If $\Im(z) > 0$then $\Im (-1/(z+1)) >0$ My work so far:
Letting $z = x+iy$, where $x,y$ are real,
then $\Im(z) = y > 0$
$y+1 > 0$
$1/(y+1) > 0$
But I am not sure how to proceed. I also attempted to multiply $-1/(z+1)$ by its conjugate, but that led to $(-x-1+iy)/((x+1)^2 + y^2)$, which I feel is the wrong way to approach the problem.
Thank you for your help.

$z = x + iy, \; x, y \in \Bbb R, \; y > 0; \tag 1$
$z + 1 = (1 + x) + iy; \tag 2$
$\dfrac{1}{z + 1} = \dfrac{1}{(1 + x) + iy}$
$= \dfrac{(1 + x) - iy}{(1 + x)^2 + y^2} = \dfrac{1 + x}{(1 + x)^2 + y^2} -i\dfrac{y}{(1 + x)^2 + y^2}; \tag 3$
$-\dfrac{1}{z + 1} = -\dfrac{1 + x}{(1 + x)^2 + y^2} +i\dfrac{y}{(1 + x)^2 + y^2}; \tag 4$
we see that 
$\Im \left ( -\dfrac{1}{1 + z} \right ) = \dfrac{y}{(1 + x)^2 + y^2}; \tag 5$
we are given that $y > 0$; manifestly $(1 + x)^2 + y^2 > 0$, so . . . 
