Cho các số phức \(z_{1}, z_{2}, z_{3}\) thoả mãn các điều kiện \(\left|z_{1}\right|=\left|z_{2}\right|=\left|z_{1}-z_{2}\right|=3\). Mô đun của số phức \(z_{1}+z_{2}\) bằng
Suy nghĩ và trả lời câu hỏi trước khi xem đáp án
Lời giải:
Báo sai\(\left|z_{1}\right|=\left|z_{2}\right|=\left|z_{1}-z_{2}\right|=3 \Leftrightarrow\left|\frac{z_{1}}{3}\right|=\left|\frac{z_{2}}{3}\right|=\left|\frac{z_{1}-z_{2}}{3}\right|=1 \Rightarrow\left\{\begin{array}{l} \frac{z_{1}}{3}=\cos \varphi_{1}+i \sin \varphi_{1} \\ \frac{z_{2}}{3}=\cos \varphi_{2}+i \sin \varphi_{2} \end{array}\right.\)
Suy ra \(\frac{z_{1}-z_{2}}{3}=\cos \varphi_{1}-\cos \varphi_{2}+i\left(\sin \varphi_{1}-\sin \varphi_{2}\right)\)
\(\left|\frac{z_{1}-z_{2}}{3}\right|=1 \Leftrightarrow\left(\cos \varphi_{1}-\cos \varphi_{2}\right)^{2}+\left(\sin \varphi_{1}-\sin \varphi_{2}\right)^{2}=1\)
\(\begin{array}{l} \Leftrightarrow 2-2\left(\cos \varphi_{1} \cos \varphi_{2}+\sin \varphi_{1} \sin \varphi_{2}\right)=1 \\ \Leftrightarrow 1-2 \cos \left(\varphi_{1}-\varphi_{2}\right)=0 \Leftrightarrow \cos \left(\varphi_{1}-\varphi_{2}\right)=\frac{1}{2} \end{array}\)
Vậy \(\frac{z_{1}+z_{2}}{3}=\cos \varphi_{1}+\cos \varphi_{2}+i\left(\sin \varphi_{1}+\sin \varphi_{2}\right)\)
Suy ra \(\left|\frac{z_{1}+z_{2}}{3}\right|^{2}=\left(\cos \varphi_{1}+\cos \varphi_{2}\right)^{2}+\left(\sin \varphi_{1}+\sin \varphi_{2}\right)^{2}=2+2 \cos \left(\varphi_{1}-\varphi_{2}\right)=3\)
Vậy \(\left|z_{1}+z_{2}\right|=3 \sqrt{3}\)