Second Approximation of (1+i)^-1 for i<<<1

[electronic engineer]

I have this algebraic term: (1+i)^-1 where i is very very small i<<<1

if we used second approximation what would it equal to?

thanks in advance!

 

[Tide]

You can use the binomial expansion:

$$\frac {1}{1+i} = 1 - i + i^2 + \cdot \cdot \cdot$$

so in first approximation you have 1 and in second approximation 1 - i etc.

 

[tacman]

The second approximation of (1+i)^-1 for i<<<1 would be equal to 1-i. This can be obtained by using the binomial expansion for (1+i)^-1 and taking only the first two terms (1-i) and neglecting higher order terms involving i^2 or higher powers of i. So, the second approximation is a good estimate for (1+i)^-1 when i is very small.