Question

The expression
[(1+sinπ/8 + icosπ/8)/(1+sinπ/8 -icosπ/8)]^8 =?

Answer 1

You can use Euler's formula for representing complex numbers...

\[e^{i\theta} = \cos{\theta} + i \sin{\theta}\]

Also, use the fact that, \(\cos{\pi/2 - \theta} = \sin{\theta}\)

Answer 2

Is it \[\cos(\frac{ \pi }{ 8 }) or \frac{ \cos (\pi) }{ 8 }\]

Answer 3

Sorry that is -> \(\cos(\pi/2 - \theta) = \sin{\theta}\) and \(\sin(\pi/2 - \theta) = \cos{\theta}\)

Answer 4

Former one ...

Answer 5

Can you convert it into Euler formula?

Answer 6

I used this
1+i (cos theta-isin theta)/1-i(cos theta+isintheta)

Answer 7

That is 1+ie^-itheta/1-ie^itheta

Answer 8

\[\cfrac{1 + ie^{-i \theta}}{1 - ie^{i \theta}}\]

Multiply numerator and denominator by \(e^{i\theta}\)

Answer 9

Wait don't multiply it :P Not gonna help much...

Answer 10

I got this
i(e^itheta +e^-itheta)/(1+e^itheta)

Answer 11

By rationalisation

Answer 12

Actually no..a slight mistake

Answer 13

It wud be 1+e^i2theta in denominator

Answer 14

I came up with this as ie^-itheta

Answer 15

See we need to convert it to pure e^i theta type term... Then only we would be able to apply power 8... Now if only I can think of a way of doing that.. :/

Answer 16

Got it.....thanks..

Answer 17

@hartnn any ideas?

Answer 18

My ans is -1..

Answer 19

Perfect!!

Answer 20

:P Care to share how you get it ?

Answer 21

*got

Answer 22

Okay..
Let's begin then!

Answer 23

I got till that point
i(e^itheta+e^-itheta)/(1+e^i2theta)
Now if we simplify this numerator term is nothing but 2cos theta
And denominator is 2costheta(costheta+isintheta)
So we r left with this
1/e^itheta
That is e^-itheta
Now there is a power of 8 on it..
So we get e^-i8theta
Plug in the value of theta as π/8
We get the ans as -1