Exercise: Let $f(x + iy) = x + y$, and for $z$ in $\mathbb{C}$, define $F(z)=\int _{[0 \to z]}f$. At which points is $F$ differentiable?
Is my answer correct? Here's my answer:
$$F(z+h)-F(z)=\int _{[0 \to z+h]}f-\int _{[0 \to z]}f=\int _{[z \to z+h]}f$$
(Edit: As someone has pointed out, we cannot assume $\int _{[0 \to z]}f+\int _{[z \to z+h]}f=\int _{[0 \to z+h]}f$, so this proof is wrong but will still be left here for your reference of possible mistakes)
Let $z=z_1+iz_2, h=h_1+ih_2$, parameterise $[z \to z+h]$ by $\gamma (t)=z+th$ where $t\in [0,1]$, then
$$\int _{[z \to z+h]}f=\int _{[z \to z+h]}(x+y)=\int_{0}^{1}(z_1+z_2+t(h_1+h_2))h dt$$
$$=h\left[(z_1+z_2)t+(h_1+h_2)\frac{t^2}{2}\right]^{1}_{0}=h[(z_1+z_2)+(h_1+h_2)/2]$$
So $$\frac{F(z+h)-F(z)}{h}=(z_1+z_2)+(h_1+h_2)/2\overset{h\to0}{\rightarrow}(z_1+z_2)$$
In particular, $F$ is differentiable at all points.
Also, if my answer is correct, can it be shorten? $f$ is not differentiable so I guess I can't use Cauchy's Theorem. Although $f$ is continuous, I don't think $\int_{\partial T} f=0$ for all triangle $T$, so I can't use that to say $F$ is differentiable.
Many thanks in advance!
Let $z=x+iy$. We parametrize the curve as $\gamma(t) = t(x+iy),~~t\in[0,1]$. Then the function $F(z)$ becoems $$F(z) =\int_{0}^{1} t(x+y)(x+iy)dt =C(x+y)(x+iy):=u(x,y) + iv(x,y).$$ Determine at which point the function is not differentiable by the Cauchy-Riemann equations.