Power Series Expansion?

Question

By Expanding e^(X(1+i)) in a power series, How can I show that....

a. e^x cos x = (The summation of n = o to infinity) (1/n!)(X^n)(2^(n/2))sin (3.14n/4)

3.14 = Pi

Answer 1

e^(x(1 + i)) = e^(x + ix) = e^x * e^(ix) = e^x(cosx + isinx)
then e^x cosx is the real part of e^(x(1 + i)).

e^(x(1 + i)) = sum from n = 0 to infinity of :
[x^n (1 + i)^n ] / n!


1 + i = sqrt(2) [cos(pi/4) + isin(pi/4)]
= 2^(1/2) [cos(pi/4) + isin(pi/4)]

(1 + i)^n = 2^(n/2) [cos(pi/4) + isin(pi/4)]^n
=2^(n/2) [cos(n*pi/4) + isin(n*pi/4)]

the real part is 2^(n/2) cos(n*pi/4)

and the result is :

e^x cosx = sum from n = 0 to infinity of :
x^n 2^(n/2) cos (n*pi/4) / n! (and not sin, I think !)