x^6+x^4+x^2+1, when x= 1+i 2 .

Question

$\text{x}^6+\text{x}^4+\text{x}^2+1,$ when $\text{x}=\frac{1+\text{i}}{\sqrt{2}}.$

✓

Answer

We have
$\text{x}=\frac{1+\text{i}}{\sqrt{2}}$
$\Rightarrow\sqrt{2}\text{x}=1+\text{i}$
$\Rightarrow\sqrt({2}\text{x})^2=(1+\text{i})^2$ (squaring both sides)
$\Rightarrow2\text{x}^2=1^2+(\text{i})^2+2\times1\times\text{i}$
$=1-1+2\text{i}$
$\Rightarrow2\text{x}^2=2\text{i}$
$\Rightarrow\text{x}^2=\text{i}$
$\Rightarrow(\text{x}^2)^2=(\text{i})^2$ (squaring both sides)
$\Rightarrow\text{x}^4=-1$
$\Rightarrow\text{x}^4+1=0 \ ...(1)$
Now
$\text{x}^6+\text{x}^4+\text{x}^2+1$
$=\text{x}^6+\text{x}^4+\text{x}^2+1$
$=\text{x}^2(\text{x}^4+1)+1(\text{x}^4+1)$
$=\text{x}^2\times0+1\times0$ (using(i))
$=0$

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