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Q.
Let $z_1, z_2$ be two complex numbers represented by points on the circle $\left|z_1\right|=1$ and $\left| z _2\right|=2$ respectively, then :
Complex Numbers and Quadratic Equations
Solution:
$\left|2 z _1+ z _2\right| \leq 2\left| z _1\right|+\left| z _2\right|=2 \times 1+2=4$
$\therefore$ Maximum value of $\left|2 z _1+ z _2\right|=4$ Clearly $\left| z _1- z _2\right|$ is least when $0, z _1, z _2$ are collinear. Then $\left| z _1- z _2\right|=1$
Again $\left|z_2+\frac{1}{z_1}\right| \leq\left|z_2\right|+\left|\frac{1}{z_1}\right|=2+\frac{1}{\left|z_1\right|}=2+\frac{1}{1}=3$
$\Rightarrow\left| z _2+\frac{1}{ z _1}\right| \leq 3$