JEE MAIN - Mathematics (2019 - 10th January Evening Slot - No. 6)
Let $$z = {\left( {{{\sqrt 3 } \over 2} + {i \over 2}} \right)^5} + {\left( {{{\sqrt 3 } \over 2} - {i \over 2}} \right)^5}.$$ If R(z) and 1(z) respectively denote the real and imaginary parts of z, then :
R(z) = $$-$$
3
R(z) < 0 and I(z) > 0
I(z) = 0
R(z) > 0 and I(z) > 0
Explanation
$$z = {\left( {{{\sqrt 3 + i} \over 2}} \right)^5} + {\left( {{{\sqrt 3 - i} \over 2}} \right)^5}$$
$$z = {\left( {{e^{i\pi /6}}} \right)^5} + {\left( {{e^{ - i\pi /6}}} \right)^5}$$
$$ = {e^{i5\pi /6}} + {e^{ - i5\pi /6}}$$
$$ = \cos {{5\pi } \over 6} + i{{\sin 5\pi } \over 6} + \cos \left( {{{ - 5\pi } \over 6}} \right) + i\sin \left( {{{ - 5\pi } \over 6}} \right)$$
$$ = 2\cos {{5\pi } \over 6} < 0$$
$${\rm I}(z) = 0$$ and $${\mathop{\rm Re}\nolimits} (z) < 0$$
$$z = {\left( {{e^{i\pi /6}}} \right)^5} + {\left( {{e^{ - i\pi /6}}} \right)^5}$$
$$ = {e^{i5\pi /6}} + {e^{ - i5\pi /6}}$$
$$ = \cos {{5\pi } \over 6} + i{{\sin 5\pi } \over 6} + \cos \left( {{{ - 5\pi } \over 6}} \right) + i\sin \left( {{{ - 5\pi } \over 6}} \right)$$
$$ = 2\cos {{5\pi } \over 6} < 0$$
$${\rm I}(z) = 0$$ and $${\mathop{\rm Re}\nolimits} (z) < 0$$
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