Express (1+i)2n+1(1−i)2n−1,n∈N in the modulus-amplitude form
A
{2(cos0o+isin0o)if n is even2(cosπ−isinπ)if n is odd
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B
{2(cos0o−isin0o)if n is even2(cosπ+isinπ)if n is odd
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C
{2(cos0o+isin0o)if n is even2(cosπ+isinπ)if n is odd
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D
{2(cos0o−isin0o)if n is even2(cosπ−isinπ)if n is odd
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Solution
The correct option is C{2(cos0o+isin0o)if n is even2(cosπ+isinπ)if n is odd The equation can be simplified to: (1+i1−i)2n.(1+i).(1−i)=((1+i)(1+i)(1−i)(1+i))2n.2=(2i2)2n.2=2.i2n=2.(−1)nneven:2=2(cos0o+isin0o)nodd:−2=2(cosπ+isinπ) Hence, (c) is correct.