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Best answer
5 5 votes
$\frac{(1-i\sqrt 3)^{30}}{ (1+i)^{60}}\\
=\frac{(1-2i\sqrt 3-3)^{15}}{(1+2i-1)^{30}}\\
=\frac{(-2-2i\sqrt 3)^{15}}{(2i)^{30}}\\
=\frac{(-2)^{15}(1+i\sqrt 3)^{15}}{(2i)^{30}}\\
=\frac{(-1)^{15}(2)^{15}(1+i\sqrt 3)^{15}}{(-1)(2)^{30}}\\
=\frac{(1+3i\sqrt 3-9-3i\sqrt 3)^5}{2^{15}}\\
=\frac{(-8)^5}{2^{15}}\\
=-1$
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0 0 votes
option c

-1 is right ans .

very simple but lengthy

do it by simplification

first break both square form

than break numerator in qubic for it will easily crack the qus
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