If a- 1-i - 2 - find the value of a 6 - a 4 - a 2 -1

a ^ 6 + a ^ 4 + a ^ 2 +1= a ^ 4 ( a ^ 2 +1 ) +1( a ^ 2 +1 ) = ( a ^ 2 +1 ) ( a ^ 4 +1 ) ..............eq ( 1 ) Here, a= 1+i ...

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Complex Numbers

If a= 1+i √...

Question

If $a = \frac{1 + i}{\sqrt{2}}$ , find the value of $a^{6} + a^{4} + a^{2} + 1$

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a = \frac{1 + i}{\sqrt{2}}

a^{6} + a^{4} + a^{2} + 1 = 0

{(1 + x - 2 x^{2})}^{10} = 1 + a_{1} x + a_{2} x^{2} + . . . + a_{20} x^{20}

a_{2} + a_{4} + a_{6} + . . . + a_{20}

1 + \frac{a^{2}}{2!} + \frac{a^{4}}{4!} + \frac{a^{6}}{6!} + \dots \infty

e^{2} - 1 = \frac{1 + \frac{a^{2}}{2!} + \frac{a^{4}}{3!} + \frac{a^{6}}{4!} + . . .}{1 + \frac{1}{2!} + \frac{2}{3!} + \frac{2^{2}}{4!} + . . .}

a

(1 + x - 2 x^{2})^{6} = 1 + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + . . .

a_{2} + a_{4} + a_{6} + . . . + a_{12}

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