Question
If complex numbers $z_{10} z_{2}, z_{y}$ are the vertices of a triangle, then the complex number $z$ which makes the triangle a parallelogram is given by(a) $z_{1}+z_{2}-z_{5}$(b) $z_{1}+z_{y}-z_{2}$(c) $z_{2}+z_{3}-z_{1}$(d) $z_{1}+z_{2}+z_{3}$
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We want to find a complex number $z$ that will make this triangle a parallelogram. Show more…
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Complex numbers $\mathrm{z}_{1}, \mathrm{z}_{2}, \mathrm{z}_{3}$ are the vertices $\mathrm{A}, \mathrm{B}, \mathrm{C}$ respectively of an isoceles right angled triangle right angled at $\mathrm{A} .$ Then, $\left(\mathrm{z}_{2}-\mathrm{z}_{3}\right)^{2}$ equals (a) $2\left(z_{1}-z_{2}\right)\left(z_{1}-z_{3}\right)$ (b) $2\left(z_{2}-z_{1}\right)\left(z_{1}-z_{3}\right)$ (c) $3\left(z_{1}+z_{2}+z_{3}\right)$ (d) $\left(z_{2}-z_{1}\right)\left(z_{1}-z_{3}\right)$
If $z_{1}$ and $z_{2}$ are two complex numbers such that $\left|z_{1}+z_{2}\right|^{2}=\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}$, then (a) $z_{1} \bar{z}_{2}$ is purely imaginary (b) $z_{1} / t_{2}$ is purely imuginary (c) $z_{1} \bar{z}_{2}+\bar{z}_{1} z_{2}=0$ (d) $O, z_{1}, z_{2}$ are vertices of a right triangle
The complex numbers $z_{1}, z_{2}, z_{3}$ and $z_{4}$ represent the vertices of a parallelogram taken in order if and only if (a) $\mathrm{z}_{1}+\mathrm{z}_{3}=\mathrm{z}_{2}+\mathrm{z}_{4}$ (b) $\mathrm{z}_{1}+\mathrm{z}_{2}+\mathrm{z}_{3}=\mathrm{z}_{4}$ (c) $\mathrm{z}_{1}+\mathrm{z}_{2}=\mathrm{z}_{3}+\mathrm{z}_{4}$ (d) $\mathrm{z}_{1}-\mathrm{z}_{2}=\mathrm{z}_{3}-\mathrm{z}_{4}$
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