(C) We have $(1+i)^2 = 1+i^2+2i = 1-1+2i = 2i$.Similarly,$(1-i)^2 = 1+i^2-2i = 1-1-2i = -2i$.Now,$(1+i)^{10} = ((1+i)^2)^5 = (2i)^5 = 2^5 \times i^5 =

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(C) We have $(1+i)^2 = 1+i^2+2i = 1-1+2i = 2i$.Similarly,$(1-i)^2 = 1+i^2-2i = 1-1-2i = -2i$.Now,$(1+i)^{10} = ((1+i)^2)^5 = (2i)^5 = 2^5 \times i^5 = 32i$.And $(1-i)^{10} = ((1-i)^2)^5 = (-2i)^5 = (-2)^5 \times i^5 = -32i$.Therefore,$(1+i)^{10} + (1-i)^{10} = 32i + (-32i) = 0$.

Class 11 Mathematics 1 Complex Numbers and Quadratic Equations: Complex Numbers Integral power of iota, Algebraic operations and Equality of complex numbers

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If $i=\sqrt{-1}$,then $(1+i)^{10}+(1-i)^{10}=$

TS EAMCET 2023 All TS EAMCET

A
$32$
B
$64$
C
$0$
D
$-64i$

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1 Complex Numbers and Quadratic Equations: Complex Numbers

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Integral power of iota, Algebraic operations and Equality of complex numbers

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If $i=\sqrt{-1}$,then $(1+i)^{10}+(1-i)^{10}=$

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