MHB Analyzing $f(z) = e^{-\frac{1}{z}}$: Analytic Region & Derivative

[ognik]

Given $ f(z) = e^{-\frac{1}{z}} $, find f'(z) and identify the maximal region within which f(z) is analytic

I found [math] f'(z) = \frac{e^{-\frac{1}{z}}}{z^2} [/math], is that right?

I think I should be using the Cauchy-Riemann Conditions to check if analytic, but this function is not in the form u+iv? A hint please?

 

[ThePerfectHacker]

Let $z = x+iy\not = 0$.

Verify $\frac{1}{z} = \frac{x}{x^2+y^2} - \frac{y}{x^2+y^2}i$ and use $e^{a+bi} = e^a(\cos b + i\sin b)$.
With that you can find the $u$ and $v$.

 

[ognik]

Thanks TPH, I followed that approach to get $$U= e^{-\frac{x}{x^2 + y^2}}Cos \frac{y}{x^2+y^2} $$ and $$V= e^{-\frac{x}{x^2 + y^2}} Sin \frac{y}{x^2+y^2} $$

To do those 4 far-from-easy partial derivatives seems excessive for an exam question - which counts only 7%; also why did they ask to find the derivative of the function first? Is there not perhaps a simpler approach to the question?

 

[ThePerfectHacker]

Thanks TPH, I followed that approach to get $$U= e^{-\frac{x}{x^2 + y^2}}Cos \frac{y}{x^2+y^2} $$ and $$V= e^{-\frac{x}{x^2 + y^2}} Sin \frac{y}{x^2+y^2} $$

To do those 4 far-from-easy partial derivatives seems excessive for an exam question - which counts only 7%; also why did they ask to find the derivative of the function first? Is there not perhaps a simpler approach to the question?

There is a simpler approach, but it does not use CR-equations. I was only doing it that way because I thought you wanted to see it by CR equation.

The easier method is that if $f(z)$ is analytic on an open set $U$ then $e^{f(z)}$ is analytic on that same set. If $g(z)$ is analytic on $U$ and not equal to zero at any point then $f(z)/g(z)$ is analytic on $U$ as well.

Pick $U$ to be the complex plane minus the origin. Choose $f(z) = -\frac{1}{z}$ and $g(z) = z^2$. Do you accept the fact that $f$ is analytic on $U$ and $g$ is analytic on $U$ and not equal to zero? Therefore, $e^{f(z)}/g(z)$ will be analytic on $U$.

Then we need to show that $U$ is maximal such region. The only region bigger than $U$ would be $\mathbb{C}$ itself by adjoining the origin to this larger region. So you need to give an argument why it is not possible to extend $e^{-1/z}/z^2$ analytically to the entiretly of the complex plane.

 

[ognik]

I would say that both f(z) and g(z) are differentiable and smooth, but have a singularity at z=0. I would conclude that they are analytic everywhere - which includes the complex plane - except at z=0?

I still don't see the reason we find f'(z) and thereby g(z)?

Thanks

 

[ognik]

Further to my previous point, I would say that we are asked to find the 1st derivative because if the function is infinitely identifiable and is smoothly continuous (except at z=0), then the function is analytic everywhere except at z=0.

Is this argument valid please?