Is this correct working out for e^(i * pi/4)?

Question

I have to solve

-2 ( e^(i*pi/4) - e^(i*pi/4))/(e^(i*pi/4) + e^-(i*pi/4))

I know that

e^(i*pi/4) = cos(pi/4) + isin(pi/4)
and I know that

e^-(i*pi/4) = cos(pi/4) - isin(pi/4) = 0

So for the first one, I get 1/sqrt(2) + i/sqrt(2)

Now it's more of a fraction question I guess, because I know the answer to the whole thing is

-2 ( sqrt(2)i / (sqrt(2)i)^2 = sqrt(2) i

My question is... what is the working out to go from

1/sqrt(2) + i/sqrt(2)

to get

sqrt(2) i

I must be missing something obvious, I just can't see it

Answer 1

Leave out the value of pi/4 for the moment

e^(ix) = cos(x) + i.sin(x)
e^(-ix) = cos(x) - i.sin(x)

Combining these two values:

e^(ix) + e^(-ix) = 2cos(x)
e^(ix) - e^(-ix) = 2i.sin(x) -------- note the i here

[e^(ix) - e^(-ix)] / [e^(ix) + e^(-ix)]
= 2i.sin(x) / 2cos(x)
= i.tan(x)

-2 ( e^(i*pi/4) - e^(-i*pi/4))/(e^(i*pi/4) + e^-(i*pi/4))
= -2 *i tan(pi/4)
= -2i * 1
= -2i

Answer 2

One error is that your partial answer:
cos(pi/4) - isin(pi/4) = 0
... is incorrect.

cos(pi/4) does equal sin(pi/4), but it seems that the "i" was dropped.

It's actually 1/sqrt(2) - i/sqrt(2)

Also, I'm thinking that the numerator is not written out correctly. You have:
e^(i*pi/4) - e^(i*pi/4)
... which is zero.
I'm guessing you wanted:
e^(i*pi/4) - e^(-i*pi/4)

Substituting, that is:
-2 ((1/sqrt(2) + i/sqrt(2) - 1/sqrt(2) + i/sqrt(2)) / (1/sqrt(2) + i/sqrt(2) + 1/sqrt(2) - i/sqrt(2))

Which simplifies to:

-2 ( 2i/sqrt(2) ) / ( 2/sqrt(2) )

And that further simplifies to:

-2i

That's not the desired answer, either, so there must be something else that is wrong.

Answer 3

If I read this right the middle term divides out to 1, so
-2(e^(pi*i/4) - 1 + e^-(pi*i/4)) =
2 - 2(e^(pi*i/4) + e^-(pi*i/4)) =
2 - 2(cos(pi/4) + i*sin(pi/4) + cos(pi/4) - i*sin(pi/4)) =
2 - 4cos(pi/4) = 2 - 4(sqrt(2)/2) = 2 - 2*sqrt(2)
a real number. I don't know why my answer is different that everyone else's. Maybe I am reading your problem incorrectly.

Answer 4

for pi/4 I shall use t for readbility
I hink you have made a mistake because

-2 ( e^(i*pi/4) - e^(i*pi/4))/(e^(i*pi/4) + e^-(i*pi/4))
becuase numerator is zero

I think you mean -2 ( e^(i*pi/4) - e^-(i*pi/4))/(e^(i*pi/4) + e^-(i*pi/4))

= 2(e^it - e-^it)/(e^it + e^it) where t -= pi/4


we have e^it = cos t + i sin t
e^-it = cos t - i sin t
add e^it + e^-it = 2 cos t ..1
subtract e^it- e^-it = 2 i sin t ..2

devide 2 by 1 (e^it-e^-it)/(e^it+e^-it) = i tan t
multiply by -2 to get result = -2 i tan t = - 2i as tan t = tan pi/4 =1

secondly
e^-(i*pi/4) = cos(pi/4) - isin(pi/4) = 0
is incorrect
correct is 1/sqrt(2) - i/sqrt(2)

Answer 5

McFate is right.
But in this problem, you'll be way better off utilizing the following definitions.

cos(x) = 1/2 * (e^ix + e^-ix)
sin(x) = 1/(2i) * (e^ix - e^-ix)

The whole expression will simplify very easily to
-2*i