Derivative of Complex Function

[mancini0]

Hi guys, I was hoping someone can check my work finding the following complex derivative:

Homework Statement

Find where the function is differentiable / holomorphic:

f(z) = e^-x * e^-iy

Homework Equations

I know I must satisfy the Cauchy - Reimann equations.

The Attempt at a Solution

I'm worried that I am not separating f(z) = e^-x * e^-iy into real and imaginary components (u,v) correctly.
In fact, I did not attempt to separate f(z) into the form f(z) = u(x,y) + v(x,y). Instead, I differentiated with
respect to f.

I found dx/ df = -e^(-x) * e^(-iy)
I found dy/ df = e^(-iy) * e^(-x)

Since dx/df = -i dy/df , the function is complex differentiable.

If that is incorrect, how would I go about seperating the original function into real and imaginary components?

Also, I noticed that f(z) = e^(-x)*e^(-iy) is equivalent to f(z) = e^-(x+iy) = f(z) = e^(-z). What can I do with this tidbit?

 

[vela]

Use the fact that e^iy = cos y + i sin y.

 

[mancini0]

Use the fact that e^iy = cos y + i sin y.

Then would e^-iy = -(cos y + i sin y)?

 

[vela]

No, you can't just pull the negative sign out of the exponent. Just follow the usual rules of algebra.

 

[mancini0]

Ahh, a negative exponent just gets pulled to the denominator... So e^-x * e^-iy = 1 / (e^x * e^iy)

= (e^x * e^iy)^-1

Which would lead to a complicated derivative... Would u = (e^x)^-1 , v =( e^iy)^-1 in this case? Or must I use the chain rule at this point?

 

[tedbradly]

$$f(z) = e^{-x}e^{-iy}$$
$$f(z) = e^{-x}cos(-y)+ie^{-x}sin(-y)$$
And using the common signed argument tricks for cosine and sine,
$$f(z) = e^{-x}cos(y)-ie^{-x}sin(y)$$

 

[mancini0]

Okay, thank you very much.