Let $f:\mathbb{C} \to \mathbb{C} \\ \,\ \ \ \ \ z \to (|z|^2-2)\overline{z}$
Determine the points where f is differentiable, presenting the due expression $f'(z)$.
$u(x,y)=Re((x^2+y^2-2)(x-yi))=x^3+xy^2-2x; \\ v(x,y)=Im((x^2+y^2-2)(x-yi))=-yx^2-y^3+2y $
$u_x=3x^2+y^2-2; \,u_y=2xy \\
v_x= \ \ -2xy;\qquad \ \ v_y=-x^2-3y^2+2$
The CR equality happens when:
$$4(x^2+y^2)=4$$
And as all partial derivatives are continuous, f is differentiable at $\partial{D(0,1)}$
And now, for $z_{0} \in \partial{D(0,1)}$ and knowing $\textit{'a priori'}$ that the limit exists:
$$ lim_{z \to z_0; z \in \partial{D(0,1)}} \frac{f(z)-f(z_0)}{z-z_0} = lim_{z \to z_0; z \in \partial{D(0,1)}}\frac{(|z|^2-2)\overline{z}+\overline{z_0}}{z-z_0}=lim_{z \to z_0; z \in \partial{D(0,1)}}-\frac{z-z_0}{z-z_0}=-1$$ So $f'(z_0)=-1$ for $z_0 \in \partial{D(0,1)}$
Is this correct? Any further suggestions?
Best Answer
You are right about the points at which $f\require{cancel}$ is differentiable, but not about the computation of $f'$. If $x^2+y^2=1$, then\begin{align}f'(x+yi)&=\frac{\partial u}{\partial x}(x,y)+\frac{\partial v}{\partial x}(x,y)i\\&=3x^2+y^2-2-2xyi\\&=\cancel{2x^2+2y^2-2}+x^2-y^2-2xyi\\&=(x-yi)^2.\end{align}In other words, if $|z|=1$, then $f'(z)=\overline z^2$.