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\(\eqalign{
& a)\,\,{(\sqrt 3 + i)^2} – {(\sqrt 3 – i)^2} \cr
& b)\,{(\sqrt 3 + i)^2} + {(\sqrt 3 – i)^2} \cr
& c)\,{(\sqrt 3 + i)^3} – {(\sqrt 3 – i)^3} \cr
& d)\,{{{{(\sqrt 3 + i)}^2}} \over {{{(\sqrt 3 – i)}^2}}} \cr} \)
a)
\(\eqalign{
& {(\sqrt 3 + i)^2} – {(\sqrt 3 – i)^2} \cr&= {\rm{[}}\sqrt 3 + i + \sqrt 3 – i{\rm{][}}\sqrt 3 + i – \sqrt 3 + i{\rm{]}} \cr
& {\rm{ = 4}}\sqrt 3 i \cr} \)
b)
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\({(\sqrt 3 + i)^2} + {(\sqrt 3 – i)^2} = 2 + 2\sqrt 3 i + 2 – 2\sqrt 3 i = 4\)
c)
\(\eqalign{
& {(\sqrt 3 + i)^2} – {(\sqrt 3 – i)^2} = {\rm{[}}\sqrt 3 + i – \sqrt 3 + i{\rm{][}}{(\sqrt 3 + i)^2} + {(\sqrt 3 )^2} – {i^2} + {(\sqrt 3 – i)^2}{\rm{]}} \cr
& = 2i(4 + 4) = 16i \cr} \)
d) \({{{{(\sqrt 3 + i)}^2}} \over {{{(\sqrt 3 – i)}^2}}} = {{2 + 2\sqrt 3 i} \over {2 – 2\sqrt 3 i}} = {{1 + \sqrt 3 i} \over {1 – \sqrt 3 i}} = {{ – 1 + \sqrt 3 i} \over 2}\)