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The centre of a square \( A B C D \) is at \( z=0, A \) is \( z_{1} \). Then, the centroid of \( \triangle A B C \) is
(a) \( z_{1}(\cos \pi \pm i \sin \pi) \)
(b) \( \frac{z_{1}}{3}(\cos \pi \pm i \sin \pi) \)
(c) \( z_{1}\left(\cos \frac{\pi}{2} \pm i \sin \frac{\pi}{2}\right) \)
(d) \( \frac{z_{1}}{3}\left(\cos \frac{\pi}{2} \pm i \sin \frac{\pi}{2}\right) \)
(where, \( i=\sqrt{-1} \) )
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