i বিষয়ক

(1+i)8+(1i)8=(1 + i)^8 + (1 -i)^8 =

JCC 24

(1+i)8+(1i)8{ \left( 1+i \right) }^{ 8 }+{ \left( 1-i \right) }^{ 8 }
(1+i)8=[(1+i)2]4=[1+2i+i2]4=[1+2i1]4\Rightarrow { \left( 1+i \right) }^{ 8 }={ \left[ { { \left( 1+i \right) }^{ 2 } } \right] }^{ 4 }={ \left[ { 1+2i+{ i }^{ 2 } } \right] }^{ 4 }={ \left[ { 1+2i-1 } \right] }^{ 4 }
=[(2i)2]2={ \left[ { { \left( 2i \right) }^{ 2 } } \right] }^{ 2 }
=(4)2={ \left( -4 \right) }^{ 2 }
=16=16

Similarely (1i)8=16{ \left( 1-i \right) }^{ 8 }=16

(1+i)8+(1i)8\therefore { \left( 1+i \right) }^{ 8 }+{ \left( 1-i \right) }^{ 8 }

=16+16=16+16

=32.=32.

Hence, the answer is 32.32.

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