Showcasing f(z)=u(x,y)+i*v(x,y) Properties

[neginf]

Trying to show any f(z)=u(x,y) + i*v(x,y) with the 3 properties:
1. f(x+0*i)=e^x,
2. f(z) is entire,
3. f ' (z)=f(z) for all z,
has to be f(z)=(e^x) * (cos y + i*siny).

The hint in the book (Complex Variables and Applications, 6 ed., Churchill+Brown) says:
(a) Get u=u sub x and v=v sub y, show there are g(y), h(y) such that u(x,y)=(e^x) * g(y)
and v(x,y)=(e^x) * h(y). did this
(b) Use u(x,y) being harmonic to get g''(y)+g(y)=0 and so g(y)=A*cos y+B*sin y for real
constants A and B. did this
(c) Point out why h(y) = A*sin y-B*cos y and note g(0)+i*h(0)=1, find A and B and conclude
that f(z)=(e^x) * (cos y + i*siny). did the h(y)=-g'(y)=A*sin y-B*cos y
part but cannot get the g(0)+ i*h(0)=1 part. I get g(0)+i*h(0)=A-i*B and get stuck.

 

[Some Pig]

$$f(z)=u(x,y)+iv(x,y)$$
$$=e^x(g(y)+ih(y))$$
$$f(x+0i)=e^x(g(0)+ih(0))=e^x$$
Compare real and imaginary parts, g(0)=1, h(0)=0.

 

[neginf]

Thank you very much.

 

[Mark44]

That really was an "e^z" question!