\(\text{consider }\tilde{z} = z-2\\ |\tilde{z}|=2 \text{ and }\arg(\tilde{z}) = -\dfrac{\pi}{3}\\ \tilde{z} = 2 \cos\left(-\dfrac{\pi}{3}\right) + i 2 \sin\left(-\dfrac{\pi}{3}\right)\\ \tilde{z} = 1-i \sqrt{3},~\text{ and thus}\\ z = \tilde{z}+2 = 3 - i \sqrt{3}\)
.
