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

• A

  \(-i \mathrm{z}_1-(1+i) z_2\)
• B

  \(\mathrm{z}_1-(1+\mathrm{i}) \mathrm{z}_2\)
• C

  \(z_1+(1+i) z_2\)
• Correct answer
  D

  \(-i \mathrm{z}_1+(1+\mathrm{i}) \mathrm{z}_2\)

Solution

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

Hence, the correct answer is option: 'D'.