WBJEE · Maths · Complex Number
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 \(z_1+(1+i) z_2\)
- D \(-i \mathrm{z}_1+(1+\mathrm{i}) \mathrm{z}_2\)
Answer & Solution
Correct Answer
(D) \(-i \mathrm{z}_1+(1+\mathrm{i}) \mathrm{z}_2\)
Step-by-step Solution
Detailed explanation
Hint : In \(\triangle A B C \frac{z_1-z_2}{z_3-z_2}=\left|\frac{z_1-z_2}{z_3-z_2}\right|\mathrm{i}^{\pi / 2}=\left|\frac{z_1-z_2}{z_3-z_2}\right| \cdot i=\frac{A B}{B C} \cdot i=i\)…
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