Gọi \(z_{1}, \quad z_{2}, \quad z_{3}, \quad z_{4}\) là các nghiệm của phương trình \(z^{4}+4 z^{3}+3 z^{2}-3 z+3=0\) . Tính \(T=\left(z_{1}^{2}+2 z_{1}+2\right)\left(z_{2}^{2}+2 z_{2}+2\right)\left(z_{3}^{2}+2 z_{3}+2\right)\left(z_{4}^{2}+2 z_{4}+2\right)\)
Hãy suy nghĩ và trả lời câu hỏi trước khi xem đáp án
Lời giải:
Báo sai\(\begin{array}{l} \text { Đặt } f(z)=z^{4}+4 z^{3}+3 z^{2}-3 z+3 \Rightarrow f(z)=\left(z-z_{1}\right)\left(z-z_{2}\right)\left(z-z_{3}\right)\left(z-z_{4}\right) \\ \text { Do } z_{1}^{2}+2 z_{1}+2=\left(z_{1}+1-i\right)\left(z_{1}+1+i\right) \text { nên } \\ T=\left(z_{1}^{2}+2 z_{1}+2\right)\left(z_{2}^{2}+2 z_{2}+2\right)\left(z_{3}^{2}+2 z_{3}+2\right)\left(z_{4}^{2}+2 z_{4}+2\right) \\ =\left(z_{1}+1-i\right)\left(z_{1}+1+i\right)\left(z_{2}+1-i\right)\left(z_{2}+1+i\right)\left(z_{3}+1-i\right)\left(z_{3}+1+i\right)\left(z_{4}+1-i\right)\left(z_{4}+1+i\right) \\ =\left[-\left(-1+i-z_{1}\right)\right]\left[-\left(-1-i-z_{1}\right)\right]\left[-\left(-1+i-z_{2}\right)\right]\left[-\left(\left(-1-i-z_{1}\right)-z_{2}\right)\right]\left[-\left(-1+i-z_{3}\right)\right]\left[-\left(\left(-1-i-z_{1}\right)-z_{3}\right)\right]\left[-\left(-1+i-z_{4}\right)\right]\left[-\left(\left(-1-i-z_{1}\right)\right.\right. \\ =\left((-1+i)-z_{1}\right)\left(\left(\left(-1-i-z_{1}\right)\right)-z_{1}\right)\left((-1+i)-z_{2}\right)\left(\left(\left(-1-i-z_{1}\right)\right)-z_{2}\right)\left((-1+i)-z_{3}\right)\left(\left(\left(-1-i-z_{1}\right)\right)-z_{3}\right)\left((-1+i)-z_{4}\right)\left(\left(\left(-1-i-z_{1}\right)\right)-z_{4}\right) \\ =\left((-1+i)-z_{1}\right)\left((-1+i)-z_{2}\right)\left((-1+i)-z_{3}\right)\left((-1+i)-z_{4}\right)\left((-1-i)-z_{1}\right)\left((-1-i)-z_{2}\right)\left((-1-i)-z_{3}\right)\left((-1-i)-z_{4}\right) \\ =f(-1+i) f(-1-i) \\ =(10-i)(10+i)=101 \end{array}\)