Many texts give an alternative definition for exp(z), starting with Identity (5.1) as the definition for f(z)=exp(z). Recall that this identity states that exp(z)=exp(x+iy)=ex(cosy+isiny). This exercise shows such a definition is a natural approach in terms of differential equations. We start by requiring f(z) to be the solution to an initial-value problem satisfying three conditions: (1) f is entire, (2) f′(z)=f(z) for all z, and (3) f(0)=1. Suppose that f(z)=f(x+iy)=u(x,y)+iv(x,y) satisfies conditions (1), (2), and (3).
((a)) Use the result f′(z)=ux(x,y)+ivx(x,y) and the requirement f′(z)=f(z) from condition (2) to show that ux(x,y)−u(x,y)=0, for all z=(x,y).
((b)) Show that the result in part (a) implies that ∂x∂[u(x,y)e−x]=0. This means u(x,y)e−x is constant with respect to x, so u(x,y)e−x=p(y), where p(y) is a function of y alone.
((c)) Using a similar procedure for v(x,y), show we wind up getting a pair of solutions u(x,y)=p(y)ex, and v(x,y)=q(y)ex where p(y) and q(y) are functions of y alone.
((d)) Now use the Cauchy-Riemann equations to conclude from part (c) that p(y)=q′(y) and p′(y)=−q(y).
((e)) Use part (d) to show that p′′(y)+p(y)=0 and q′′(y)+q(y)=0.
((f)) Identify the general solutions to part (e). Then, given the initial conditions f(0)=f(0+0i)=u(0,0)+iv(0,0)=1+0i, find the particular solutions and conclude that Identity (5.1) follows.