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z=-1-i \sqrt{3} \\ \\|z|= \sqrt{(-1)^2+( \sqrt{3})^2 }= \sqrt{1+3}= \sqrt{4}=2\\ \\ \phi=cos^{-1}( \frac{-1}{2} )= \frac{2 \pi }{3}\ or\  \frac{4 \pi }{3}

But z is in the third quadrant. So \phi= \frac{4 \pi }{3}
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