$$ z^2+(a+bi)z+(c+di)=0 $$
首先了解一下复数开平方的运算公式
- $ B>0 $时
$$ \sqrt{A+Bi}=\pm\left(\sqrt{\frac{\sqrt{A^2+B^2}+A}{2}}+i\sqrt{\frac{\sqrt{A^2+B^2}-A}{2}}\right) $$
- $ B<0 $时
$$ \sqrt{A+Bi}=\pm\left(\sqrt{\frac{\sqrt{A^2+B^2}+A}{2}}-i\sqrt{\frac{\sqrt{A^2+B^2}-A}{2}}\right) $$
令\( \Delta_1=ab-2d,\Delta_2=a^2-b^2-4c \)
\(\Delta_1=0 \)时
- \( \Delta_2>0 \)时
\begin{cases} \displaystyle z_1=-\frac{1}{2}(a-\sqrt{\Delta_2})-\frac{1}{2}bi\\[2mm] \displaystyle z_2=-\frac{1}{2}(a+\sqrt{\Delta_2})-\frac{1}{2}bi\end{cases}
- \( \Delta_2=0 \)时
$$ z_1=z_2=-\frac{1}{2}a-\frac{1}{2}bi $$
- \( \Delta_2<0 \)时
\begin{cases} \displaystyle z_1=-\frac{1}{2}a-\frac{1}{2}(b-\sqrt{-\Delta_2})i\\[2mm]\displaystyle z_2=-\frac{1}{2}a-\frac{1}{2}(b+\sqrt{-\Delta_2})i\end{cases}
\(\Delta_1>0 \)时
\begin{cases}\displaystyle z_1=-\frac{1}{2}\left(a-\sqrt{\frac{\sqrt{\Delta_2^2+4\Delta_1^2}+\Delta_2}{2}}\right)-\frac{1}{2}\left(b-\sqrt{\frac{\sqrt{\Delta_2^2+4\Delta_1^2}-\Delta_2}{2}}\right)i \\[2mm]\displaystyle z_2=-\frac{1}{2}\left(a+\sqrt{\frac{\sqrt{\Delta_2^2+4\Delta_1^2}+\Delta_2}{2}}\right)-\frac{1}{2}\left(b+\sqrt{\frac{\sqrt{\Delta_2^2+4\Delta_1^2}-\Delta_2}{2}}\right)i\end{cases}
\(\Delta_1<0 \)时
\begin{cases}\displaystyle z_1=-\frac{1}{2}\left(a-\sqrt{\frac{\sqrt{\Delta_2^2+4\Delta_1^2}+\Delta_2}{2}}\right)-\frac{1}{2}\left(b+\sqrt{\frac{\sqrt{\Delta_2^2+4\Delta_1^2}-\Delta_2}{2}}\right)i \\[2mm]\displaystyle z_2=-\frac{1}{2}\left(a+\sqrt{\frac{\sqrt{\Delta_2^2+4\Delta_1^2}+\Delta_2}{2}}\right)-\frac{1}{2}\left(b-\sqrt{\frac{\sqrt{\Delta_2^2+4\Delta_1^2}-\Delta_2}{2}}\right)i\end{cases}
