A RAPID INTRODUCTION TO COMPLEX ANALYSIS

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1 A RAPID INTRODUCTION TO COMPLEX ANALYSIS AKHIL MATHEW ABSTRACT. These notes give a rapid introduction to some of the basic results in complex analysis, assuming familiarity from the reader with Stokes theorem on manifolds. CONTENTS. Complex Differentiability 2. Complex integration 2 3. Representability as power series 3 4. Laurent series 4 5. Residues 5 6. Examples of these techniques 5 References 6. COMPLEX DIFFERENTIABILITY So, let s consider an open set O C, and a C 2 function f : O C. We can consider the differential df := f x dx + f y dy which is a complex-valued -form on O. It is also convenient to write the differential using the z and z-derivatives I talked about earlier, i.e. f z := ( 2 x i ) f, f z := ( y 2 x + i ) f. y The reason these are important is that if w 0 O, we can choose A,B C with f(w 0 + h) = f(w 0 ) + Ah + B h + o( h ), h C by differentiability, and it is easy to check that A = f z (w 0 ),B = f z (w 0 ). So we can define a function f to be holomorphic if it satisfies the differential equation f z = 0, which is equivalent to being able to write f(w 0 + h) = f(w 0 ) + Ah + o( h ) Date: January 0, 200.

2 2 AKHIL MATHEW for each w 0 O and a suitable A C. In particular, it is equivalent to a difference quotient definition. The derivative f z of a holomorphic function thus satisfies all the usual algebraic rules, under which holomorphic functions are closed. Recall now that f z z is a multiple of the Laplacian. In particular, it follows that a holomorphic function is harmonic, and thus infinitely differentiable. The other reason we care about the z-derivative is that we can write, for any f, df = f z dz + f z d z where naturally enough dz = dx + idy,d z = dx idy. It is easy to check that dz,d z and z, z are dual to each other as complex-valued elements of the cotangent and tangent spaces, respectively. So, as a result, we can compute on (complex-valued) -forms: d(f(z)dz) = df dz = (f z dz + f z d z) dz = f z dz d z. This will lead to the Cauchy theorem. Note that dz d z = 2idx dy is a multiple of the usual area element. 2. COMPLEX INTEGRATION Cauchy s theorem is now a simple corollary of Stokes theorem. Given a - dimensional manifold O, we can consider the integral of a complex-valued -form dz, which, given a parametrization, can of course be computed using the change-ofvariables formula, and by taking real and imaginary parts. More interestingly, when is the boundary of a compact 2-dimensional submanifold-with-boundary X we can compute it using Stokes theorem: dz = d(dz) = g z dz d z. In particular, if g is holomorphic on O, we find: Theorem (Cauchy). For the boundary of X O, we have dz = 0. Using a similar approach, we can establish the following: X Theorem 2 (Cauchy formula). If w is in the interior of X and g is holomorphic in O, then dz = g(w). 2πi z w In the proof, we will actually show something more general. So let us not assume g holomorphic, only differentiable. To prove this, let s consider the manifold-with-boundary Y := X N r (w) for r > 0 very small, as in the figure. X

3 A RAPID INTRODUCTION TO COMPLEX ANALYSIS 3 X gamma w The boundary Y is the disjoint union of X with the usual orientation and the circle C r (w) with the opposite orientation. Also, z w is holomorphic in a neighborhood of Y. This is the key fact, and we will use this together with a limiting argument on r. So ( ) = Y z w dz C r(w) z w dz = d X N r(w) z w dz. First, we can compute the term C r(w) z wdz by using the parametrization t w + re it,0 t < 2π of the circle. Then dz = d(re it ) = ire it dt, so 2π z w dz = g(w + re it ) 2π re it ire it dt = i g(w + re it )dt C r(w) 0 and as r 0, this approaches 2πig(w). The integrand in the second term becomes dz d z times ( ) = g z z z w z w + ( ) = g z z z w z w because the z-derivative clearly satisfies the product rule and z w (by the quotient rule). So, when we let r 0, using the integrability of z w, we find: dz = 2πig(w) z w X 0 g z dz d z. z w is holomorphic This is a more general analog of the Cauchy formula, and it clearly implies the usual one when g is holomorphic. 3. REPRESENTABILITY AS POWER SERIES Fix a holomorphic function f on O. In a neighborhood N r (z 0 ) with N r (x 0 ) O; then for any z N r (z 0 ) we have f(z) = 2πi w z dw C r(z 0 ) In particular, we have represented f(z) as a weighted average of the functions g (w) (z) := w z for w ranging over the circle C r(z 0 ). Each function g (w) = w z is represented by a power series in N r (z 0 ), the geometric series. Thus, the weighted

4 4 AKHIL MATHEW average f has the same property. We now prove this rigorously; by a translation assume z 0 = 0. So, writing out the geometric series: f(z) = ( z ) n 2πi n C r(0) w w dw = c n z n n ( where c n is the constant ) n 2πi C r(0) w w dw. It is immediate that c n = O(r n ), so convergence questions don t arise when z < r, and the interchange of summation and integration was justified. In fact the convergence is uniform on compact subsets (where z is bounded by something less than r) by the same argument. We have thus proved that f can be represented as a power series converging in the whole disk. Differentiation of this power series can be done term-by-term as one can check directly, though the justification is slightly technical. 4. LAURENT SERIES The Taylor series just discussed works for functions in a disk. In an annulus, we can t necessarily do that, but we end up with terms of negative order. As usual, let f : O C be holomorphic. Suppose the closed annulus A R r (0) O, where A R r (0) = {z : r < z < R}. Then by Cauchy s formula applied to X = A R r (0), we have for z A R r (0), ( ) 2πif(z) = w z dw = I (z) I 2 (z). C R C r Now I is holomorphic in N R (0) so can be represented as a power series as before. I 2 is holomorphic in the disk at infinity {z : z > r}, and we can write for such z I 2 (z) = z C r w/z dw = z n+ w n dw. n C r The convergence issues can be handled as before because the integrands are O(r n ) and z > r. In particular, we can represent f in the annulus as f(z) = n Zc n z n, where the c n are uniquely determined constants. The uniqueness is seen because one can choose r between r,r, and get f(z)z k dz = c n z n k dz = 2πc k, C r n Z C r because all but one of the integrals vanish by direct computation. So f determines the c k.

5 A RAPID INTRODUCTION TO COMPLEX ANALYSIS 5 5. RESIDUES We re now going to allow singularities. Suppose given a region O C as before, a finite set F O, and a holomorphic function f : O F C. Given a -submanifold not passing through F, we want to compute f(z)dz. Since f is holomorphic in a deleted neighborhood of each z 0 F, we can write f(z) = n Z c n(z z 0 ) n around each z 0, the series converging in some deleted neighborhood. Then c is called the residue of f at z 0. The residue theorem states that the residues determine the integrals. Theorem 3. Let X O be a compact 2-dimensional submanifold with O =. Let F = F X, and suppose does not pass through F, so the following integral is defined. Then f(z)dz = Res(f,z 0 ), 2πi z 0 F with obvious notation for the residue. The idea is very similar to the proof of the Cauchy formula. Choose small neighborhoods N,...,N k for each element of F that do not overlap, and consider the compact manifold-with-boundary X N i. Now f is holomorphic on this, so by Cauchy s theorem X f(z)dz = N i f(z)dz. We compute a sample N i f(z)dz and show that it is 2πi times the residue at the center. Indeed, we locally expand in N i minus the center (at, say, z 0 ) the function f as a Laurent series n c n(z z 0 ) n, and integrating this shows f(z)dz = c n (z z 0 ) n dz = 2πic. N i N i Indeed, it is easily checked: N i (z z 0 ) n dz = 2πiδ i, because N i is parametrized by t z 0 + re it for r chosen as the radius. 6. EXAMPLES OF THESE TECHNIQUES We now know how to compute complex integrals, so let s use these facts to prove a few things. First, I claim that the uniform limit of holomorphic functions is holomorphic, which is a change from the real line (cf. the Weierstrass function). This is a local question, so we ll consider f i : O C holomorphic and converging uniformly to f : O C, and prove f is holomorphic in a neighborhood of each point z 0. We can write f i (w) 2πi C r(z 0 ) w z dw = f i(z),

6 6 AKHIL MATHEW and taking limits shows the same is satisfied for f. Thus f has a nice integral interpretation, which means in particular that f is holomorphic indeed, differentiating under the integral w.r.t. z yields 2πi C r(z 0 ) (w z) 2dw = f z. This proves the claim. Moreover, it is clear that the differentiating under the integral sign procedure works for any holomorphic f. So the boxed identity is always true if N r (z 0 ) is contained in O. Incidentally, repeated differentiation under the integral gives another proof that a holomorphic function is C. As a result, let s prove a theorem of Liouville: Theorem 4. A bounded holomorphic function on C (i.e., an entire function) is constant. We need to show f (z) = f z 0. By translation, we may just prove f (0) = 0. Now choose R > 0; then the boxed formula gives f (0) = 2πi w 2 dw = 2π 2πi R f(re it )ie it dt C R (0) by the usual parametrization. As R, the integrand is bounded by the hypothesis on f, and the R tends to 0, so it follows f (0) is arbitrarily small. As a side remark, let s note that the procedure used in the last theorem of bounding the countour integral is part of a well-known fact: if is a curve and f any continuous function, bounded by M, then f(z)dz ML(). This follows from parametrizing appropriately and using the definition of the length. A simple corollary of Liouville s theorem is the fundamental theorem of algebra, i.e. that C is algebraically closed. Indeed, for any zerofree polynomial P(z), P is holomorphic on C and bounded because as z, P(z) = cz n + o( z n ). So P, hence P, is constant. REFERENCES. Lars Ahlfors, Complex analysis, McGraw-Hill, Lars Hormander, An introduction to complex analysis in several variables, third edition, North Holland, Michael Spivak, Calculus on manifolds, Westview Press, 97. 0