JEE Main 2025 — 3 April, Evening Shift — Question 37

$$Z_{0}=Z_{1}+Z_{2}+Z_{3}$$

$\sum_{k=1}^{3}\left(z_{k}-z_{0}\right)^{2}=\left(z_{1}-z_{0}\right)^{2}+\left(z_{2}-z_{0}\right)^{2}+\left(z_{3}-z_{0}\right)^{2}$

Let $z_{0}$ is origin $\Rightarrow z_{1}, z_{2}, z_{3}$ lies on a circle

$$\text { having }\left|z_{0}-z_{i}\right|=R$$

$\therefore z_{1}=R e^{i 2 \pi / 3} Z_{2}=R e^{i 4 \pi / 3} z_{3}=R e^{i \sigma \pi / 3}$

$\Rightarrow z_{1}^{2}+z_{2}^{2}+z_{3}^{2}=R^{2}\left[e^{i 4 \pi / 3}+e^{i 8 \pi / 3}+e^{i 12 \pi / 3}\right]$

$=0$

$\therefore \sum_{k=1}^{3}\left(z_{k}-z_{0}\right)^{2}=0$