MATH 215A NOTES MOOR XU NOTES FROM A COURSE BY KANNAN SOUNDARARAJAN Abstract. These notes were taken during Math 215A (Complex Analysis) taught by Kan- nan Soundararajan in Fall 2011 at Stanford University. They were live-TEXed during lectures in vim and compiled using latexmk. Each lecture gets its own section. The notes are not edited afterward, so there may be typos; please email corrections to [email protected]. 1. 9/27 We will think about functions defined on either the complex numbers C or some region Ω ⊆ C. Throughout, a region will be an open and connected subset of C. It may have some holes. We are interested in considering functions f :Ω ! C. We'd like to understand functions that are nice. Definition 1.1. f is said to be holomorphic at a point z0 2 Ω if it is differentiable there, i.e. f(z + h) − f(z ) lim 0 0 h!0 h 0 exists. If it exists, we denote it as f (z0). f is holomorphic on all of Ω if it is holomorphic at every point in Ω. We can also say that f is complex differentiable. Example 1.2. Polynomials in z are holomorphic on C. P (z) Example 1.3. Rational functions f = Q(z) are holomorphic except at the zeros of Q. You can check that if f and g are holomorphic at some point z0, then f + g and f · g are holomorphic, and f=g is too so long as g(z0) 6= 0. 1 Example 1.4. f(z) = z is holomorphic on the punctured unit disc = f0 < jzj < 1g. Another basic example is the idea of a power series. To start, let's center the power series P1 n at 0. We are interested in f(z) = n=0 anz for some sequence an 2 C. A priori, this might not converge. We can ask when this converges. It should be familiar that there is a number −1=n R, called the radius of convergence, defined by R = lim infn!1 janj . This means that f converges absolutely in jzj < R, and the series will diverge for jzj > R. It could do anything on jzj = R. P1 n Example 1.5. Consider n=0 z , which converges in jzj < 1 and diverges as jzj ! 1. Or P1 zn P1 n! n=0 n2 , which converges on jzj = 1. Or n=0 z , which also has R = 1, but it ! 1 as z ! e2πia=q. 1 P1 zn Example 1.6. For the exponential function n=0 n! , we have R = 1. P1 n Proposition 1.7. In the region fjzj < Rg, the power series f(z) = n=0 anz is holo- 0 P1 n−1 morphic, and its derivative is given by term-by-term differentiation f (z) = n=1 nanz . Since n1=n ! 1 as n ! 1, this also has radius of convergence R. Therefore, f is infinitely differentiable. Then 1 (k) X n−k f (z) = n(n − 1) ··· (n − k + 1)anz : n=k f (k)(0) This shows that ak = k! , and this is just the Taylor series. Proof. We'll sketch the proof. It suffices to consider the first derivative. Consider z such that jzj < R, and consider h very small, so jz + hj < R. We compute N f(z + h) − f(z) X (z + h)n − zn = lim an : h N!1 h n=0 We can now use the binomial theorem to work out what this is. The numerator is (z + h)n − zn = h(zn−1 + (z + h)zn−2 + ··· + (z + h)n−1) = h(nzn−1 + stuff that will go to zero as h ! 0); and the rest should be easy. This is certainly true for real-valued power series too. This is what we do with Taylor series expansions; the point here is that f is represented by its Taylor series. In complex analysis, there is a remarkable converse to this result. In a certain sense, holomorphic functions are exactly given by power series. Theorem 1.8. Suppose that f is holomorphic in a region Ω. If z0 2 Ω and the disc with center z0 and radius r is contained in Ω, then in that disc fjz − z0j < Rg we have f(z) = P1 n n=0 an(z − z0) . In particular, f is infinitely differentiable on Ω and locally has a Taylor expansion. Example 1.9. From real analysis, we remember bad examples like ( 2 e−1=x x 6= 0 0 x = 0; which is infinitely differentiable but not analytic at zero. This cannot happen in the complex case. The point is that being holomorphic is more restrictive than simply being differentiable in the real variable case. Think about what a complex differentiable function looks like. Let z = x + iy. We have f(x + iy) = u(x; y) + iv(x; y), as a function in two real variables, but the partial derivatives interact as in the Cauchy-Riemann equations. Consider h being purely real. Then u(x + h; y) + iv(x + h; y) − u(x; y) − iv(x; y) = u (x; y) + iv (x; y): h x x 2 But if we want f to be complex differentiable, this must be the same as in the case where h is complex imaginary. Then u(x; y + ih) + iv(x; y + h) − u(x; y) − iv(x; y) 1 = (u (x; y) + iv (x; y)): ih i y y Comparing yields the Cauchy-Riemann equations: ux = vy vx = −uy: We know that this is necessary to be complex differentiable; it is also sufficient. Theorem 1.10 (Looman and Menchoff). The Cauchy-Riemann equations imply that f is holomorphic. This is kind of technical and we won't prove it. Here's what we will prove; Proposition 1.11. If u(x; y) and v(x; y) have continuous partial derivatives and satisfy the Cauchy-Riemann equations, then f(z) = u(x; y) + iv(x; y) is holomorphic. Proof. We want to compute f(z + h) − f(z) h where h = k + il. This requires that we understand quantities like u(x + k; y + l) − u(x; y) and v(x + k; y + l) − v(x; y). Note that u(x + k; y + l) − u(x; y) = kux(x; y) + luy(x; y) + o(jkj + jlj): Here, o(jkj + jlj) means that a quantity that is < "(jkj + jlj) if jkj + jlj is sufficiently small. This is where we use the continuity of partial derivatives. Similarly, v(x + k; y + l) − v(x; y) = kvx(x; y) + lvy(x; y) + o(jkj + jlj): Adding these two equations and dividing by k + il gives ku (x; y) + ilv (x; y) + lu (x; y) + ikv (x; y) + o(jkj + jlj) x y y x ; k + il and applying the Cauchy-Riemann equations finishes the proof. Instead of thinking of f(z) as a function of the real and imaginary parts x and y, we can z+z z−z think of it as a function of z and z. Here, x = 2 and y = 2i . We want to consider d d differential operators dz f and dz f. Then we have d df dx df dy 1 df 1 df f = + = + dz dx dz dy dz 2 dx i dy d df dx df dy 1 df 1 df f = − = − : dz dx dz dy dz 2 dx i dy d 0 d Then the Cauchy-Riemann equations simply say that dz f = f (z) and dz f = 0, which means that f is only a function of z and not of z. 3 If f is holomorphic and f(x; y) = u(x; y) + iv(x; y) then the real and imaginary parts are harmonic, which means that uxx + uyy = 0 and vxx + vyy = 0. We'll come back to this later. This operator is called the Laplacian and is denoted 4. We will now define path integrals, or integrals over curves. We will only deal with curves that are piecewise smooth. Think of a parametrized curve as z :[a; b] ! C, connecting z(a) to z(b). We also get an orientation, pointing in the direction of increasing time. We are going to assume that z is continuously differentiable except at the endpoints, where we want one-sided limits, i.e. z(a + h) − z(a) lim : h!0+ h We also insist that z0(t) 6= 0 for all t, so the curve doesn't get stuck at any point. This is not strictly necessary. We can imagine some seemingly nice curves that are piecewise smooth, and we will allow these curves. This means that we divide [a; b] into intervals [a; a1]; [a1; a2];:::; [an; b], and the curve is smooth on each interval. Given a parametrized curve γ (which is the curve in C together with an orientation), we R R b 0 can think of γ f(z) dz = a f(z(t))z (t) dt. R R b 0 0 We also have path integrals γ(p dx + q dy) = a (p(x; y)x + q(x; y)y ) dt. This is independent of parametrization. This means that if we have z1 :[a; b] ! C and z2 :[c; d] ! C, they are equivalent if there exists a continuously differentiable function t :[c; d] ! [a; b] with z2(s) = z1(t(s)). The point is that the curve looks exactly the same in C, but it is parametrized differently. We require that t0(s) > 0 in order to preserve the orientation. Example 1.12.