Cauchy's integral (therom aand formula)

[ajayguhan]

let D be simple connected Domain and C be simple close curve in D.

then by cauchy integral theorm ∫(z) dz over C is zero.→let this be my 1st equation.

but by cauchy integral formula for a point (a) inside C, we can say

f(a)=(1/2∏i)[closed integral over c]∫f(z)dz/(z-a)

NOTE f(z) is analytic function in D.

but substituting [closed integral over c]∫f(z)dz=0 from equation 1,

we get f(a)=0, for all a belonging to C but which is not true.

now where i am wrong ?

i don't get the intuition behind cauchy integral therom and formula ,

would be glad if someone helped me.

 

[SteamKing]

It's 't-h-e-o-r-e-m'.

 

[HallsofIvy]

let D be simple connected Domain and C be simple close curve in D.

then by cauchy integral theorm ∫(z) dz over C is zero.→let this be my 1st equation.

You mean $$\int f(z)dz$$ over C is zero when is analytic inside C? Of course $\int z dz= 0$ but that is irrelevant to what you have below.

but by cauchy integral formula for a point (a) inside C, we can say

f(a)=(1/2∏i)[closed integral over c]∫f(z)dz/(z-a)

NOTE f(z) is analytic function in D.

but substituting [closed integral over c]∫f(z)dz=0 from equation 1,

we get f(a)=0, for all a belonging to C but which is not true.

now where i am wrong ?

i don't get the intuition behind cauchy integral therom and formula ,

would be glad if someone helped me.

Apparently you forgot about the $$\frac{1}{z- a}$$ part! If f(z) is analytic inside C, $$\frac{f(z)}{z- a}$$ is not- it has a pole of order 1 at z= a.

 

[homeomorphic]

If you want intuition for this stuff, you should read Visual Complex Analysis. I could hardly do it justice here.