Let $$z = \,\cos \,\theta \, + i\,\sin \,\theta $$ . Then the value of $$\sum\limits_{m = 1}^{15} {{\mathop{\rm Im}\nolimits} } ({z^{2m - 1}})\,at\,\theta \, = {2^ \circ }$$ is

A particle P stats from the point $${z_0}$$ = 1 +2i, where $$i = \sqrt { - 1} $$. It moves horizontally away from origin by 5 unit and then vertically away from origin by 3 units to reach a point $${z_1}$$. From $${z_1}$$ the particle moves $$\sqrt 2 $$ units in the direction of the vector $$\hat i + \hat j$$ and then it moves through an angle $${\pi \over 2}$$ in anticlockwise direction on a circle with centre at origin, to reach a point $${z_2}$$. The point $${z_2}$$ is given by

Let A, B, C be three sets of complex numbers as defined below :

$$A = \left\{ {z:\,{\mathop{\rm Im}\nolimits} \,\,z\,\, \ge \,1} \right\}$$

$$B = \left\{ {z:\,\,\left| {z - 2 - i} \right| = 3} \right\}$$

$$C = \left\{ {z:\,{\mathop{\rm Re}\nolimits} (1 - i)z) = \sqrt 2 \,} \right\}$$

Let z be any point in $$A \cap B \cap C$$

Then, $${\left| {z + 1 - i} \right|^2} + {\left| {z - 5 - i} \right|^2}$$ lies between :