If the vertices of a square are $z_1, z_2, z_3$ and $z_4$ taken in the anti-clockwise order, then $z_3=$

Question

If the vertices of a square are $z_1, z_2, z_3$ and $z_4$ taken in the anti-clockwise order, then $z_3=$, $-i \mathrm{z}_1-(1+i) z_2$, $\mathrm{z}_1-(1+\mathrm{i}) \mathrm{z}_2$, $z_1+(1+i) z_2$, $-i \mathrm{z}_1+(1+\mathrm{i}) \mathrm{z}_2$

MARKS App · Accepted Answer

Hint : In $\triangle A B C \frac{z_1-z_2}{z_3-z_2}=\left|\frac{z_1-z_2}{z_3-z_2}\right|\mathrm{i}^{\pi / 2}=\left|\frac{z_1-z_2}{z_3-z_2}\right| \cdot i=\frac{A B}{B C} \cdot i=i$ $\Rightarrow \mathrm{z}_1-\mathrm{z}_2=\mathrm{i}\left(\mathrm{z}_3-\mathrm{z}_2\right) \Rightarrow-i \mathrm{z}_1+\mathrm{i} \mathrm{z}_2=\mathrm{z}_3-\mathrm{z}_2 \Rightarrow \mathrm{z}_3=-\mathrm{i} \mathrm{z}_1+(\mathrm{i}+1) \mathrm{z}_2=-\mathrm{i} \mathrm{z}_1+(1+\mathrm{i}) \mathrm{z}_2$

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