Answer:
[tex]\qquad\qquad\qquad\boxed{ \bf{ \:(b) \: \: \: 2 \: \: }} \\ \\ [/tex]
Explanation:
Given that,
[tex]\qquad\sf \: {(1 + i)}^{2n} = {(1 - i)}^{2n} \\ \\ [/tex]
can be further rewritten as
[tex]\qquad\sf \: \dfrac{{(1 + i)}^{2n}}{{(1 - i)}^{2n}} = 1 \\ \\ [/tex]
[tex]\qquad\sf \: {\bigg(\dfrac{1 + i}{1 - i} \bigg) }^{2n} = 1 \\ \\ [/tex]
On rationalizing the denominator, we get
[tex]\qquad\sf \: {\bigg(\dfrac{1 + i}{1 - i} \times \dfrac{1 + i}{1 + i} \bigg) }^{2n} = 1 \\ \\ [/tex]
[tex]\qquad\sf \: {\bigg(\dfrac{(1 + i)^{2} }{ {1}^{2} - {i}^{2} } \bigg) }^{2n} = 1 \\ \\ [/tex]
[tex]\qquad\boxed{ \sf{ \: \because \: (x + y)(x - y) = {x}^{2} - {y}^{2} \: }} \\ \\ [/tex]
[tex]\qquad\sf \: {\bigg(\dfrac{ {1}^{2} + {i}^{2} + 2i}{ 1 - ( - 1) } \bigg) }^{2n} = 1 \: \: \: \: \boxed{ \sf{ \: \because \: {i}^{2} = - 1 \: }}\\ \\ [/tex]
[tex]\qquad\boxed{ \sf{ \: \because \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \: }} \\ \\ [/tex]
[tex]\qquad\sf \: {\bigg(\dfrac{ 1 - 1+ 2i}{2 } \bigg) }^{2n} = 1 \: \: \: \: \boxed{ \sf{ \: \because \: {i}^{2} = - 1 \: }}\\ \\ [/tex]
[tex]\qquad\sf \: {\bigg(\dfrac{ 2i}{2 } \bigg) }^{2n} = 1 \: \\ \\ [/tex]
[tex]\qquad\sf \: {(i)}^{2n} = 1 \: \\ \\ [/tex]
[tex]\qquad\sf \: {(i)}^{2n} = {(i)}^{4} \: \\ \\ [/tex]
[tex]\qquad\sf \: 2n = 4 \: \\ \\ [/tex]
[tex]\qquad\sf\implies \sf \: n = 2 \: \\ \\ [/tex]