Find all $z$ for which $z^2+2z+2$ is real positive. Find all $z$ for which $z^2+2z+2$ is real positive. 
My Attempt :
Let $z=x+iy,  (x,y\in R)$:
$$z^2+2z+2= (x^2-y^2+2ixy) +2(x+iy)+2$$
$$=(x^2-y^2+2x+2)+i(2xy+2y)$$
 A: You want $2xy+2y = 2y(x+1)=0$ and $x^2-y^2+2x+2 = (x+1)^2-y^2+1>0$. From the first equation, $y=0$ or $x=-1$; if $y=0$, you want $(x+1)^2+1>0$, which is always true; if $x=-1$, you want $-y^2+1>0$, which is true when $|y|<1$. So the solutions are 
\begin{equation*}
  \{x+iy\,|\,y=0\text{ or }(x=-1\text{ and }|y|<1)\}.
\end{equation*}
A: Your calculations are correct. Now, a complex number $a+ib$ (with $a, b\in\Bbb R$) is real and positive iff
$$a>0\text{ and }b=0$$
Thus you have to find all $z$ s.t.
$$x^2-y^2+2x+2>0\text{ and }2xy+2y=0$$
A: You're almost there! All you need to do is solve the conditions required for your quantity to be real positive, I.e.:
$$
x^2 - y^2 + 2x + 2 > 0
\text{ and }
2xy + 2y = 0
$$
Hint: start with the equality and factor. You should get two cases; use these to establish the requirements given by the inequality.
A: $z^2+2z+2 = (z+1)^2+1$.
So this is real (positive or negative) if $z+1$ is in $\{ci; c \in \mathbb{R} \} \cup \{c; c \in \mathbb{R} \}$.
This is positive if $z+1$ is in $\{ci; c \in \mathbb{R}; -1< c < 1 \}$ or if
$z+1$ is in $\cup \{c; c \in \mathbb{R} \setminus \{0\}\}$.
So $z^2+2z+2 = (z+1)^2+1$ is positive if and only if $z$ is in $\{ci-1; -1< c < 1\} \cup \mathbb{R} \setminus \{0\}$
A: Let $w:=u+iv:=z+1$. Then $w^2+1$ is real when $w$ is real or purely imaginary.


*

*$u^2>-1$ is always true,

*$(iv)^2>-1\iff |v|<1$.
The solutions are $w-1$.

A: You want the real part to be positive and the imaginary part to be $0$
$$ (x^2-y^2+2x+2)+i(2xy+2y)$$
$$2xy+2y=0 \implies 2y(x+1)=0$$
Thus  $y=0$ or $x=-1$
For $y=0$ we get $x^2+2x+2=(x+1)^2+1 >0$
For $ x=-1 $ we need $1-y^2>0$ which implies $|y|<1$
Thus the solution is either $ y=0$ or ($|y|<1$ and $x=-1$ )    
A: Let $p$ be a real number.
$$(z+1)^2+1=p^2\iff z=\pm\sqrt{p^2-1}-1.$$
If $|p|<1$, $z=\pm i\sqrt{1-p^2}-1$ runs from $-1-i$ to $-1+i$,
If $|p|\ge1$, $z=\pm\sqrt{p^2-1}-1$, runs from $-\infty$ to $\infty$.
A: A slightly geometric approach:
If we have $p(z) = (z-w)(z-\overline{w})$, then let $m={1 \over 2} (w+ \overline{w}) = \operatorname{re} w$ and so
(with $r \ge 0$)
\begin{eqnarray}
p(m + r e^{i \theta}) &=& (m-w+r e^{i \theta}) (m-\overline{w}+r e^{i \theta}) \\
&=& (-i\operatorname{im} w + r e^{i \theta})(i \operatorname{im} w+r e^{i \theta}) \\
&=& (\operatorname{im} w)^2+ r^2 e^{2 i \theta}
\end{eqnarray}
We see that this is real iff $\theta$ is a multiple of ${\pi \over 2}$.
If an even multiple, $p(m + r e^{i \theta}) > 0$ for any $r\ge 0$.
If an odd multiple, $p(m + r e^{i \theta}) > 0$ iff  $0 \le r < |\operatorname{im} w|$.
In this particular case, $w=-1+i$, so $m= -1$, $\operatorname{im} w = 1$.
Hence $p(z) $ is real and positive iff
$z \in \mathbb{R} \cup \{ -1+iv | v \in (-1,1)\}$.
