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Question: The complex number \(\frac{2^{n}}{(1 - i)^{2n}} + \frac{(1 + i)^{2n}}{2^{n}},(n \in Z)\) is equal to...

The complex number 2n(1−i)2n+(1+i)2n2n,(n∈Z)\frac{2^{n}}{(1 - i)^{2n}} + \frac{(1 + i)^{2n}}{2^{n}},(n \in Z) is equal to

A

0

B

2

C

[1+(−1)n].in\lbrack 1 + ( - 1)^{n}\rbrack.i^{n}

D

None of these

Answer

None of these

Explanation

Solution

Sol.(1+i)2n=((1+i)2)n=(1+i2+2i)n=(1−1+2i)n=2nin(1−i)2n=((1−i)2)n=(1+i2−2i)n=(1−1−2i)n=(−2)nin(1 + i)^{2n} = ((1 + i)^{2})^{n} = (1 + i^{2} + 2i)^{n} = (1 - 1 + 2i)^{n} = 2^{n}i^{n}\mathbf{(1 - i}\mathbf{)}^{\mathbf{2n}}\mathbf{= ((1 - i}\mathbf{)}^{\mathbf{2}}\mathbf{)}^{\mathbf{n}}\mathbf{= (1 +}\mathbf{i}^{\mathbf{2}}\mathbf{- 2i}\mathbf{)}^{\mathbf{n}}\mathbf{= (1 - 1 - 2i}\mathbf{)}^{\mathbf{n}}\mathbf{= ( - 2}\mathbf{)}^{\mathbf{n}}\mathbf{i}^{\mathbf{n}}∴2n(1−i)2n+(1+i)2n2n=2n(−2)nin+2nin2n=1(−1)nin+in\frac{2^{n}}{(1 - i)2^{n}} + \frac{(1 + i)^{2n}}{2^{n}} = \frac{2^{n}}{( - 2)^{n}i^{n}} + \frac{2^{n}i^{n}}{2^{n}} = \frac{1}{( - 1)^{n}i^{n}} + i^{n} =1+(−1)ni2n(−1)nin=1+(−1)n(i2)n(−1)nin= \frac{1 + ( - 1)^{n}i^{2n}}{( - 1)^{n}i^{n}} = \frac{1 + ( - 1)^{n}(i^{2})^{n}}{( - 1)^{n}i^{n}}

=1+(−1)n(−1)n(−1)nin=1+(−1)2n(−1)nin=1+1(−1)nin=2(−1)nin.= \frac{1 + ( - 1)^{n}( - 1)^{n}}{( - 1)^{n}i^{n}} = \frac{1 + ( - 1)^{2n}}{( - 1)^{n}i^{n}} = \frac{1 + 1}{( - 1)^{n}i^{n}} = \frac{2}{( - 1)^{n}i^{n}}.