Discussion
When a complex number Z is multiplied by i, the resulting vector iz, in the z plane, is obtained by rotating the vector z through :
Answer: D
Multiplying by i rotates the vector through \(\pi/2\), anticlockwise.
In polar form \(i = e^{i\pi/2}\) and \(z = re^{i\theta}\), so
\[ iz = re^{i(\theta + \pi/2)} \]
The modulus is unchanged and the argument increases by \(\pi/2\). For example, \(i \times 1 = i\) turns the point (1, 0) to (0, 1).
A rotation through \(\pi\) is multiplication by \(-1\), and a rotation through \(-\pi/2\) (equivalently \(3\pi/2\)) is multiplication by \(-i\), which turns 1 into \(-i\).
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