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.
 A: Consider $z=a+bi$ such that $b>0$. Then:
$$-\frac{1}{1+z}=-\frac{1}{1+a+bi}=-\frac{1+a-bi}{(1+a)^2+b^2}=-\frac{1+a}{(1+a)^2+b^2}+\frac{b}{(1+a)^2+b^2}i;\\
\text{Im}\left(-\frac{1}{1+z}\right)=\frac{b}{(1+a)^2+b^2}>0.$$
A: Let $w=z+1$. The imaginary part of $w$ is also positive; it lies above the
$x$-axis. Write $w=re^{it}$. We can take $0<t<\pi$. Then $-1/w=(1/r)(-e^{-it})$.
Can you prove $-e^{-it}$ has positive imaginary part.
A: $z = x + iy, \; x, y \in \Bbb R, \; y > 0; \tag 1$
$z + 1 = (1 + x) + iy; \tag 2$
$\dfrac{1}{z + 1} = \dfrac{1}{(1 + x) + iy}$
$= \dfrac{(1 + x) - iy}{(1 + x)^2 + y^2} = \dfrac{1 + x}{(1 + x)^2 + y^2} -i\dfrac{y}{(1 + x)^2 + y^2}; \tag 3$
$-\dfrac{1}{z + 1} = -\dfrac{1 + x}{(1 + x)^2 + y^2} +i\dfrac{y}{(1 + x)^2 + y^2}; \tag 4$
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 . . . 
A: Alt. hint:   using that $\Im(z) = \dfrac{1}{2i}\left(z - \bar z\right)\,$:
$$\require{cancel}
2i \;\Im\left(\frac{1}{z+1}\right) = \frac{1}{z+1} - \frac{1}{\bar z+1} = \frac{\bar z + \cancel{1} - z - \cancel{1}}{(z+1)(\bar z+1)} = \frac{\;-2i\;\Im(z)\;\;}{|z+1|^2}
$$
