Question

From where does come this "y"? Electromagnetic field. The complete question is attached.

Answer 1

the image:

Answer 2

There is no y in the problem. You are being told that (1+a)^-1 is approximately 1-a when a<<1.

Answer 3

and similarly (1-a)^-1 = 1+a for a<<1.

Answer 4

nice douglas

Answer 5

Glad to have helped and to be thanked.

Answer 6

$$
\cfrac{1}{1-y}=\cfrac{1+y}{(1-y)(1+y)}=\cfrac{1+y}{1-y^2}
$$

When y << 1, the \(y^2\) term is very very small and can be ignored compared to \(y\)
So,
$$
\cfrac{1+y}{1-y^2}\approx 1+y
$$
For small \(y\).