Question

Prove that: $\sf{(1+i)^{4}\times(1+\frac{1}{i})^{4}=16}$​

Answer 1

Simple question

take LHS

LHS=(1+i)^4 ×(i+1/i)^4

LHS=(1+i)^4×(i+1)^4/i^4

we know that i^4=1

LHS=(1+i)^4×(1+i)^4

we know,(1+i)^2=1+i^2+2i

=2i

{(1+I)^2}^2=(2i)^2=-4

putting this value we get

LHS=-4×-4=16

NOTE=>i=√-1,so i^2=-1

Answer 2

Given :

$\sf{(1+i)^{4}\times(1+\dfrac{1}{i})^{4}=16}$

where i = √-1

To prove :

• LHS = RHS

Proof :

LHS :-

$\sf{(1+i)^{4}\times(\dfrac{i + 1}{i})^{4}}$

$\sf{(1+i)^{4}\times\dfrac{ {(i + 1)}^{4} }{ {(i)}^{4} }}$

we know that :-

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

Therefore we get :-

$\sf{(1+i)^{4}\times\dfrac{ {(i + 1)}^{4} }{ 1 }}$

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

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

Now :-

$\sf{(1 + i)}^{2} = {1}^{2} + {i}^{2} + 2i$

• now we know that i² = -1

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

$\sf{(1 + i)}^{2} = 2i$

$\sf{(2i)^{2}\times(2i)^{2}}$

$\sf{4 \times ( - 1)\times4 \times ( - 1)}$

$\sf{16}$

Therefore LHS = RHS

hence proved