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

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The discussion focuses on the second approximation of the algebraic term (1+i)-1 where i is a very small value (i<<<1). The binomial expansion is utilized to derive the approximation, resulting in the expression 1 - i + i2 + ... for the series. The first approximation yields 1, while the second approximation provides 1 - i, effectively simplifying the calculation for small values of i.

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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!
 
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You can use the binomial expansion:

[tex]\frac {1}{1+i} = 1 - i + i^2 + \cdot \cdot \cdot[/tex]

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

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