Question

if \[x =\frac{ \sqrt{2}+1 }{ 2\sqrt{2} } + i \frac{ \sqrt{2}-1 }{ 2\sqrt{2} }\] ,then find x^8-ix^4+x^2

Answer 1

@waterineyes @tkhunny

Answer 2

Do you have DeMoivre?

Answer 3

Yeah!

Answer 4

for that we have to find the angle....but i am not getting fancy values for that

Answer 5

Ah, that's annoying. Plug away, then. I'll get you started.

\(x^{2} = \dfrac{1}{4}(2\sqrt{2} + i)\)

Fortunately, we have only x^4 and x^8, so only two more squarings.

Answer 6

\[x^4 = \frac{1}{4}(7 + 4 \sqrt{2})\]

Answer 7

*i with 4root(2)..

Answer 8

\[x^4 = \frac{1}{4}(7 + 4 \sqrt{2}i)\]

Answer 9

?? Should be 1/16.

Answer 10

Yep... Sorry my mistake there..

Answer 11

\[x^4 = \frac{1}{16}(7 + 4 \sqrt{2}i)\]

Answer 12

x^8 will harm you more..

Answer 13

\[x^8 = \frac{1}{256}(17 + 56\sqrt{2}i)\]

Answer 14

Yeah!!! One more thing.

Answer 15

There must be an easy solution for this, may be we can reduce x^8 - x^4i .... whatever it is , so that we can use value of x efficiently..

Answer 16

Not really seeing it, but I'm open to the idea. What if it were a "nice" angle? How would you do it with DeMoivre?

Answer 17

@Yahoo! Jump in..

Because I don't know much of De-Moivre's Theorem..

Answer 18

De-Moivre's Theorem.. will be useful if we know the angle.....

Answer 19

like u said... i think we have to do something with x^8 -ix^4 + x^2

Answer 20

Wait I think cos of or sin of something should be :

\[\frac{\sqrt{2} + 1}{2 \sqrt{2}}\]

Answer 21

is it 15

Answer 22

Check it for sin(15) cos(15), cos(75) sin(75)..

Can you check the values for these??

Answer 23

I don't have something to check, so check it yourself, may be it is 15..

Answer 24

@hartnn Congratulations For MODERATOR..

Answer 25

No it is not matching with any of the values..

Answer 26

Have you tried by substituting the values we provided??