If `z!=0` upper bound of `|z^(i)|` is A) `|z|` B) 1 C) `e^(-pi)` D) `e^(pi)`

If z_1 and z_2 are two complex numbers for which |(z_1-z_2)(1-z_1z_2)|=1 and |z_2|!=1 then (A) |z_2|=2 (B) |z_1|=1 (C) z_1=e^(itheta) (D) z_2=e^(itheta)

If A+B+C, = pi and e^(i theta) = cos theta+i sin theta and z = det [[e^(-iC), e^(2iB), e^(-iA) e^ (-iB), e^(-iA), e^(2iC)]], then

int_ (0) ^ ((pi) / (3)) (dz) / (sqrt (e ^ (z)))

Let z_(1)=e^((i pi)/(5))

If z+sqrt(2)|z+1|+i=0 , then z= (A) 2+i (B) 2-i (C) -2-i (D) -2+i

If z=6e^(i(pi)/(3)) then |e^(iz)=1

If z_1 and z_2 are two non zero complex number such that |z_1+z_2|=|z_1|+|z_2| then arg z_1-argz_2 is equal to (A) - pi/2 (B) 0 (C) -pi (D) pi/2

For a comple number Z, if |Z-1+i|+|Z+i|=1 , then the range of the principle argument of Z is (where principle arg (Z) in (-pi, pi] )

Find the area bounded by |arg z|<=pi/4 and |z-1|<|z-3|