Question

Solution

Correct option is

                               

                               

                               equals I m open square brackets fraction numerator minus 1 over denominator 2 space i space sinθ end fraction open curly brackets 1 minus cos space 30 to the power of straight o minus i space sin space 30 to the power of straight o close curly brackets close square brackets

                               

SIMILAR QUESTIONS

Q1

then area of the triangle whose vertices are z1, z2, zis

Q2
Q5

 then absolute value of 

Q6
Q7

Shaded region is given by

                                                         

Q8

For complex numbers z1, z2 and z3 satisfying  are the vertices of a triangle which is 

Q9

If the points z, – iz and 1 are collinear then z lies on

 

Q10

Let a, b, c be three points lying on the circle