1. ## [SOLVED] Complex Problem

Consider the complex function

f(z) = $e^{-iz}$.

(a) Express f(z) in the form u(x, y) + iv(x, y), give u and v.

(b) Use the Cauchy-Riemann equations and the existence and continuity of the partial derivatives of u and v to show that f '(z) exists for all z.

(c) Using your results from b), find an expression for f '(z) in terms of z.

Help with this question please, Thanks.

2. $e^{iz} = e^{\left( { - y + ix} \right)} = \left[ {e^{ - y} \cos (x)} \right] + i\left[ {e^{ - y} \sin (x)} \right]$
Now, you finish the other parts.

3. Sorry I made a mistake its ie^{-iz}

4. Originally Posted by ronaldo_07
Sorry I made a mistake its ie^{-iz}
So? If you know what $e^{-iz}$ is (what Plato did) then it is easy to find $ie^{-iz}$.

5. $ie^{iz} = e^{\left( { - y + ix} \right)} = \left[ {e^{ - iy} \cos (x)} \right] + i\left[ {e^{ - iy} \sin (x)} \right]$

is this correct?

6. Originally Posted by Plato
$e^{iz} = e^{\left( { - y + ix} \right)} = \left[ {e^{ - y} \cos (x)} \right] + i\left[ {e^{ - y} \sin (x)} \right]$
Now, you finish the other parts.

should it not be y-xi?

7. Originally Posted by ronaldo_07
$ie^{iz} = {\color{red}\large i}e^{\left( { - y + ix} \right)} = \left[ {{\color{red}\large i}e^{ - iy} \cos (x)} \right] {\color{red}\large -} \left[ {e^{ - iy} \sin (x)} \right]$
You missed a sign and two i's. See the red above.

Originally Posted by ronaldo_07
should it not be y-xi?
No indeed!
$i(x + yi) = ix - y$

8. Originally Posted by ronaldo_07
$ie^{-iz} = e^{\left( { - y + ix} \right)} = \left[ {e^{ - iy} \cos (x)} \right] + i\left[ {e^{ - iy} \sin (x)} \right]" alt="ie^{-iz} = e^{\left( { - y + ix} \right)} = \left[ {e^{ - iy} \cos (x)} \right] + i\left[ {e^{ - iy} \sin (x)} \right]" />

is this correct?
The - sign was there when i wrote the question origionally.