Complex Analysis - Is the function f : $z → $ $\overline z $ $(1 + z^ 2 )$ complex differential on C?

Question

So what I've done is

$ \overline Z = x-iy $

= $ (x-iy)(1+(x+iy)^2) $

= $ (x-iy)(1+x^2+2ixy+y^2) $

= $ (x + x^3 + 2i(x^2)y +xy^2 -iy -iyx^2 - 2xy^2 + iy^3) $

= $ (x+x^3-xy^2) + i(yx^2-y-y^3) $

$ du/dx = 1 + 3x^2 - y^2 -xy^2 $

$ dv/dy = yx^2 +x^2 -1 - 3y^2 $

hence $ du/dx = dv/dy $ is not applicable, hence not differentiable .

Is this correct?

Answer 1

The answer is no, this function isn’t derivable in C.

You have some errors in proceeding:

$$ f(z) = (x-iy)(1+(x+iy)^2) $$

$$= (x-iy)(1+x^2+ 2ixy - y^2) $$

$$= (x + x^3 + 2i(x^2)y - xy^2 -iy -iyx^2 - 2xy^2 - iy^3) $$

$$= (x+x^3-xy^2) + i(yx^2-y-y^3)$$

$ du/dx = 1 + 3x^2 - y^2 $

$ dv/dy = -1 + x^2 - 3y^2 $

So they’re not equal, concluding that the function isn’t derivable.

Answer 2

The function you wrote in the title is different from the one in the text. Anyway, I think that the procedure is essentially correct. It might be easier to write the Cauchy-Riemann equations in the form $$ \frac{\partial f}{\partial \overline z} = 0, $$ because your function (the one in the title) can be rewritten as $$ f(z, \overline z)=\overline z ( 1+\overline z^2), $$ so $\partial_{\overline z}\, f = (1+\overline z^2) + 2 \overline z^2\ne 0$.