If $A (z_{1}), B (z_{2}), C (z_{3})$ be the vertices of triangle ABC in which $| A B C - ---- -$ = $\frac{π}{4}$ and $\frac{A B}{B C}$ = $\sqrt{2}$ then $z_{2}$ is equal to

Q. If

z_{1}, z_{2}, z_{3}

are the vertices of an equilateral triangle ABC such that

| z_{1} - i | = | z_{2} - i | = | z_{3} - i |

, then

| z_{1} + z_{2} + z_{3} |

equals to

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If A(z1),B(z2),C(z3) are the vertices of the triangle ABC such that (z1−z2)/(z3−z2)=(1/√2)+(i/√2), the triangle ABC is

The correct option is C Isoscelesz1−z2z3−z2=1+i√2Taking modulus we get|z1−z2z3−z2|=1|z1−z2|=|z3−z2|Therefore, two arms of the triangle have equal length.Hence the triangle is isosceles.

If $A (z_{1}), B (z_{2}), C (z_{3})$ are the vertices of the triangle ABC such that $(z_{1} - z_{2}) / (z_{3} - z_{2}) = (1 / \sqrt{2}) + (i / \sqrt{2})$ , the triangle ABC is

The correct option is C Isosceles
$\frac{z_{1} - z_{2}}{z_{3} - z_{2}} = \frac{1 + i}{\sqrt{2}}$
Taking modulus we get
$| \frac{z_{1} - z_{2}}{z_{3} - z_{2}} | = 1$
$| z_{1} - z_{2} | = | z_{3} - z_{2} |$
Therefore, two arms of the triangle have equal length.
Hence the triangle is isosceles.