Tag: complex numbers and linear inequations

Let $A$ and $B$ represent $z _{1}$ and $z _{2}$ in the Argand plane and $z _{1},z _{2}$ be the roots of the equation $z^{2}+pz+q=0$ where $p,q$ are complex numbers. If $O$ is the origin $OA=OB$ and $\angle AOB=\alpha$ then $p^{2}=$

Let  $z _ { 1 } , z _ { 2 }$  and  $z _ { 3 }$  represent the vertices  $A, B$  and  $C$  of the triangle  $A B C$  in the argand that  $\left| z _ { 1 } \right| = \left| z _ { 2 } \right| = \left| z _ { 3 } \right| = 5,$  then  $z _ { 1 } \sin 2 A + z _ { 2 } \sin 2 B + z _ { 3 } \sin 2 C = 0.$

If $\sin \frac {6\pi}5+i(1+\cos \frac {6\pi }5)$ then

If Arg $(z + i)\, -$ Arg $(z - i)$ $= \dfrac{\pi}{2}$, then $z$ lies on a ..........

If $\overline { z } $ lies in the third quadrant then $z$ lies in the

Let $z _1$ and $z _2$ are two complex numbers such that $(1-i)z _1=2z _2$ and $arg(z _1z _2)=\dfrac{\pi}{2}$ then $arg(z _2)$ is equals to:

The complex number $\dfrac{1 + 2i}{1 - i}$ lies in which quadrant of the complex plane.

If $arg(z) < 0$, then $arg(-z)-arg(z)=$

Which of the given alternatives represent a point in Argand plane, equidistant from roots of the equation $(z+1)^4= 16z^4$?

A particle starts from a point $z _0= I + i$, where $i
=\sqrt{-1}$ It moves horizontally away from origin by $2$ units and then
vertically away from origin by $3$ units to reach a point$ z _1$. From $z _1$
particle moves $\sqrt{5}$ units in the direction of $2\hat i + \hat j$ and
then it moves through an angle of $\cos e{c^{ - 1}}\sqrt 2 $ in anticlockwise
direction of a circle with centre at origin to reach a point $z _2$ . The arg $z _2$ is given by