Derivatives of $f$ doesn't match accordingly to the Cauchy-Riemann equations? For the function $f(x+iy) = 2xy+i(x+\frac2 3y^3)$, I've decided that $f$ is differentiable at the points $-1/2$ and $-1/2 + i$ by using the Cauchy-Riemann equations:
$\frac {\partial u}{\partial x} = 2y$,  $\frac {\partial u}{\partial y} = 2x$,  $\frac {\partial v}{\partial x} = 1$ and $\frac {\partial v}{\partial y} = 2y^2$.
Here $u = \Re f$ and $v = \Im f$.
Now it should hold (according to my book) that $f^{'}(z) = \frac {\partial f}{\partial x}(x,y) = \frac 1 i \frac {\partial f}{\partial y}(x,y)$ .
But $\frac {\partial f}{\partial x}(x,y) = 2y+i$ and $\frac 1 i \frac {\partial f}{\partial y}(x,y) = \frac {2x} i + 2y^2$.
Equality holds in the case $z = -1/2$ but not in the case that $z = -1/2 + i$ ?
 A: $$f_x(-1/2+i)=2(1)+i=2+i$$
Remember that $1/i=-i$
$$-if_y(-1/2+i)=-i(2(-1/2)+2i(1)^2)=-i(-1+2i)=2+i$$
A: since the Cauchy-Riemann equations are of fundamental significance in complex analysis, it is worth looking beyond the detail of this example, to take in something of the general picture.
$\mathbb{C}$ is isomorphic to $\mathbb{R}^2$ as a 2-dimensional real vector space. but $\mathbb{C}$ also has a much richer structure.
as an endomorphism of $\mathbb{R}^2$ your function is:
$$
(f,g):(x,y) \mapsto \left(2xy,x+\frac2 3y^3 \right)
$$
its derivative is the following $2 \times 2$ matrix, which denotes a linear mapping of the (2-d) tangent space:
$$
\begin{pmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y}\\
\frac{\partial g}{\partial x} & \frac{\partial g}{\partial y} 
\end{pmatrix} = \begin{pmatrix} 2y & 2x\\ 1 & 2y^2\end{pmatrix}
$$
in general, except at singular points (where the determinant vanishes), this matrix will involve a degree of shearing in the tangent plane. however at certain points you find that the tangent transform splits nicely into a rotation combined with an isotropic (direction-independent) stretching or contraction. at such a point the transformation preserves angles, so is said to be conformal
this happens, in fact, when the matrix takes the form which is the faithful representation of $\mathbb{C}$ as the subgroup of $GL(2,\mathbb{R})$ consisting of matrices:
$$
\begin{pmatrix}a & -b \\ b & a \end{pmatrix}
$$
