Question

If  a = (1+i)/2 , if the value of a^6 + a^4 + a^2 + 1 is  u(1+i)  . Find u​

Answer 1

Given:-

⇒ $a = \frac{1 + i}{2}$

and ⇒ $a^6 + a^4 + a^2 + 1 = u(1+i)$

To find:-

⇒ u = ?

Solution:-

⇒ $a^6 + a^4 + a^2 + 1 = u(1+i)$

( taking common )

⇒ $a^4 ( a^2 + 1 ) + 1( a^2 + 1 )$

⇒ $( a^4 + 1 )(* a^2 + 1 )------(1)$

@from given:

⇒ $a = \frac{ 1 + i}{2}$

squaring both side:

⇒ $a^2 = \frac{ 1 + i^2 + 2i}{2}$

⇒ $a^2 = \frac{2i}{2}$

⇒ $a^2 = i$

therefore:

⇒ $a^6 + a^4 + a^2 + 1 = u(1+i)$

⇒ $( a^4 + 1 ) ( a^2 + 1) = u(1+i) ------( from(1))$

⇒ $((a^2)^2 + 1 ) ( a^2 +1) = u(1+i)$

⇒ $( i^2 + 1 ) ( i^2 + 1 ) = u ( 1+i )$

⇒ $0 = u(1+i)$

⇒ u = 0

hence the value of u is 0..

thank you......

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