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A complex number z is rotated in anti-clockwise direction by an angle $α$ and we get $z^{1}$ and if the same complex number z is rotated by an angle $α$ in clockwise direction and we get $z^{11}$ then

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$z^{1}, z, z^{11}$ are in G.P

$z^{1}, z, z^{11}$ are in H.P

$z^{1} + z^{11} = 2 z \cos α$

${(z^{1})}^{2} + {(z^{11})}^{2} = 2 z^{2} \cos 2 α$

answer is A, C, D.

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Detailed Solution

Therefore, the number of solutions is p+q+1

$z^{'} = z e^{i α} z^{''} = z e^{- i α} ∴ z^{'} z^{''} = z^{2} \Rightarrow z^{'}, z, z^{''} a r e i n G . P A l s o \Rightarrow z^{'} + z^{''} = 2 z \cos α$

${(z^{1})}^{2} + {(z^{11})}^{2} = z^{2} (e^{i 2 α} + e^{- i 2 α}) = 2 z^{2} \cos 2 α$

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