Question

15) Prove that (1+i)^4×(1+1/i)4=16.​

Answer 1

Answer:

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Answer 2

Answer:

$(1+i)^{2} \times(1+\frac{1}{i} )^{4} =16$

Step-by-step explanation:

We know that,

The following are few important properties of iota;

$i^{2} =-1$

$i^{3} =-i$

$i^{4} =1$

Also the identity we are going to apply in this Proof is,

                                   $a^{2} -b^{2} = (a-b)\times (a+b)$

Consider the left hand side:

$(1+i)^{2} \times(1+\frac{1}{i} )^{4}$

(Rationalization )multiply i in numerator and denominator in fractional part of  $(1+\frac{1}{i})^{4}$

= $(1+i)^{2} \times(1+\frac{i}{i^{2} } )^{4}$

= $(1+i)^{2} \times(1+\frac{i}{-1} } )^{4}$

= $(1+i)^{2} \times(1- i} )^{4}$

= $[(1+i) \times (1-i)]^{4}$

The terms inside the square bracket is in the form of (a+b) × (a-b),

hence the $a^{2} -b^{2}$ identity can be applied here

= $[(1^{2}- i^{2} ) ]^{4}$

= $[(1- (-1)) ]^{4}$     ∵$i^{2} =-1$

= $(1+1)^{4}$

= $2^{4}$

= 16

=RHS

Therefore LHS= RHS

Hence   $(1+i)^{2} \times(1+\frac{1}{i} )^{4} =16$ is proved.