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quasar987

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Consider a domain D and f:D-->[itex]\mathbb{C}[/itex] a holomorphic function and C a contractible Jordan path contained in D and z1, z2, two points in the interior of C. Evaluate

[tex]\int_C \frac{f(z)}{(z-z_1)(z-z_2)}dz[/tex]

What happens as [itex]z_1 \rightarrow z_2[/itex]?

I have found that

[tex]\int_C \frac{f(z)}{(z-z_1)(z-z_2)}dz = \frac{2i \pi}{z_1-z_2}(f(z_1)-f(z_2))[/tex]

and that is correct according to the solution manual.

Of course, as [itex]z_1 \rightarrow z_2[/itex], the RHS goes to [itex]2i\pi f'(z_2)[/itex], and according to the solution manual, the LHS goes to

[tex]\int_C \frac{f(z)}{(z-z_2)^2}dz[/tex].

But considering the integral as one big function of z_1 (let's call it [itex]I(z_1)[/itex]), since it equals

[tex]\frac{2i \pi}{z_1-z_2}(f(z_1)-f(z_2))[/tex],

which is not defined, hence not continuous at [itex]z_1 = z_2[/itex], I cannot say that

[tex]\lim_{z_1\rightarrow z_2} I(z_1) = I(\lim_{z_1\rightarrow z_2} z_1)[/tex]

So how is the conclusion attained?

thx!

[tex]\int_C \frac{f(z)}{(z-z_1)(z-z_2)}dz[/tex]

What happens as [itex]z_1 \rightarrow z_2[/itex]?

I have found that

[tex]\int_C \frac{f(z)}{(z-z_1)(z-z_2)}dz = \frac{2i \pi}{z_1-z_2}(f(z_1)-f(z_2))[/tex]

and that is correct according to the solution manual.

Of course, as [itex]z_1 \rightarrow z_2[/itex], the RHS goes to [itex]2i\pi f'(z_2)[/itex], and according to the solution manual, the LHS goes to

[tex]\int_C \frac{f(z)}{(z-z_2)^2}dz[/tex].

But considering the integral as one big function of z_1 (let's call it [itex]I(z_1)[/itex]), since it equals

[tex]\frac{2i \pi}{z_1-z_2}(f(z_1)-f(z_2))[/tex],

which is not defined, hence not continuous at [itex]z_1 = z_2[/itex], I cannot say that

[tex]\lim_{z_1\rightarrow z_2} I(z_1) = I(\lim_{z_1\rightarrow z_2} z_1)[/tex]

So how is the conclusion attained?

thx!

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