Question

F(z) = z/(z^2 + 1) ;by z = x + jy determine where f'(z) exists and find it?

Answer 1

`F'(z) = ( 1-z^2) / (z^2+1)^2` provided `z != +-i`

Explanation:

We have:

`F(z) = z/(z^2+1)` where `z in CC`

The derivative exists at all points with the exception of the simple poles that `F(z)` possess, which are given by:

`z^2+1 = 0 => z=+-i`

Assuming that `z != +-i`, then we have, using the quotient rule:

`F'(z) = ( (z^2+1)(1) - (2z)(z) ) / (z^2+1)^2`
`\ \ \ \ \ \ \ \ \ = ( z^2+1 - 2z^2 ) / (z^2+1)^2`
`\ \ \ \ \ \ \ \ \ = ( 1-z^2) / (z^2+1)^2`