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$z_1$ and $z_2$ are two complex numbers such that $|z_1 + z_2|$ = $1$ and $\left| {z_1^2 + z_2^2} \right|$ = $25$ , then minimum value of $\left| {z_1^3 + z_2^3} \right|$ is
$24$
$42$
$37$
$33$
Solution
$\left|z_{1}+z_{2}\right|=1$ and $\left|z_{1}^{2}+z_{2}^{2}\right|=25$
$\because\left(z_{1}+z_{2}\right)^{3}=z_{1}^{3}+z_{2}^{3}+3 z_{1} \cdot z_{2}\left(z_{1}+z_{2}\right)$
$\Rightarrow z_{1}^{3}+z_{2}^{3}=\left(z_{1}+z_{2}\right)\left(\left(z_{1}+z_{2}\right)^{2}-3 z_{1} z_{2}\right)$
$=\left(z_{1}+z_{2}\right)\left\{\left(z_{1}+z_{2}\right)^{2}-\frac{3}{2}\left(\left(z_{1}+z_{2}\right)^{2}-\left(z_{1}^{2}+z_{2}^{2}\right)\right)\right\}$
$=\left(z_{1}+z_{2}\right)\left\{\frac{3}{2}\left(z_{1}^{2}+z_{2}^{2}\right)-\frac{1}{2}\left(z_{1}+z_{2}\right)^{2}\right\}$
$\Rightarrow\left|z_{1}^{3}+z_{2}^{3}\right|=\left|z_{1}+z_{2}\right|\left|\frac{3}{2}\left(z_{1}^{2}+z_{2}^{2}\right)-\frac{1}{2}\left(z_{1}+z_{2}\right)^{2}\right|$
$ \ge \left| {\frac{3}{2}\left| {z_1^2 + z_2^2} \right| – \frac{1}{2}\left| {{z_1} + {z_2}} \right|} \right| \ge \left| {\frac{3}{2} \cdot 25 – \frac{1}{2}} \right| \ge 37$
