3z^2-iz+(4+i)=0

asked by guest
on Jan 30, 2025 at 12:53 am



You asked:

Investigate the equation: 3z2iz+4+i=03 {z}^{2} - i z + 4 + i = 0.

MathBot Answer:

The 2 solutions to the equation are: z=25454sin(arctan(1249)2)6+i(16+25454cos(arctan(1249)2)6)0.14181331+1.3419207iz=25454sin(arctan(1249)2)6+i(25454cos(arctan(1249)2)6+16)0.141813311.0085874i\begin{aligned}z &= - \frac{\sqrt[4]{2545} \sin{\left(\frac{\arctan{\left(\frac{12}{49} \right)}}{2} \right)}}{6} + i \left(\frac{1}{6} + \frac{\sqrt[4]{2545} \cos{\left(\frac{\arctan{\left(\frac{12}{49} \right)}}{2} \right)}}{6}\right) \approx -0.14181331 + 1.3419207 i\\z &= \frac{\sqrt[4]{2545} \sin{\left(\frac{\arctan{\left(\frac{12}{49} \right)}}{2} \right)}}{6} + i \left(- \frac{\sqrt[4]{2545} \cos{\left(\frac{\arctan{\left(\frac{12}{49} \right)}}{2} \right)}}{6} + \frac{1}{6}\right) \approx 0.14181331 -1.0085874 i\end{aligned}


ii is the imaginary unit, defined as i2=1i^2 = -1.