0
$\begingroup$

Let $f(z)=\bar{z}e^{{-\lvert z\rvert}^2}$. We can show using Cauchy-Riemann equations that $f$ is differentiable at points $w$ satisfying $\lvert w \rvert=1$, that is on the unit circle. I was attempting to find $f'(w)$, and I'm now wondering how to do it. For if it weren't possible, would the function be differentiable in the first place? [By definition of differentiability, the derivatives must exist].

$$f'(w)=\lim\limits_{h\to 0} \frac{(\bar{w}+\bar{h})e^{-\left| w+h\right|^2}-\bar{w}e^{-\left| w\right|^2}}{h}$$ $$=\lim\limits_{h\to 0} \frac{(\bar{w}+\bar{h})e^{2\Re\{\bar{w}h\}}e^{-\left| h\right|^2}-\bar{w}}{eh}$$

But how can we proceed next? Do we use l'Hôpital's rule and differentiate the numerator and the denominator, or is there some other way?

  • 1
    Why you didn't use $f'(z)=u_x+iu_y$.? – 2017-01-18
  • 0
    @MyGlasses, we'll then get $e^{-1} [(1-2x^2)-i2xy]$. How can this be expressed in terms of $z=x+iy$? – 2017-01-18
  • 0
    Actually we can solve from $x^2+y^2=1$ and $z=x+iy$, but the derivative becomes quite tedious. – 2017-01-18
  • 0
    But $1-2x^2-2ixy=-z^2$. – 2017-01-18
  • 1
    $-z^2=-(x^2-y^2+2ixy)=-x^2+(1-x^2)-2ixy$ on $|z|=1$. – 2017-01-18
  • 0
    @MyGlasses You might want to consider giving this as an answer. – 2017-01-18
  • 0
    @sequence edited my answer to take this method into account (and correct a couple small errors/typos I made). The imaginary part in the formula you're using has the wrong sign. When those minor errors are fixed, the two answers match up. – 2017-01-19

1 Answers 1

2

Let $z_0 = e^{i\theta}$ on the unit circle and $h = re^{i\phi}.$ Then $|z_0+h|^2 = 1+r^2 + 2r\cos(\theta-\phi)$ and we have $$f(z_0+h) = (e^{-i\theta}+re^{-i\phi})e^{-(1+r^2+2r\cos(\theta-\phi))} \\ f(z_0) = e^{-i\theta}e^{-1}$$

We will take the limit $r\rightarrow 0 $ to approach $z_0$ from the angle $\phi.$ Expanding to lowest order in $r$, $$ f(z_0+h)-f(z_0)\approx \frac{1}{e}(e^{-i\theta}+re^{-i\phi})(1-2r\cos(\theta-\phi)) - \frac{1}{e}e^{-i\theta} \approx r\frac{e^{-i\phi}-2e^{-i\theta}\cos(\theta-\phi)}{e}.$$

Dividing through by $h=re^{i\phi}$ gives $$ \left.\frac{df}{dz}\right|_{z=e^{i\theta}}= \lim_{r\rightarrow0}\frac{f(z_0+h)-f(z_0)}{h} = \frac{e^{-2i\phi}-2e^{-i(\theta+\phi)}\cos(\theta-\phi)}{e} = -\frac{e^{-2i\theta}}{e}.$$

The answer comes out independent of $\phi$ as it must since the function is complex-differentiable at $z_0 = e^{i\theta}.$

EDIT:

I wanted to follow the method using the definition of the derivative outlined in the question and show how you get an answer that is independent of the direction the limit is taken in. However, as commenter MyGlasses noted, given that you have proven it is differentiable, there is an easier way to compute the derivative. If $f(x+iy) = u(x,y) + iv(x,y)$ and the Cauchy-Riemann equations hold, then the derivative is given by $$f'(z) = \frac{\partial u}{\partial x} -i\frac{\partial u}{\partial y}.$$ In this case,this gives $$ f'(z) = (1-2x^2)e^{-(x^2+y^2)} +2ixye^{-(x^2+y^2)}.$$ Since we're on the unit circle, $x^2 + y^2 = 1$ and $1-2x^2 = y^2-x^2$ so $$ f'(z) = \frac{1}{e}\left(y^2-x^2 + 2ixy)\right) = -\frac{\bar z^2}{e}.$$

If we write $z=e^{i\theta},$ we see that the two answers match.

As a third alternative, we can consider $z$ and $\bar z$ to be the indpendent variables rather than $x$ and $y$. Then the function can be expressed as $$ f(z) = \bar z e^{-z\bar z}$$ and we have $$\frac{\partial f}{\partial z} = -\bar z^2 e^{-z\bar z} \\ \frac{\partial f}{\partial \bar z} = (1-z\bar z)e^{-z \bar z}.$$ In this language, the Cauchy-Riemann equations are just the statement that $\frac{\partial f}{\partial \bar z} = 0,$ which means $z\bar z = 1,$ i.e. $z$ is on the unit circle like we knew all along. Under these conditions, $f$ is differentiable and we have $$f'(z) =\frac{\partial f}{\partial z} = -\bar z^2 e^{-z\bar z} = -\frac{\bar z^2}{e}$$ where in the last equality we plugged in $z\bar z = 1.$

  • 0
    What are you expanding about? Also, how do you know the derivative is precise if there is an approximation sign? – 2017-01-18
  • 1
    I am expanding about $r=0$ where $r$ is the modulus of $h$. The limit $r\rightarrow0$ takes $h\rightarrow0$ along the line at angle $\phi.$ I keep only the term proportional to $r$ in the expansion since higher order terms will go to zero when the limit $r\rightarrow 0$ is taken. This is just like when you differentiate $x^2.$ You take $(x+h)^2-x^2 \approx 2hx$ so $((x+h)^2-x)/h \rightarrow 2x.$ You could omit the $h^2$ term since it vanishes in the limit. – 2017-01-18
  • 0
    Are you expanding $f(z+h)$ in Taylor series around $r$? I'm a bit confused as I don't see how this expansion exactly works. – 2017-01-18
  • 1
    @sequence Yes. So keeping terms up to linear in $r,$ the $\exp(-(1+r^2+2r\cos(\theta-\phi))) factor$ becomes approximately $(1-2r\cos(\theta-\phi))/e$ and multiplying this by $(e^{-i\theta} + re^{-i\phi})$ gives approximately $re^{-i\theta}+re^{-i\phi}-2re^{-i\theta}\cos(\theta-\phi)$ – 2017-01-18
  • 1
    @sequence edited again to do things in terms of $z$ and $\bar z.$ It's probably the quickest and easiest way. – 2017-01-19
  • 0
    @sequence Cause it's (one of) the correct expression(s) and the other isn't. http://mathworld.wolfram.com/Cauchy-RiemannEquations.html – 2017-01-20