Question

A complex number $z$ is rotated in anticlockwise direction by an angle $\alpha$ and we get $z^{\prime }$, and if the same complex number $z$ is rotated by an angle $\alpha$ in clockwise direction and we get $z^{\prime \prime }$, then

• A. $z^{\prime },z,z^{\prime \prime }$ are in $G.P$.
• B. $z^{\prime },z,z^{\prime \prime }$ are in $H.P$.
• C. $z^{\prime }+z^{\prime \prime }=2z\cos \alpha$
• D. $(z^{\prime }{)}^2+(z^{\prime \prime }{)}^2=2z^2\cos 2\alpha$

Answer 1

Answer: (A), (C), (D)

$z^{\prime }+z^{\prime \prime }=2z\cos \alpha$
$(z^{\prime }{)}^2+(z^{\prime \prime }{)}^2=2z^2\cos 2\alpha$
$z^{\prime },z,z^{\prime \prime }$ are in $G.P$.

If a complex number $z$ is rotated by an angle $\alpha$ in the anticlockwise direction, we get the answer to be $z{e}^{i\alpha }.$
So, when it is rotated in clockwise direction, the answer would be $z{e}^{-i\alpha }$.
So, $z^{\prime }=z{e}^{i\alpha },z^{\prime \prime }=z{e}^{-i\alpha }$.
Thus, $z^{\prime }{z}^{\prime \prime }=z^2$, implying the three are in G.P.
Also, $z^{\prime }+z^{\prime \prime }=z(e^{i\alpha }+e^{-i\alpha })=2zcos\alpha$.
$z^{\prime 2}+z^{\prime \prime 2}=z^2{e}^{2i\alpha }+e^{-2i\alpha }=2z^{2}cos2\alpha$.