Hard complex numbers question (De Moivre's)

A) Express e^ix/2 - e^-ix/2 in terms of sin x/2

Fairly simple

B) Hence, or otherwise, show that 1/(e^(ix) -1) = -1/2 - i/2 (cot (x/2)).

Have no clue, obviously you can simplify the cot to give i/2tan(x/2) but other than that not sure really, especially with the hence.

Help appreciated

NOTE: I know I post a lot but it's only because 1. I love maths and 2. I do a lot of questions

Reply 1

ghostwalker

17

Original post by psycholeo

A) Express e^ix/2 - e^-ix/2 in terms of sin x/2

Fairly simple

B) Hence, or otherwise, show that 1/(e^(ix) -1) = -1/2 - i/2 (cot (x/2)).

Have no clue, obviously you can simplify the cot to give i/2tan(x/2) but other than that not sure really, especially with the hence.

Help appreciated

NOTE: I know I post a lot but it's only because 1. I love maths and 2. I do a lot of questions

I'd start with the lefthand side.

How can you you use part A? Clearly it relates to the denominator. So, what could you multiply the top and bottom of the fraction by to get the denominator in the form of part A?...

(edited 5 years ago)

Reply 2

testgeek2

1

rearrange your answer to part A to get an expression for e^ix in terms sin(x/2), substitute this into B and replace i/2tan(x/2) by i/2sin(x/2)/cos(x/2) and simplify.

Reply 3

psycholeo

OP

9

Original post by ghostwalker

I'd start with the lefthand side.

How can you you use part A? Clearly it relates to the denominator. So, what could you multiply the top and bottom of the fraction by to get the denominator in the form of part A?...