Evaluate the mentioned below-iv-6-4-2-1-when -1-2

Given that 𝑥=(1+𝑖)/ √(2) ⇒√(2) 𝑥=1+𝑖 ⇒(√(2) 𝑥)^2=(1+𝑖)^2 nbsp; nbsp; nbsp; nbsp; nbsp; squaring on both side ⇒2𝑥^2=1+𝑖^2+2𝑖 ^2=𝑖 …(1) ^6+𝑥^4+𝑥^2+1 =( ...

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Standard XII

Mathematics

Modulus of a Complex Number

Evaluate the ...

Question

Evaluate the mentioned below:
(iv) $x^{6} + x^{4} + x^{2} + 1, w h e n x = (1 + i) / \sqrt{2}$

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Solution

Given that-
$x = (1 + i) / \sqrt{2}$
$\Rightarrow \sqrt{2} x = 1 + i$
$\Rightarrow (\sqrt{2} x)^{2} = (1 + i)^{2} s q u a r i n g o n b o t h s i d e$
$\Rightarrow 2 x^{2} = 1 + i^{2} + 2 i$
$\Rightarrow x^{2} = i \dots (1)$
$\Rightarrow x^{6} + x^{4} + x^{2} + 1$
$= (x^{2})^{3} + (x^{2})^{2} + x^{2} + 1$
$= (i)^{3} + (i)^{2} + i + 1 f r o m e q n (1)$
$= - i - 1 + i + 1$
=0
Therefore, $x^{6} + x^{4} + x^{2} + 1 = 0$

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Modulus of a Complex Number

Standard XII Mathematics

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Evaluate the mentioned below: (iv)x6+x4+x2+1,whenx=(1+i)/√2

Given that- x=(1+i)/√2 ⇒√2x=1+i ⇒(√2x)2=(1+i)2squaringonbothside ⇒2x2=1+i2+2i ⇒x2=i…(1) ⇒x6+x4+x2+1 =(x2)3+(x2)2+x2+1 =(i)3+(i)2+i+1fromeqn(1) =−i−1+i+1 =0 Therefore, x6+x4+x2+1=0

Evaluate the mentioned below:
(iv) $x^{6} + x^{4} + x^{2} + 1, w h e n x = (1 + i) / \sqrt{2}$