Let O be the origin and A be the point z_{1} | vedclass

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(D) Since $	riangle OAB$ is a right-angled isosceles triangle with $OB$ as the hypotenuse, the vector $\vec{AB}$ is obtained by rotating $\vec{OA}$ b

Vedclass · Accepted Answer

$|2z_{1} - z_{2}| = 5$

Vedclass · Answer

(D) Since $	riangle OAB$ is a right-angled isosceles triangle with $OB$ as the hypotenuse, the vector $\vec{AB}$ is obtained by rotating $\vec{OA}$ by $90^{\circ}$ ($i$ or $-i$).Given $z_{1} = 1 + 2i$, the vector $\vec{OA} = 1 + 2i$.Case $1$: $z_{2} - z_{1} = i(z_{1} - 0) = i(1 + 2i) = -2 + i$.Then $z_{2} = z_{1} + (-2 + i) = (1 + 2i) + (-2 + i) = -1 + 3i$.Here $\operatorname{Re}(z_{2}) = -1 Case $2$: $z_{2} - z_{1} = -i(z_{1} - 0) = -i(1 + 2i) = 2 - i$.Then $z_{2} = z_{1} + (2 - i) = (1 + 2i) + (2 - i) = 3 + i$.Here $\operatorname{Re}(z_{2}) = 3 > 0$, which is rejected.Thus, $z_{2} = -1 + 3i$.Now, $|z_{2}| = \sqrt{(-1)^{2} + 3^{2}} = \sqrt{10}$. (Option $C$ is true)$\arg z_{2} = \pi - 	an^{-1} 3$. (Option $A$ is true)$|2z_{1} - z_{2}| = |2(1 + 2i) - (-1 + 3i)| = |2 + 4i + 1 - 3i| = |3 + i| = \sqrt{3^{2} + 1^{2}} = \sqrt{10}$.Since $|2z_{1} - z_{2}| = \sqrt{10} 
eq 5$, option $D$ is $NOT$ true.

Let $O$ be the origin and $A$ be the point $z_{1} = 1 + 2i$. If $B$ is the point $z_{2}$ with $\operatorname{Re}(z_{2}) < 0$, such that $\triangle OAB$ is a right-angled isosceles triangle with $OB$ as the hypotenuse, then which of the following is $NOT$ true?

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