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I am quoting the solution of this post (https://math.stackexchange.com/a/2924208/603303)

I have no idea why the function $f(z)$ is broken down to $f(x)$ and $f(iy)$ to find the partial derivatives. Assuming we can do this then the rest is clear, but, for example,

if $f(x+iy) = yx$

then $f_x = y$ and $f_y = x$

However, the solution above plugs in 0 and then computes the partials. Doing so here, we get that $f_x(x) = 0 = f_y(iy)$

Whereas in reality, $f_x = y$ and $f_y = x.$

So, can someone please help explain why his method works (even if it only works for the specific function he is considering at the point he is considering).

The function he is considering is $f(z) = z^5/\mathrm{abs}(z)^4$ where $\mathrm{abs}(z)$ is the norm.

In polar coordinates, this turns into

$$f(r*\exp(iw)) = r * \exp(i * 5w) = r * (\cos(5*w) + i * \sin(5*w))$$ (where * is multiplication)

Hence, we have $f = u + i * v$ where

$u(r,w) = r * \cos(5w)$ and $v(r,w) = r * \sin(5w)$

Hence,

$u_r = \cos(5w)$ and $v_r = \sin(5w)$

and

$u_w = -5r * \sin(5w)$ and $v_w = 5w * \cos(5w)$

Now, because of this factor of 5, we get that the Cauchy Riemann equations don't hold.

Where is my error?

Presumably, it is because the denominator $\exp( i * 4w)$ should cancel out 4 of the $iw$ multiples in the numerators argument, however, since we have absolute value, the denominator should equal to

$$r^4 * \mathrm{abs}(\exp( i * 4w))$$

and since $\mathrm{abs}(exp(z)) = 1$ for all complex numbers z on the imaginary axis (eg $z = iw$), this confuses me greatly.

Can someone please help me understand what is going on here?

dsh
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Jonathan Valenzuela
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  • Use MathJax. ${}{}$ – Arturo Magidin Oct 18 '23 at 04:38
  • https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – geetha290krm Oct 18 '23 at 04:43
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    They are only interested in the partial derivatives at a specific point, because they want to show that CR-equations$\Rightarrow$complex-differentiable fails at that specific point. Not necessarily at all points. – Vercassivelaunos Oct 18 '23 at 04:50
  • I actually apologize now for this question. I’m simply not used to a technique like letting one variable be the coordinate (0) of the point (0,0) we’re interested in, but the logic is sound now that I view it with fresh eyes – Jonathan Valenzuela Oct 18 '23 at 21:49

1 Answers1

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The CR equations are criteria which needs to be satisfied by any function, for the derivative of the function to exist at any given point. But, they are not sufficient since we are not sure of whether the partial derivatives participating in the equations themselves exist at the given point.

The answerer as you quoted, has taken the cases of $z = x +i0$ and $z = 0+iy$ and elucidated that $f(z)=z^5/(abs(z))^4$ satisfies CR equations at the point $(0,0$). But, that ain't sufficient since the limit does not exist. So, f is not differentiable at $(0,0)$.

In your polar form for $f$, The CR equations are satisfied at the point $(0,0)$ i.e They are solutions of the equations $ru_r - v_\omega = 0$ and $rv_r + u_\omega = 0$.