2z^2-iz+2i=0
You asked:
Investigate the equation: \(2 {z}^{2} - i z + 2 i = 0\).
MathBot Answer:
The 2 solutions to the equation are: \[\begin{aligned}z &= - \frac{\sqrt[4]{257} \sin{\left(\frac{\arctan{\left(16 \right)}}{2} \right)}}{4} + i \left(\frac{1}{4} + \frac{\sqrt[4]{257} \cos{\left(\frac{\arctan{\left(16 \right)}}{2} \right)}}{4}\right) \approx -0.68536531 + 0.97953794 i\\z &= \frac{\sqrt[4]{257} \sin{\left(\frac{\arctan{\left(16 \right)}}{2} \right)}}{4} + i \left(- \frac{\sqrt[4]{257} \cos{\left(\frac{\arctan{\left(16 \right)}}{2} \right)}}{4} + \frac{1}{4}\right) \approx 0.68536531 -0.47953794 i\end{aligned}\]
\(i\) is the imaginary unit, defined as \(i^2 = -1\).