Let a-1-i2 then prove that a6-a4-a2-1-0-

Given #160; a=1+i2 .Now a^2=(1+i)^22=2i2=i .This gives a^4= 1 and a^6= i .Then #160; a^6+a^4+a^2+1 =( i)+( 1)+i+1 [ Using the ...

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Complex Numbers

Let a=1+i2 ...

Question

Let $a = \frac{1 + i}{2}$ then prove that $a^{6} + a^{4} + a^{2} + 1 = 0$ .

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Solution

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

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

a_{1}, a_{2}, . . . . . .

\frac{a_{4}}{a_{6}} = \frac{1}{4}

a_{2} + a_{5} = 216

a_{1}

a_{n}

\frac{a_{4}}{a_{6}} = \frac{1}{4}

a_{2} + a_{5} = 216

a_{1} =

a_{1}, a_{2},

a_{n}

a_{1} = 0

\frac{a_{3}}{a_{2}} + \frac{a_{4}}{a_{3}} + \frac{a_{5}}{a_{4}} +

+ \frac{a_{n}}{a_{n - 1}} - a_{2} (\frac{1}{a_{2}} + \frac{1}{a_{3}} + . . . + \frac{1}{a_{n - 2}}) = \frac{a_{n - 1}}{a_{2}} + \frac{a_{2}}{a_{n - 1}}

a < 1, (1 + a x) (1 + a^{3} x) (1 + a^{5} x) . . . . . .

= 1 + \frac{a x}{1 - a^{2}} + \frac{a^{4} x^{2}}{(1 - a^{2}) (1 - a^{4})} + \frac{a^{9} x^{3}}{(1 - a^{2}) (1 - a^{4}) (1 - a^{6})} + . . . . . .

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