Polynomial (quartic) with complex coefficient: root localization For all $u\in \mathbb C$, let $P_u(X)= X^4+4X+u$. I know that the roots of $P_u$ could be found explicitly, but it seems to lead to inextricably complex calculations.
Let $H= \{z \in \mathbb C  ~|~ \mathrm{Re}(z)<1/2\}$,
$Q_a^+ =  \{z \in \mathbb C  ~|~ \mathrm{Re}(z)>a, \mathrm{Im}(z)>a\}$,
$Q_a^- =  \{z \in \mathbb C  ~|~ \mathrm{Re}(z)>a, \mathrm{Im}(z)<-a\}$.
I would like to show that there exists $a>1/2$ such that for all $u$, $P_u$ always has at least one root in $Q_a^+$, one in $Q_a^-$ and one in $H$.
I found some interesting results about polynomial roots localization, but often they show some bound which don't seem to help here (even if I take the reciprocal polynomial).
Any help would be useful. 
Here is an example of roots with $u=1+I$, and $a=0.7$.


This is false.  For instance, if $u=-1000$, then according to WolframAlpha the roots are approximately $-5.65$, $5.59$, and $0.03\pm 5.62i$ and so none of them are in $Q_a^+$ or $Q_a^-$ for any $a>1/2$.  As motivation for trying this example, note that when $u$ is large, you can expect the roots of $P_u$ to be close to the fourth roots of $-u$, so when $-u$ is large and positive the roots will be close to the axes and thus avoid $Q_a^+$ and $Q_a^-$.  (In fact, $-u=4$ is already large enough.)
It is true that there is always a root in $H$, since the roots must add up to $0$ (since $P_u$ has no $x^3$ term) and in particular at least one must have nonpositive real part.
