`lim_(z rarr-3)[(sqrt(z+6))/(z)]`

int(dz)/(z sqrt(z^(2)-1))

(lim)_(z->1)(z^(1/3)-1)/(z^(1/6)-1)

If x=int_(0)^(t^(2))e^(sqrt(z)){(2tan sqrt(z)+1-tan^(2)sqrt(z))/(2sqrt(z)sec^(2)sqrt(z))}dz and y=int_(0)^(t^(2))e^(sqrt(z)){(1-tan^(2)sqrt(z)-2tan sqrt(z))/(2sqrt(z)sec^(2)sqrt(z))}dz : Then the inclination of the tangent to the curve at t=(pi)/(4) is :

z is such that a r g ((z-3sqrt(3))/(z+3sqrt(3)))=pi/3 then locus z is

If z=cos(pi/4)+isin(pi/6) , then a. |z|=1, arg(z)=\ pi/4 b. |z|=1, arg(z)=\ pi/6 c. |z|=(sqrt(3))/2, arg(z)=(5pi)/(24) d. |z|=(sqrt(3))/2, arg(z)=tan^(-1)(1/sqrt(2))

If log_(sqrt3) |(|z|^2 -|z| +1|)/(|z|+2)|<2 then locus of z is

Let z_1, z_2, z_3 be the three nonzero complex numbers such that z_2!=1,a=|z_1|,b=|z_2|a n d c=|z_3|dot Let |a b c b c a c a b|=0 a r g(z_3)/(z_2)=a r g((z_3-z_1)/(z_2-z_1))^2 orthocentre of triangle formed by z_1, z_2, z_3, i sz_1+z_2+z_3 if triangle formed by z_1, z_2, z_3 is equilateral, then its area is (3sqrt(3))/2|z_1|^2 if triangle formed by z_1, z_2, z_3 is equilateral, then z_1+z_2+z_3=0

For any two complex numbers, z_(1),z_(2) |1/2(z_(1)+z_(2))+sqrt(z_(1)z_(2))|+|1/2(z_(1)+z_(2))-sqrt(z_(1)z_(2))| is equal to

The complex numbers z_1, z_2 and z_3 satisfying (z_1-z_3)/(z_2-z_3) =(1- i sqrt(3))/2 are the vertices of triangle which is (1) of area zero (2) right angled isosceles(3) equilateral (4) obtuse angled isosceles

The point z_1=3+sqrt(3)i and z_2=2sqrt(3)+6i are given on a complex plane. The complex number lying on the bisector of the angel formed by the vectors z_1a n dz_2 is z=((3+2sqrt(3)))/2+(sqrt(3)+2)/2i z=5+5i z=-1-i none of these