3z^2-2z+4i=0
You asked:
Investigate the equation: \(3 \cdot {z}^{2} - 2 z + 4 i = 0\).
MathBot Answer:
The 2 solutions to the equation are: \[\begin{aligned}z &= \frac{1}{3} + \frac{\sqrt[4]{145} \cos{\left(\frac{\arctan{\left(12 \right)}}{2} \right)}}{3} - \frac{\sqrt[4]{145} i \sin{\left(\frac{\arctan{\left(12 \right)}}{2} \right)}}{3} \approx 1.1845284 -0.78321256 i\\z &= - \frac{\sqrt[4]{145} \cos{\left(\frac{\arctan{\left(12 \right)}}{2} \right)}}{3} + \frac{1}{3} + \frac{\sqrt[4]{145} i \sin{\left(\frac{\arctan{\left(12 \right)}}{2} \right)}}{3} \approx -0.51786173 + 0.78321256 i\end{aligned}\]
\(i\) is the imaginary unit, defined as \(i^2 = -1\).