The correct option is C 32 ( 1+i ) #160; ^ 8 +( 1 i ) #160; ^ 8 ⇒( 1+i ) #160; ^ 8 =[ ( 1+i ) #160; ^ 2 ] #160; ^ 4 =[ 1+2i+ i ^ 2 ] ...

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Complex Numbers

1 + i8 + 1 -i...

Question

$(1 + i)^{8} + (1 - i)^{8} =$

16

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- 16

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32

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- 32

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Solution

The correct option is C $32$
${(1 + i)}^{8} + {(1 - i)}^{8}$
$\Rightarrow {(1 + i)}^{8} = {[{(1 + i)}^{2}]}^{4} = {[1 + 2 i + i^{2}]}^{4} = {[1 + 2 i - 1]}^{4}$
$= {[{(2 i)}^{2}]}^{2}$
$= {(- 4)}^{2}$
$= 16$

Similarely ${(1 - i)}^{8} = 16$

$∴ {(1 + i)}^{8} + {(1 - i)}^{8}$

$= 16 + 16$

$= 32.$

Hence, the answer is $32.$

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