Find Derivative of f(z) Where z = a+ib

[8daysAweek]

Find all points where the function has a derivative. At which of these points the function is analytical.$$<br /> f(z) = \left\{<br /> \begin{array}{ll}<br /> {z^2}sin(\frac{1}{|z^2|}) & z \neq 0 \\<br /> 0 & z = 0}<br /> \end{array}<br /> \right.<br />$$

I have tried deriving directly using the limit and also tried using Cauchy-Riemann, both tries led to complicated formulas.

For example Cauchy-Riemann approach:
$$f(a+ib) = \underbrace{ (a^2-b^2)sin(1/\sqrt{a^2+b^2})}_{u} +<br /> i\underbrace{{2ab}\cdot{sin(1/\sqrt{a^2+b^2})}}_{v}$$
Now I need to calculate dv/da ,du/da ,dv/db,du/db, but this seems like a headache.

 

[snipez90]

Why do you have square roots over the a^2 + b^2 expressions? Isn't there a |z^2| = |z|^2 = a^2 + b^2 term in the original function?

Also taking partial derivatives is easy. Just check the Cauchy-Riemann equations for z = a + ib =/= 0. At the origin, you'll probably have to resort to the definition anyways. You might already know the answer at least for the origin since there is a real-valued function that is similarly defined and often used as a counterexample.

 

[8daysAweek]

Why do you have square roots over the a^2 + b^2 expressions? Isn't there a |z^2| = |z|^2 = a^2 + b^2 term in the original function?

Also taking partial derivatives is easy. Just check the Cauchy-Riemann equations for z = a + ib =/= 0. At the origin, you'll probably have to resort to the definition anyways. You might already know the answer at least for the origin since there is a real-valued function that is similarly defined and often used as a counterexample.

Yes, my mistake, it should be a^2+b^2 without the root.
But, still it does not make finding partial derivatives simple. For example, here is du/da calculated by WolframAlpha:

Can you explain the real-valued function method? How do you use it as a counter example?