z^2-3iz+2i=0
You asked:
Investigate the equation: \({z}^{2} - 3 i z + 2 i = 0\).
MathBot Answer:
The 2 solutions to the equation are: \[\begin{aligned}z &= - \frac{\sqrt[4]{145} \sin{\left(\frac{\arctan{\left(\frac{8}{9} \right)}}{2} \right)}}{2} + i \left(\frac{3}{2} + \frac{\sqrt[4]{145} \cos{\left(\frac{\arctan{\left(\frac{8}{9} \right)}}{2} \right)}}{2}\right) \approx -0.61660305 + 3.1217889 i\\z &= \frac{\sqrt[4]{145} \sin{\left(\frac{\arctan{\left(\frac{8}{9} \right)}}{2} \right)}}{2} + i \left(- \frac{\sqrt[4]{145} \cos{\left(\frac{\arctan{\left(\frac{8}{9} \right)}}{2} \right)}}{2} + \frac{3}{2}\right) \approx 0.61660305 -0.12178893 i\end{aligned}\]
\(i\) is the imaginary unit, defined as \(i^2 = -1\).