Re: Complex Analysis

nonton wrote:
> How would i go about showing that if each w and z are complex numbers
> then E(w)E(z)=E(w+z)
-snip-
> How do i show that f(0)=1 without using trig. identities or demoivres
> law and just by definition.
>
> This is by the way how an exponential functionwas defined in my
> notes.i.e. An exponential function E is that entire function f such
> that f'=f and f(0)=1.
-snip-

Well if it's defined as being the function that has f(0) = 1, then you
can't "prove" it as it's true by definition. If you use the series
definition and show that d/dt (exp(tz)) = z exp(tz), (from the series
def it's clear that exp(0) = 1, then I offer the following proof:

f(t) = exp(tw)exp(tz) - exp(t(z+w))

Now since exp(0) = 1, then f(0) = 0, further differentiate:

f'(t) = w(exp(tw)exp(tz)) + z(exp(tw)exp(tz)) - (z+w)exp(t(z+w))
= (z+w)(exp(tw)exp(tz) - exp(t(z+w)))
= (z+w) f(t)

Now this is an ordinary differential equation and thus a solution is
unique by picards theorem. (it may seem like solving the f'=f equation,
but note here that f(0) = 0!), now f(t) = 0 is a solution and thus it
is the unique solution and thus in particular we must have that f(1) =
0 and so

exp(w)exp(z) - exp(z+w) = 0

which is what you wanted. Not sure if that helps. It all really
depends on how you define the exponential function for complex numbers.
I suppose most books use the series definition.

Jiri

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