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5. The Exponential Function and its Friends The exponential function. Define exp z = ∞ ∑ zn . n! n=0 This has radius of convergence R where 1/R = lim sup(1/n!)1/n = 0, n→∞ so R = ∞ and the series converges absolutely for all z ∈ C. We may differentiate term by term (Theorem!) to get ∞ ∞ ∞ ∑ ∑ ∑ d z n−1 z n−1 zn exp z = n = = = exp z. dz n! (n − 1)! n! n=1 n=1 n=0 Exponential of a sum. Set f (z) = exp z exp(c − z), c ∈ C. Then d d f ′ (z) = exp z exp(c−z)+exp z exp(c−z) = exp z exp(c−z)−exp z exp(c−z) = 0. dz dz By Theorem 3.5, this tells us f (z) is constant. This constant must be f (0) = exp c, so exp z exp(c − z) = exp c. Set c = z1 + z2 , z = z1 , then exp(z1 + z2 ) = exp z1 exp z2 . From this it is easy to deduce (induction) that for n ∈ Z+ , exp nz = (exp z)n . The number e. Define e = exp 1(= 2.7182818 . . . ). The above shows exp n = en , n ∈ Z+ . Clearly exp 0 = 1 = e0 . Hence exp n exp(−n) = exp(n − n) = exp 0 = 1 so exp(−n) = (exp n)−1 = e−n . So exp(n) = en , n ∈ Z. For a rational number m/n (n > 0), (exp(m/n))n = exp(nm/n) = exp m = em , so exp(m/n) = (em )1/n = em/n . It is then consistent with this to use the notation exp z = ez , for all z ∈ C. The formula (*) becomes ez1 +z2 = ez1 ez2 . If z = x + iy, we have ez = ex eiy . Notice also that exp z exp(−z) = exp 0 = 1, so exp z ̸= 0 for all z ∈ C. 1 Trigonometric functions. Define cos z = ∞ ∑ (−1)n n=0 ∞ ∑ z 2n , (2n)! z 2n+1 sin z = (−1) . (2n + 1)! n=0 n By the Ratio Test, these converge absolutely for all z ∈ C. Substituting −z, we see cos is an even function and sin is an odd function, i.e., sin(−z) = − sin z. cos(−z) = cos z, Also cos 0 = 1, sin 0 = 0. Differentiating term by term, d cos z = − sin z, dz For example, d sin z = cos z. dz ∞ ∑ d z 2n sin z = (−1)n · (2n + 1) dz (2n + 1)! n=0 = ∞ ∑ (−1)n n=0 z 2n = cos z. (2n)! Theorem. (Euler’s Formula) eiz = cos z + i sin z. Proof. z2 iz 3 z4 iz 5 eiz = 1 + iz − − + + + ··· 2! 3! 4! 5! ( ) ( ) z2 z4 z3 z5 = 1− + + ··· + i z − + + ··· 2! 4! 3! 5! = cos z + i sin z. 2