Prove the integral is analytic I have this question which I`m not sure at all how to finish the proof. Please help me: 
Let $f$ be an analytic function on $B_1(0)$: For $z \in B_1(0)$; let
$z : [0; 1] \to \mathbb C$; 
$\gamma(t) = tz$:
Show that $F(z) =\int_\gamma f dt$ is an analytic function satisfying $F' = f$.
I tried to write the integral in terms of $\gamma$ and this will give a function of 2 variables $z$ and $t$. Should I then try to prove Cauchy-Riemman equations by taking derivative using Leibniz rule? 
Please help me out... 
 A: Fix $z$ with $|z| < 1$ and let $h$ be a complex number small enough so that $|z+h| < 1$.  Denote by $C(z)$ the straight-line path from $0$ to $z$, so that $C(z) = \gamma$ in your notation.
Consider the difference
$$
F(z+h) - F(z) = \int_{C(z+h)} f(t)\,dt - \int_{C(z)} f(t)\,dt = \int_{C(z+h)-C(z)} f(t)\,dt,
$$
where in the last integral the line segment $C(z)$ has the reverse orientation, as in the image below.

If we also integrate twice over the line segment connecting $z$ and $z+h$, once from $z$ to $z+h$ then once from $z+h$ back to $z$, the contributions from these traversals will cancel each other and the integral will remain unchanged.

Since $f$ is analytic in the disk, Cauchy's theorem for triangles tells us that the integral around the triangle we have constructed is $0$, which leaves us with only the integral along the line segment from $z$ to $z+h$.  Let's call this line segment $E$.

We have shown that
$$
F(z+h) - F(z) = \int_E f(t)\,dt.
$$
Since $f$ is analytic at $z$ it is also continuous at $z$, so when $t$ is close to $z$ we can write
$$
f(t) = f(z) + \epsilon(t),
$$
where $\epsilon(t) \to 0$ as $t \to z$.  Thus, for $h$ small enough, we have
$$
\begin{align}
F(z+h) - F(z) &= \int_E f(z)\,dt + \int_E \epsilon(t)\,dt \\
              &= f(z) \int_E dt + \int_E \epsilon(t)\,dt \\
              &= f(z)h + \int_E \epsilon(t)\,dt.
\tag{1}
\end{align}
$$
By the triangle inequality we have
$$
\left|\int_E \epsilon(t)\,dt\right| \leq L(E) \max_{t \in E} |\epsilon(t)| = |h| \max_{t \in E} |\epsilon(t)|,
$$
where $L(E)$ is the length of the line segment $E$.  Since $\epsilon(t) \to 0$ for all $t \in E$ as $h \to 0$, we know that $\max_{t \in E} |\epsilon(t)| \to 0$ as $h \to 0$.  Thus, by dividing equation $(1)$ by $h$, we may conclude that
$$
\lim_{h \to 0} \frac{F(z+h)-F(z)}{h} = f(z),
$$
which shows that $F' = f$ on $|z| < 1$.  Note that $F$ is analytic on $|z| < 1$ since it is differentiable there.
