Question

I^104+i^109+i^114+i^119=0​

Answer 1

Answer:

To Find :

Prove that I^104+i^109+i^114+i^119=0

Solution:

LHS :

$\sf \: {i}^{104} \\ \\ \tt \: we \: know \: that \:$

$\sf \: {i}^{4} = 1 \\ \\ therefore$

$\sf \: {i}^{104} = 1$

$\sf \: {i}^{109} = {i}^{108 + 1}$

$\tt \: we \: know \: that$

$\sf \: {i}^{108 } = 1 \: \: and \: {i}^{1} = 1$

$therefore$

$\sf \: {i}^{109} = i$

$\: \sf \: {i}^{114} = {i}^{112 + 2}$

$\tt \: we \: know \: \: that \:$

$\sf \: {i}^{112} = 1 \: \: and \: \: {i}^{2} = - 1$

$therefore$

$\sf \: {i}^{114} = - 1$

$\sf \: {i}^{119} = {i}^{116 + 3}$

$\tt \: we \: know \: that$

$\sf \: {i}^{116} = 1 \: \: and \: \: {i}^{3} = - 1$

$therefore$

$\sf \: {i}^{119} = - i$

$\sf \longrightarrow \: {i}^{114} + {i}^{109} + {i}^{114} + {i}^{119}$

$\sf \longrightarrow \: (1) + (i) + ( - 1) + ( - i)$

$\sf \longrightarrow \: 1 - 1 + i - i$

$\sf \longrightarrow \: 0$

LHS = RHS

Hence Proved!!!

Answer 2

Given :   i¹⁰⁴ + i¹⁰⁹  + i¹¹⁴ + i¹¹⁹  = 0

To Find :  Show

Solution:

i² = - 1

i³ = - i

i⁴  =  1

xᵃ⁺ᵇ = xᵃxᵇ

i¹⁰⁴ + i¹⁰⁹  + i¹¹⁴ + i¹¹⁹  = 0

LHS =

i¹⁰⁴  (1  + i⁵ ) + i¹¹⁴(1  + i⁵)

i⁵ = i⁴.i = i

=  i¹⁰⁴  (1  + i ) + i¹¹⁴(1  + i)

= (1 + i) (  i¹⁰⁴ +  i¹¹⁴)

=  (1 + i) i¹⁰⁴ ( 1 + i¹⁰)

=   (1 + i) i¹⁰⁴ ( 1 + i⁸.i²)

i⁸ = (i⁴)² = 1

=   (1 + i) i¹⁰⁴ ( 1 + i²)

= (1 + i)i¹⁰⁴ ( 1 -1)

= (1 + i)i¹⁰⁴ (0)

= 0

= RHS

Hence LHS = RHS

i¹⁰⁴ + i¹⁰⁹  + i¹¹⁴ + i¹¹⁹  = 0

Shown

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