Complex Variables: Expressing f(z)=cos(z)

[sara_87]

Homework Statement

let:
f(z)=u(x,y)+iv(x,y)
I want to express the following function like the one above:
$$f(z)=cos(z)\equiv=\frac{1}{2}(e^{iz}+e^{-iz})$$

Homework Equations

(i=sqrt(-1))
f(z) is a complex function

The Attempt at a Solution

$$f(z)=cos(z)\equiv=\frac{1}{2}(e^{iz}+e^{-iz})$$
$$=\frac{1}{2}(e^{i(x+iy)}+e^{-i(x+iy)})=\frac{1}{2}(e^{x-iy}+e^{-xi+y})$$

this is where i stopped because i got stuck. any help please?

 

[Mark44]

Homework Statement

let:
f(z)=u(x,y)+iv(x,y)
I want to express the following function like the one above:
$$f(z)=cos(z)\equiv=\frac{1}{2}(e^{iz}+e^{-iz})$$

Homework Equations

(i=sqrt(-1))
f(z) is a complex function

The Attempt at a Solution

$$f(z)=cos(z)\equiv=\frac{1}{2}(e^{iz}+e^{-iz})$$
$$=\frac{1}{2}(e^{i(x+iy)}+e^{-i(x+iy)})=\frac{1}{2}(e^{x-iy}+e^{-xi+y})$$

this is where i stopped because i got stuck. any help please?

Check your work in the expression before the last =.
i(x + iy) = ix + i²y = -y + ix. The other one is OK.

So one of your expressions will be e^{-y + ix} = e^-ye^ix = e^-y(cos x + i sin x) = e^-y cos x + i e^-y sin x. Do about the same thing to the other expression and add the two together.