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

• fishturtle1
In summary: Otherwise, you can simplify the calculation by using the original form for e-z.In summary, the function is notholomorphic, not differentiable, and not entire.
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

∂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 ux = -(-sin(-x)cos(-y)) + cos(-x)sin(-y)

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

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

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

so ux =/= vy

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.

Last edited:
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.

FactChecker said:
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?

fishturtle1 said:
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 said:
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.

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.

Have you already established in class that ez is an entire function?
If so, the easiest proof is:
1) g(z) = -z and h(z)=ez 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.

## What is a complex function?

A complex function is a mathematical function that takes a complex number as an input and produces a complex number as an output. It can be written in the form f(z) = u(x,y) + iv(x,y), where u and v are real-valued functions of the variables x and y, and i is the imaginary unit.

## What does it mean for a function to be holomorphic?

A function is holomorphic if it is complex differentiable at every point in its domain. This means that the function is smooth and has a well-defined derivative at every point. In other words, it is a function that can be locally approximated by a linear function at each point.

## What is the difference between a holomorphic function and a differentiable function?

A holomorphic function is a special case of a differentiable function, where the derivative exists and is complex at every point in its domain. This means that a complex differentiable function is always holomorphic, but a holomorphic function is not necessarily complex differentiable at every point.

## What are some properties of holomorphic functions?

Some important properties of holomorphic functions include the Cauchy-Riemann equations, which describe the relationship between the real and imaginary parts of a holomorphic function, and the Cauchy integral theorem, which states that the integral of a holomorphic function along a closed contour is equal to 0.

## What are some applications of complex function holomorphic?

Complex function holomorphic has many applications in physics, engineering, and mathematics. It is used to model physical phenomena such as fluid flow, electrical circuits, and quantum mechanics. It is also used in signal processing, image processing, and data analysis. In mathematics, it plays a crucial role in complex analysis and number theory.

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