LECTURE 4: DERIVATIVE OF POWER SERIES AND COMPLEX EXPONENTIAL The reason of dealing with power series is that they provide examples of analytic functions. P1 n Theorem 1. If n=0 anz has radius of convergence R > 0; then the function P1 n F (z) = n=0 anz is di®erentiable on S = fz 2 C : jzj < Rg; and the derivative is P1 n¡1 f(z) = n=0 nanz : Proof. (¤) We will show that j F (z+h)¡F (z) ¡ f(z)j ! 0 as h ! 0 (in C), whenever h ¡ ¢ n Pn n k n¡k jzj < R: Using the binomial theorem (z + h) = k=0 k h z we get F (z + h) ¡ F (z) X1 (z + h)n ¡ zn ¡ hnzn¡1 ¡ f(z) = a h n h n=0 µ ¶ X1 a Xn n = n ( hkzn¡k) h k n=0 k=2 µ ¶ X1 Xn n = a h( hk¡2zn¡k) n k n=0 k=2 µ ¶ X1 Xn¡2 n = a h( hjzn¡2¡j) (by putting j = k ¡ 2): n j + 2 n=0 j=0 ¡ n ¢ ¡n¡2¢ By using the easily veri¯able fact that j+2 · n(n ¡ 1) j ; we obtain µ ¶ F (z + h) ¡ F (z) X1 Xn¡2 n ¡ 2 j ¡ f(z)j · jhj n(n ¡ 1)ja j( jhjjjzjn¡2¡j) h n j n=0 j=0 X1 n¡2 = jhj n(n ¡ 1)janj(jzj + jhj) : n=0 P1 n¡2 We already know that the series n=0 n(n ¡ 1)janjjzj converges for jzj < R: Now, for jzj < R and h ! 0 we have jzj + jhj < R eventually. It thus follows from F (z+h)¡F (z) above that j h ¡ f(z)j ! 0 as h ! 0; whenever jzj < R: ¤ We are now going to de¯ne the complex analogue of the exponential function, that is, ex: 1 2 LECTURE 4: DERIVATIVE OF POWER SERIES AND COMPLEX EXPONENTIAL z P1 zn De¯nition 2. (Exponential function) We de¯ne e = n=0 n! for all z 2 C: Since j an+1 j = 1 ! 0; the series converges for all z 2 C: The following theorem an (n+1) summarizes important properties of the exponential. Theorem 3. The function f(z) = ez is analytic on C and satis¯es the following properties d i) (ez) = ez ii) ez1+z2 = ez1 ez2 iii) eiθ = cos θ + i sin θ; θ 2 R: dz Proof. By the previous result ez is an analytic function on C and d X1 n X1 1 (ez) = zn¡1 = zn¡1 = ez: dz n! (n ¡ 1)! n=1 n=1 We de¯ne g(z) = ezez1+z2¡z: Then g is analytic on C and g0(z) = 0 for all z 2 C: It follows from CR equations that g(z) = ® for some ® 2 C: Since g(0) = ® = ez1+z2 z z1+z2¡z z1+z2 we get that e e = e : By choosing z = z1 the result follows. For the last part we have, X1 (iθ)n X1 1 1 eiθ = = [ (iθ)2n + (iθ)2n+1] n! (2n)! (2n + 1)! n=0 n=0 X1 θ2n(i2)n θ2n+1(i2)n = [ + i ] = cos θ + i sin θ: (2n)! (2n + 1)! n=0 ¤ We have the following observations to make. (1) Since eze¡z = 1 it follows that ez 6= 0 for all z 2 C: (2) ez = ex+iy = ex(cos y + i sin y) and ez = ez¹ = exe¡yi: (3) ez+2n¼i = ex(cos(2n¼ + y) + i sin(2n¼ + y)) = ex(cos y + i sin y) = ez: Thus complex exponential is a periodic function with period 2¼i and hence it is not injective, unlike the real exponential. (4) It follows now easily that ez = 1 () z = 2n¼i for some n 2 Z and hence z1 z2 e = e () z2 = z1 + 2n¼i; for some n 2 Z: Surjectivity of Exponential: We know that complex exponential is not an injec- tive function however it is surjective from C to Cnf0g: If w 2 Cnf0g then using polar coordinates we can write w = jwjeiθ where θ 2 (¡¼; ¼]: If we de¯ne z = log jwj + iθ then ez = elog jwj+iθ = elog jwjeiθ = w: In fact, by our previous observation, it follows LECTURE 4: DERIVATIVE OF POWER SERIES AND COMPLEX EXPONENTIAL 3 that elog jwj+i(θ+2n¼i) = w (not surprising as exponential is not injective). Thus we have that elog jwj+iArgw = w: It follows from the above discussion that if we restrict the domain of the expo- nential then it becomes injective. In fact, if H = fz = x + iy : ¡¼ < y · ¼g then z ! ez is a bijective function from H to C n f0g: We can also understand now image of certain subsets of H under the exponential. For example, for a ¯xed y0 2 (¡¼; ¼] if A = fx+iy0 : x 2 Rg (which is a line parallel to the real axis) then its image under exponential is fexeiy0 : x 2 Rg which is a one sidded ray with angle y0: If for a ¯xed x0 2 R;B = fx0 + iy : y 2 (¡¼; ¼]g (which is part of a line parallel to the imaginary axis) then its image under exponential is fex0 eiy : y 2 (¡¼; ¼]g which is a circle about origin with radius ex0 :.