Where is f(z) = e-xe-iy differentiable and holomorphic?

[fishturtle1]

Homework Statement

Suppose z = x + iy. Where are the following functions differentiable? Where are they holomorphic? Which are entire?

the function is f(z) = e^-xe^-iy

Homework Equations

∂u/∂x = ∂v/∂y

∂u/∂y = -∂v/∂x

The Attempt at a Solution

f(z) = e^-xe^-iy

I convert it to polar form:

f(z) = cos(xy) + isin(xy)

then i set u as the real part of the equation, and v as the complex part of the equation.

u = cos(xy)
v = isin(xy)

∂u/∂x = -ysin(xy)
∂v/∂y= ixcos(xy)

then ∂u/∂x =/= ∂v/∂y so the function is not holomorphic.

I think i did this wrong because I've done this for the majority of the problem set, and every function is coming up as not holomorphic. Is my work correct?

EDIT: I did it again and here is my new work, my other work was wrong because i multiplied exponents wrong

f(z) = e^-xe^-iy

= (cos(-x) + isin(-x)) * (cos(-y) + isin(-y))
= cos(-x)cos(-y) + icos(-y)sin(-x) + (-1)sin(-x)(sin-y) + isin(-x)cos(-y)

so Re(f(z)) = cos(-x)cos(-y) - sin(-x)sin(-y) = u
Im(f(z)) = cos(-y)sin(-x) + cos(-x)sin(-y) = v

then u_x = -(-sin(-x)cos(-y)) + cos(-x)sin(-y)

u_x = sin(-x)cos(-y) + cos(-x)sin(-y)

and v_y = - (-sin(-y)sin(-x)) + cos(-x)cos(-y)

v_y = sin(-y)sin(-x) - cos(-x)cos(-y)

so u_x =/= v_y

so f(z) is not holomorphic, and is not differentiable at all points( but may be differentiable at some points ), and is not entire because it is not differentiable at all points.

 

[FactChecker]

It is holomorphic. There must be an error in your calculation. The calculations might be simpler if you leave e^-x as it is, since it is a real function whose derivative you know.

 

[fishturtle1]

It is holomorphic. There must be an error in your calculation. The calculations might be simpler if you leave e^-x as it is, since it is a real function whose derivative you know.

I just want to confirm though, that if a complex function satisfies the Cauchy Riemann equations, then the complex function is also holomorphic. So my logic is right, but I probably just differentiated incorrectly somewhere?

 

[FactChecker]

I just want to confirm though, that if a complex function satisfies the Cauchy Riemann equations, then the complex function is also holomorphic.

With one caveat. The first partial derivatives of u and v must be continuous.

So my logic is right, but I probably just differentiated incorrectly somewhere?

Yes.

 

[fishturtle1]

I just want to confirm though, that if a complex function satisfies the Cauchy Riemann equations, then the complex function is also holomorphic. So my logic is right, but I probably just differentiated incorrectly somewhere?

Thank you for the help. I think my error is when I'm converting from exponential to polar form.

f(z) = e^-xe^-iy

= e^-iy-x

and now I need to rewrite -iy-x so that i can pull out the i, and have a real angle for my polar form.

So I know z = x+iy, then I can say -z = -iy - x

so I can rewrite: f(z) = e^-z

i'm unsure of whether I can treat z as a single variable and then say that e raised to any single variable will be differentiable at all points.

Am I on the right path here? I feel like I've made a couple assumptions in differentiating e^-z where z ∈ ℂ.

 

[FactChecker]

I don't think you need to make that big a change. It takes you straight into complex analysis and complex derivatives.

Just to make the calculations of your original approach easier, try:
f(z) = e^-xe^-iy
= e^-x×(cos(-y) + isin(-y))
= e^-x × cos(-y) + i*e^-x ×sin(-y)
The partial derivatives of the real and imaginary parts are probably easier in that form.

 

[FactChecker]

Have you already established in class that e^z is an entire function?
If so, the easiest proof is:
1) g(z) = -z and h(z)=e^z are both entire functions.
2) The composition of entire functions is entire. So f(z) = h(g(z)) is entire.

If those facts have not already been established, your original approach is the right thing to do.