If a-1-i-2 then show that a6-a4-a2-1-0-

Given, #160; a=1+i√(2) .Now #160; a^2=(1+i)^22 or, #160; a^2=(2i)2 or, a^2 =i .Then a^4= 1, a^6= i .Then a^6+a^4+a^2+1= i 1+i+1 ...

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Standard XII

Mathematics

Complex Numbers

If a=1+i√2 ...

Question

If $a = \frac{1 + i}{\sqrt{2}}$ then show that $a^{6} + a^{4} + a^{2} + 1 = 0$ .

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Solution

Given, $a = \frac{1 + i}{\sqrt{2}}$ .
Now
$a^{2} = \frac{(1 + i)^{2}}{2}$
or, $a^{2} = \frac{(2 i)}{2}$
or, $a^{2} = i$ .
Then $a^{4} = - 1, a^{6} = - i$ .
Then $a^{6} + a^{4} + a^{2} + 1 = - i - 1 + i + 1 = 0$ .

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Similar questions

Q. If

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

, find the value of

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

Q. Consider the following statements :
A number

a_{1} a_{2} a_{3} a_{4} a_{5} a_{6}

is divisible by 11 if
1.

(a_{1} + a_{3} + a_{5}) - (a_{2} + a_{4} + a_{6}) = 0

(a_{2} + a_{4} + a_{6}) - (a_{1} + a_{3} + a_{5}) = 0

Which of these statements is/are correct?

Q. If

{(1 + x - 2 x^{2})}^{6} = 1 + a_{1} x + a_{2} x^{2} + . . . + a_{12} x^{12}

, then

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

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

Suppose $a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}$ are integers such that $\frac{5}{7} = \frac{a_{2}}{2!} + \frac{a_{3}}{3!} + \frac{a_{4}}{4!} + \frac{a_{5}}{5!} + \frac{a_{6}}{6!} + \frac{a_{7}}{7!}$ where 0 $\leq$ a< j for j= 2,4,5,6,7.

The sum $a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}$ is

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If a=1+i√2 then show that a6+a4+a2+1=0.

Given, a=1+i√2.Now a2=(1+i)22or, a2=(2i)2or, a2=i.Then a4=−1,a6=−i.Then a6+a4+a2+1=−i−1+i+1=0.

If $a = \frac{1 + i}{\sqrt{2}}$ then show that $a^{6} + a^{4} + a^{2} + 1 = 0$ .

Given, $a = \frac{1 + i}{\sqrt{2}}$ .
Now
$a^{2} = \frac{(1 + i)^{2}}{2}$
or, $a^{2} = \frac{(2 i)}{2}$
or, $a^{2} = i$ .
Then $a^{4} = - 1, a^{6} = - i$ .
Then $a^{6} + a^{4} + a^{2} + 1 = - i - 1 + i + 1 = 0$ .