Complex Numbers
This lecture notes and solved problems on complex numbers is designed for advanced high school and intermediate students preparing for competitive examinations.
If \(z_1,z_2,z_3\in \mathbb{C}\) forms an equilateral triangle. Prove that \[\frac{1}{z_1-z_2}+\frac{1}{z_2-z_3}+\frac{1}{z_3-z_1}=0\]
If \(z_1,z_2,z_3\in \mathbb{C}\) such that \(|z_1|=|z_2|=|z_3|=\lambda\) and \(|z_1+z_2+z_3|=\alpha\), find the value of \(\left|\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}\right|\) in terms of \(\alpha,\lambda\).
\[|z_1+z_2+z_3|=\alpha\] \[\implies\left|z_1\frac{\bar{z_1}}{\bar{z_1}}+z_2\frac{\bar{z_2}}{\bar{z_2}}+z_3\frac{\bar{z_3}}{\bar{z_3}}\right|=\alpha\] \(\because{|z|^2=z\bar{z}}\) \[\implies\left|\frac{|z_1|^2}{\bar{z_1}}+\frac{|z_2|^2}{\bar{z_2}}+\frac{|z_3|^2}{\bar{z_3}}\right|=\alpha\] \[\implies\lambda^2\left|\frac{1}{\bar{z_1}}+\frac{1}{\bar{z_2}}+\frac{1}{\bar{z_3}}\right|=\alpha\] \(\because |z|=|\bar{z}|\) \[\implies\left|\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}\right|=\frac{\alpha}{\lambda^2}\]
Author
Anurag Gupta is an M.S. graduate in Electrical and Computer Engineering from Cornell University. He also holds an M.Tech degree in Systems and Control Engineering and a B.Tech degree in Electrical Engineering from the Indian Institute of Technology, Bombay.
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