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i^{2}\left(\sqrt{3}\right)^{2}+1-i\sqrt{3}
Expand \left(i\sqrt{3}\right)^{2}.
-\left(\sqrt{3}\right)^{2}+1-i\sqrt{3}
Calculate i to the power of 2 and get -1.
-3+1-i\sqrt{3}
The square of \sqrt{3} is 3.
-2-i\sqrt{3}
Add -3 and 1 to get -2.