Contour question in complex analysis Let $C_1$ be the line segment from $-1-i$ to $3-i$, and $C_2$ be the portion of the parabola $x=y^2+2y$ joining the above points $-1-i$ and $3-i$.  Show that 
$$\int_{C_1}zdz=\int_{C_2}zdz=4+2i.$$
So I know I can parametrize the segment by getting the slope intercept form and writing it in terms of x, i.e., $x=2y+1$ but how do I parametrize it so that i have a $z$ and a $dz$ so I can integrate?  Same problem with the parabola.  How do you pick your parametrization?
 A: As you said, the points $(x,y$) on the line segment joining the two points $A(-1-i)$ and $B(3+i)$ verify $x=2y+1$. Notice that $y$ goes from $-1$ to $1$ if you go from $A$ to $B$. 
Here is a parametrization of $C_1$ : $$\begin{array}{rcl}\gamma_1:[-1,1]&\longrightarrow&\Bbb C\\ t&\longmapsto& 2t+1+it\end{array}$$
From this, you get : $$\begin{array}{rcl}\int_{C_1} zdz&=&\int_{-1}^1 \gamma_1(t)\gamma_1'(t)dt\\&=&\int_{-1}^1(2t+1+it)(2+i)dt\\&=&(2+i)\left[t^2+t+i\dfrac{t^2}2\right]_{-1}^1\\&=&(2+i)\left[2+\dfrac i2-\dfrac i2\right]\\&=&4+2i\end{array}$$
You probably can do the other calculus easily if you understand this one !
A: firstly,by parametrizing the parabola.$X= Y^2 + 2Y
(X+1) = (Y+1)^2$  complete square. to get
$y = t-1   x=t^2 -1$
substitute the coordinates i.e $(-1, -1)$ to $(3, 1)$.
$t=0$ for $y=-1$   and $t=0$ for $x=-1$
$t=2$ for $y=1$   $t=2$ for $x=3$.
now integrate over $0$ to $2$.
$$Zdz= (X + iY)(dx + idy)
    =[t^2 -1 + i(t-1)][2tdt + idt]
    =[2t^3 -3t+1]dt +i[3t^2-2t-1]dt
  =[t^4/2 - 3t^2/2 +t] +i[t^3-t^2-t]$$
pass limit $0$ to $2$
$   =[8-6+2] + i[8-4-2]
   =4 + 2i $
