The vector z = 3 - 4i is turned anticlockwise | vedclass

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Question

(B) Let the given complex number be $z = 3 - 4i$.Rotating a complex number $z$ anticlockwise by an angle $	heta$ is equivalent to multiplying it by $

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$\frac{-15}{2} + 10i$

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(B) Let the given complex number be $z = 3 - 4i$.Rotating a complex number $z$ anticlockwise by an angle $	heta$ is equivalent to multiplying it by $e^{i	heta}$.For $	heta = 180^{\circ} = \pi$ radians,the rotation factor is $e^{i\pi} = \cos(\pi) + i\sin(\pi) = -1$.After rotation,the number becomes $z' = z 	imes (-1) = -(3 - 4i) = -3 + 4i$.Stretching the vector by $2.5$ times is equivalent to multiplying the complex number by the scalar $2.5 = \frac{5}{2}$.Thus,the final complex number is $z'' = 2.5 	imes (-3 + 4i) = \frac{5}{2}(-3 + 4i) = \frac{-15}{2} + 10i$.

The vector $z = 3 - 4i$ is turned anticlockwise through an angle of $180^{\circ}$ and stretched $2.5$ times. The complex number corresponding to the newly obtained vector is

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