JEE MAIN - Mathematics (2021 - 26th August Morning Shift - No. 12)
Let $$z = {{1 - i\sqrt 3 } \over 2}$$, $$i = \sqrt { - 1} $$. Then the value of $$21 + {\left( {z + {1 \over z}} \right)^3} + {\left( {{z^2} + {1 \over {{z^2}}}} \right)^3} + {\left( {{z^3} + {1 \over {{z^3}}}} \right)^3} + .... + {\left( {{z^{21}} + {1 \over {{z^{21}}}}} \right)^3}$$ is ______________.
Answer
13
Explanation
$$z = {{1 - \sqrt {3i} } \over 2} = {e^{ - i{\pi \over 3}}}$$
$${z^r} + {1 \over {{z^r}}} = 2\cos \left( { - {\pi \over 3}} \right)r = 2\cos {{r\pi } \over 3}$$
$$ \Rightarrow 21 + \sum\limits_{r = 1}^{21} {{{\left( {{z^r} + {1 \over {{z^r}}}} \right)}^3} = 8\left( {{{\cos }^3}{{r\pi } \over 3}} \right) = 2\left( {\cos r\pi + 3\cos {{r\pi } \over 3}} \right)} $$
$$ \Rightarrow 21 + {\left( {z + {1 \over 2}} \right)^3} + {\left( {{z^2} + {1 \over {{z^2}}}} \right)^3} + ....{\left( {{z^{21}} + {1 \over {{z^{21}}}}} \right)^3}$$
$$ = 21 + \sum\limits_{r = 1}^{21} {{{\left( {{z^r} + {1 \over {{z^r}}}} \right)}^3}} $$
$$ = 21 + \sum\limits_{r = 1}^{21} {\left( {2\cos r\pi + 6\cos {{r\pi } \over 3}} \right)} $$
$$ = 21 - 2 - 6$$
$$ = 13$$
$${z^r} + {1 \over {{z^r}}} = 2\cos \left( { - {\pi \over 3}} \right)r = 2\cos {{r\pi } \over 3}$$
$$ \Rightarrow 21 + \sum\limits_{r = 1}^{21} {{{\left( {{z^r} + {1 \over {{z^r}}}} \right)}^3} = 8\left( {{{\cos }^3}{{r\pi } \over 3}} \right) = 2\left( {\cos r\pi + 3\cos {{r\pi } \over 3}} \right)} $$
$$ \Rightarrow 21 + {\left( {z + {1 \over 2}} \right)^3} + {\left( {{z^2} + {1 \over {{z^2}}}} \right)^3} + ....{\left( {{z^{21}} + {1 \over {{z^{21}}}}} \right)^3}$$
$$ = 21 + \sum\limits_{r = 1}^{21} {{{\left( {{z^r} + {1 \over {{z^r}}}} \right)}^3}} $$
$$ = 21 + \sum\limits_{r = 1}^{21} {\left( {2\cos r\pi + 6\cos {{r\pi } \over 3}} \right)} $$
$$ = 21 - 2 - 6$$
$$ = 13$$
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