiable The derivative: preliminariesIn calculus we defined the derivative as a limit. In complex analysis we will do the (z) = lim z 0 f z= lim z 0f(z+ z) f(z) giving the derivative our full attention we are going to have to spend some timeexploring and understanding limits.

2 To motivate this we ll first look at two simple examples one positive and one the derivative off(z) = :We compute using the definition of the derivative as a z 0(z+ z)2 z2 z= lim z 0z2+ 2z z+ ( z)2 z2 z= lim z 02z+ z= was a positive example. Here s a negative one which shows that we need a carefulunderstanding of (z) =z. Show that the limit forf (0) does not :Let s try to computef (0) using a limit:f (0) = lim z 0f( z) f(0) z= lim z 0 z z= x i y x+i we used z= x+i , z 0 means both xand yhave to go to 0. There are lots of ways to do example, if we let zgo to 0 along thex-axis then, y= 0 while xgoes to 0.

3 In thiscase, we would havef (0) = lim x 0 x x= the other hand, if we let zgo to 0 along the positivey-axis thenf (0) = lim y 0 i yi y= Analytic FUNCTIONS2 The limitsdon tagree! The problem is that the limit depends on how zapproaches 0. Ifwe came from other directions we d get other values. There s nothing to do, but agree thatthe limit does not , there is something we can do: explore and understand limits. Let s do that Open disks, open deleted disks, open open disk of radiusraroundz0is the set of pointszwith|z z0|< r, all points within open deleted disk of radiusraroundz0is the set of pointszwith 0<|z z0|< r.

4 Thatis, we remove the centerz0from the open disk. A deleted disk is also called a : an open disk aroundz0; right: a deleted open disk open region in the complex plane is a setAwith the property that everypoint inAcan be be surrounded by an open disk that lies entirely inA. We will often dropthe word open and simply callAa the figure below, the setAon the left is an open region because for every point inAwecan draw a little circle around the point that is completely inA. (The dashed boundaryline indicates that the boundary ofAis not part ofA.) In contrast, the setBis not anopen region.

5 Notice the pointzshown is on the boundary, so every disk aroundzcontainspoints : an open regionA; right:Bis not an open Limits and continuous (z) is defined on a punctured disk aroundz0then we saylimz z0f(z) =w0iff(z) goes tow0no matter what figure below shows several sequences of points that approachz0. If limz z0f(z) =w0thenf(z) must go tow0along each of these Analytic FUNCTIONS3 Sequences going toz0are mapped to sequences going functions have obvious limits. For example:limz 2z2= 4andlimz 2(z2+ 2)/(z3+ 1) = 6 is an example where the limit doesn t exist because different sequences give (No limit) Show thatlimz 0zz= limz 0x+iyx iydoes not :On the real axis we havezz=xx= 1,so the limit asz 0 along the real axis is contrast, on the imaginary axis we havezz=iy iy= 1,so the limit asz 0 along the imaginary axis is -1.

6 Since the two limits do not agree thelimit asz 0 does not exist! Properties of limitsWe have the usual properties of limits. Supposelimz z0f(z) =w1and limz z0g(z) =w2then limz z0f(z) +g(z) =w1+w2. limz z0f(z)g(z) =w1 Analytic FUNCTIONS4 Ifw26= 0 then limz z0f(z)/g(z) =w1/w2 Ifh(z) is continuous and defined on a neighborhood ofw1then limz z0h(f(z)) =h(w1)(Note: we will give the official definition of continuity in the next section.)We won t give a proof of these properties. As a challenge, you can try to supply it usingthe formal definition of limits given in the can restate the definition of limit in terms of functions of (x,y).

7 To this end, let s writef(z) =f(x+iy) =u(x,y) +iv(x,y)and abbreviateP= (x,y), P0= (x0,y0), w0=u0+ z0f(z) =w0iff{limP P0u(x,y) =u0limP P0v(x,y) = term iff stands for if and only if which is another way of saying is equivalentto . Continuous functionsA function is continuous if it doesn t have any sudden jumps. This is the gist of the the functionf(z) is defined on an open disk aroundz0and limz z0f(z) =f(z0)then we sayfis continuous atz0. Iffis defined on an open regionAthen the phrase fiscontinuous onA means thatfis continuous at every point usual, we can rephrase this in terms of functions of (x,y) (z) =u(x,y) +iv(x,y) is continuous iffu(x,y) andv(x,y) are continuous asfunctions of two (Some continuous functions )(i) A polynomialP(z) =a0+a1z+a2z2+.}

8 +anznis continuous on the entire plane. Reason: it is clear that each power (x+iy)kis continuousas a function of (x,y).(ii) The exponential function is continuous on the entire plane. Reason:ez= ex+iy= excos(y) +iexsin(y).So the both the real and imaginary parts are clearly continuous as a function of (x,y).(iii) The principal branch Arg(z) is continuous on the plane minus the non-positive realaxis. Reason: this is clear and is the reason we defined branch cuts for arg. We have toremove the negative real axis because Arg(z) jumps by 2 when you cross it.

9 We also haveto removez= 0 because Arg(z) is not even defined at Analytic FUNCTIONS5(iv) The principal branch of the function log(z) is continuous on the plane minus the non-positive real axis. Reason: the principal branch of log haslog(z) = log(r) +iArg(z).So the continuity of log(z) follows from the continuity of Arg(z). Properties of continuous functionsSince continuity is defined in terms of limits, we have the following properties of (z) andg(z) are continuous on a regionA. Then f(z) +g(z) is continuous onA. f(z)g(z) is continuous onA. f(z)/g(z) is continuous onAexcept (possibly) at points whereg(z) = 0.

10 Ifhis continuous onf(A) thenh(f(z)) is continuous these properties we can claim continuity for each of the following functions : ez2 cos(z) = (eiz+ e iz)/2 IfP(z) andQ(z) are polynomials thenP(z)/Q(z) is continuous except at roots ofQ(z). The point at infinityBy definition the extended complex plane =C { }. That is, we haveonepoint at infinityto be thought of in a limiting sense described as sequence of points{zn}goes to infinity if|zn|goes to infinity. This point at infinity is approached in any direction we go. All of the sequences shown in the figure below aregrowing, so they all go to the (same) point at infinity.