Show that the function is $C^1$. I need help with the following,
Given $f=u(x,y) + iv(x,y)$ a complex valued function, where $u$ and $v$ have continuous first order partial derivatives on some open set containing the image of the $C^1$-curve $\gamma :[a,b]\rightarrow \mathbb{C}$. (Why open set? Why not any set containing the image of the $C^1$ curve?)
Show that $f\circ \gamma$ is a $C^1$-function. (How?)
And hence, $\int_{a}^{b}(f\circ \gamma)'(t)dt=f(\gamma(b))-f(\gamma(a)).$
Why  can the Fundemental theorem of calculus for real-valued function be applied to the case of the complex valued function $f\circ \gamma$ ?
Please help!
 A: Let $x(t)=\Re(\gamma(t))$ and $y(t)=\Im(\gamma(t))$. That is, $\gamma(t)=x(t)+iy(t)$.
Then that $\gamma$ is a $C^{1}$-curve really means that $x$, $y$
are $C^{1}$-functions. Hence $f\circ\gamma(t)=u(x(t),y(t))+iv(x(t),y(t))$.
By a result in advanced calculus, the functions $t\mapsto u(x(t),y(t))$,
$t\mapsto v(x(t),y(t))$ are $C^{1}$-functions. 
For, the derivative of the first function, evaluated at $t$, is given
by $u_{x}(x(t),y(t))x'(t)+u_{y}(x(t),y(t))y'(t)$ and clearly $t\mapsto u_{x}(x(t),y(t))x'(t)+u_{y}(x(t),y(t))y'(t)$
is continuous because partial derivatives $u_{x}$, $u_{y}$ and derivative
$x'$, $y'$ are continuous.
Hence, $f\circ\gamma:[a,b]\rightarrow\mathbb{C}$ is a $C^{1}$-function.
For the second part involving integral, we simply separate the real
and imaginary parts and the result will follow. For example, 
\begin{eqnarray*}
 &  & \int_{a}^{b}\left(f\circ\gamma\right)'(t)dt\\
 & = & \int_{a}^{b}\left\{ \frac{d}{dt}\left[u(x(t),y(t))\right](t)+i\frac{d}{dt}\left[v(x(t),y(t))\right](t)\right\} dt\\
 & = & \int_{a}^{b}\frac{d}{dt}\left[u(x(t),y(t))\right](t)dt+i\int_{a}^{b}\frac{d}{dt}\left[v(x(t),y(t))\right](t)dt\\
 & = & \left[u(x(b),y(b))-u(x(a),y(a))\right]+i\left[v(x(b),y(b))-v(x(a),y(a))\right]\\
 & = & f\circ\gamma(b)-f\circ\gamma(a).
\end{eqnarray*}
In the above, we use the famous fact that if $g:[a,b]\rightarrow\mathbb{R}$
is $C^{1}$, then $\int_{a}^{b}g'(t)dt=g(b)-g(a).$
