asked by guest
on Jan 30, 2025 at 12:53 am
You asked:
Investigate the equation:
3 z 2 − i z + 4 + i = 0 3 {z}^{2} - i z + 4 + i = 0 3 z 2 − i z + 4 + i = 0 .
MathBot Answer:
The 2 solutions to the equation are:
z = − 2545 4 sin ( arctan ( 12 49 ) 2 ) 6 + i ( 1 6 + 2545 4 cos ( arctan ( 12 49 ) 2 ) 6 ) ≈ − 0.14181331 + 1.3419207 i z = 2545 4 sin ( arctan ( 12 49 ) 2 ) 6 + i ( − 2545 4 cos ( arctan ( 12 49 ) 2 ) 6 + 1 6 ) ≈ 0.14181331 − 1.0085874 i \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} z z = − 6 4 2545 sin ( 2 a r c t a n ( 49 12 ) ) + i 6 1 + 6 4 2545 cos ( 2 a r c t a n ( 49 12 ) ) ≈ − 0.14181331 + 1.3419207 i = 6 4 2545 sin ( 2 a r c t a n ( 49 12 ) ) + i − 6 4 2545 cos ( 2 a r c t a n ( 49 12 ) ) + 6 1 ≈ 0.14181331 − 1.0085874 i
i i i is the imaginary unit , defined as i 2 = − 1 i^2 = -1 i 2 = − 1 .