//-2 - |1 + y| for |1 + y| <= 0\ //-2 - |1 + y| for |1 + y| <= 0\
x1 = I*im|< | + re|< |
\\ nan otherwise / \\ nan otherwise /
$$x_{1} = \operatorname{re}{\left(\begin{cases} - \left|{y + 1}\right| - 2 & \text{for}\: \left|{y + 1}\right| \leq 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)} + i \operatorname{im}{\left(\begin{cases} - \left|{y + 1}\right| - 2 & \text{for}\: \left|{y + 1}\right| \leq 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)}$$
x1 = re(Piecewise((-Abs(y + 1 - 2, |y + 1| <= 0), (nan, True))) + i*im(Piecewise((-|y + 1| - 2, |y + 1| <= 0), (nan, True))))
Suma y producto de raíces
[src]
//-2 - |1 + y| for |1 + y| <= 0\ //-2 - |1 + y| for |1 + y| <= 0\
I*im|< | + re|< |
\\ nan otherwise / \\ nan otherwise /
$$\operatorname{re}{\left(\begin{cases} - \left|{y + 1}\right| - 2 & \text{for}\: \left|{y + 1}\right| \leq 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)} + i \operatorname{im}{\left(\begin{cases} - \left|{y + 1}\right| - 2 & \text{for}\: \left|{y + 1}\right| \leq 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)}$$
//-2 - |1 + y| for |1 + y| <= 0\ //-2 - |1 + y| for |1 + y| <= 0\
I*im|< | + re|< |
\\ nan otherwise / \\ nan otherwise /
$$\operatorname{re}{\left(\begin{cases} - \left|{y + 1}\right| - 2 & \text{for}\: \left|{y + 1}\right| \leq 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)} + i \operatorname{im}{\left(\begin{cases} - \left|{y + 1}\right| - 2 & \text{for}\: \left|{y + 1}\right| \leq 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)}$$
//-2 - |1 + y| for |1 + y| <= 0\ //-2 - |1 + y| for |1 + y| <= 0\
I*im|< | + re|< |
\\ nan otherwise / \\ nan otherwise /
$$\operatorname{re}{\left(\begin{cases} - \left|{y + 1}\right| - 2 & \text{for}\: \left|{y + 1}\right| \leq 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)} + i \operatorname{im}{\left(\begin{cases} - \left|{y + 1}\right| - 2 & \text{for}\: \left|{y + 1}\right| \leq 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)}$$
/-(2 + |1 + y|) for |1 + y| <= 0
<
\ nan otherwise
$$\begin{cases} - (\left|{y + 1}\right| + 2) & \text{for}\: \left|{y + 1}\right| \leq 0 \\\text{NaN} & \text{otherwise} \end{cases}$$
Piecewise((-(2 + |1 + y|), |1 + y| <= 0), (nan, True))