Discussion
The point \(z = \frac{1}{\sqrt{2}}(1+i)\) in the complex plane is rotated about the origin through an angle \(\frac{\pi}{4}\) in the clockwise direction, then the new position of \(z\) is
Answer: B
The new position is 1.
Write z in polar form: \(\frac{1}{\sqrt{2}}(1 + i) = \cos\frac{\pi}{4} + i\sin\frac{\pi}{4} = e^{i\pi/4}\), a point at distance 1 from the origin at angle \(\frac{\pi}{4}\).
A clockwise rotation through \(\frac{\pi}{4}\) multiplies by \(e^{-i\pi/4}\):
\[ e^{i\pi/4} \cdot e^{-i\pi/4} = e^{0} = 1 \]
The point lands on the positive real axis at 1. \(\frac{1}{\sqrt{2}}(1 - i)\) is the result of rotating clockwise through \(\frac{\pi}{2}\) instead, and 2 and \(1 - i\) change the distance from the origin, which a rotation cannot do.
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