# Complex Analysis Question

1. Apr 7, 2006

### Edwin

Consider the function:

g(z(t)) = i*f '(c+it)/(f(c+it) - a)

Where {-d <= t < d}

If we let z = c+it

By change of variables don't we get:

Line integral of g(z(t)) = i ln[f(c+it) - a]

evaluated from t = - d to t = d?

note: ln is the natural log.

Inquisitively,

Edwin G. Schasteen

2. Apr 8, 2006

### leach

You may consider the line integral as the complex path integral

$$\int_\gamma\, \frac{f'(z)}{f(z)-a}\,dz$$

where $$\gamma(t) = c + it$$, for $$-d\le t\le d$$.

Since the integrand has the trivial primitive $$G(z) = \ln(f(z)-a)$$, you may indeed consider that:

$$\int_\gamma\, \frac{f'(z)}{f(z)-a}\,dz \quad = \quad G(c+id) - G(c-id)$$

You should be cautious anyway, since for this integral to be right, it must be $$f(z) \ne a$$ along $$\gamma$$. Assuming that g(z) has no singularities in a domain containing the path $$\gamma$$, I think you can safely consider the primitive $$G(z)$$ as correct to compute the integral.

3. Apr 8, 2006

### Edwin

That makes sense.

Thanks! I appreciate your help!

Best Regards,

Edwin