Question

Iota^107+iota^112+ iota^117+iota^122=0​

Answer 1

Step-by-step explanation:

i^107 + i^112 + i^117 + i^122

i^107 ( 1 + i^5 + i^10 + i^15 )

(i²)^53 × i { 1 + (i²)² × i + (i²)^5 + (i²)^7 × i }

(-1)^53 × i { 1 + (-1)² × i + (-1)^5 + (-1)^7 × i }

(-1) × i { 1 + i + (-1) + (-1)i }

-i { 1 + i - 1 - i }

-i × 0 = 0

Answer 2

Step-by-step explanation:

iota ^107 + iota^112 + iota^117 + iota^122 = 0

L.H.S :-

1. iota^107 ( 1+ iota^5 + iota^10 + iota^15 )

= (iota^2)^53 × iota [ 1+(i^2)^2×i+(i^2)^5+(i^2)^7×i ]

we know that :- i^2 = (-1) ......

= (-1)^53×i( 1+(-1)^2×i + (-1)^5+(-1)^7×i )

= (-1)×i[ 1+i+(-1)+(-1)×i ]

= -i ( 1 + i - 1 - i )

= -i × 0

= 0

L.H.S = R.H.S

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(hope it helps you) ...