JEE MAIN - Mathematics Hindi (2019 - 10th January Morning Slot - No. 16)
यदि z1 और z2 कोई दो शून्य से भिन्न जटिल संख्याएँ हों और $$3\left| {{z_1}} \right| = 4\left| {{z_2}} \right|.$$ यदि $$z = {{3{z_1}} \over {2{z_2}}} + {{2{z_2}} \over {3{z_1}}}$$ तो :
$${\rm I}m\left( z \right) = 0$$
$$\left| z \right| = \sqrt {{17 \over 2}} $$
$$\left| z \right| =$$ $${1 \over 2}\sqrt {9 + 16{{\cos }^2}\theta } $$
Re(z) $$=$$ 0
Comments (0)
