why if f(z) is analytic in a multiply connected domain D then we cannot Why is it that if $f(z)$ is analytic in a multiply connected domain $D$, we cannot conclude that $\oint_C f(z)dz=0$ for every simple closed contour $C$ in $D$?
 A: One perfectly good answer is to say: look, consider the function $f(z) = \frac{1}{z}$, $D = \mathbb{C} \setminus \{0\}$ and the path $\gamma(t): [0,1] \rightarrow D$ given by $t \mapsto e^{2 \pi i t}$.  The integral is $2\pi i$, not zero.
This should cause you to look back at the proof of Cauchy's Theorem and see where the simple connectivity is used.  This is a good opportunity to try to think about what the theorem is really saying, including geometrically and physically.  I recommend thinking about other versions of the Fundamental Theorem of Calculus as well.
For instance if I have a smooth function defined on $\mathbb{R} \setminus \{0\}$ but blowing up at zero, that singularity is going to cause problems.  Consider for instance $f(x) = \frac{1}{|x|}$.  If we naively tried to compute $\int_{-1}^1 f(x)dx$ using FTC, we get $\log(|1|) - \log(|-1|) = 0$.  This is obviously the wrong answer.  (In fact the area under the curve is infinite.)  This is a somewhat crude analogy (in topological terms, having dropped a dimension we are now dealing with a domain which is not even connected, rather than a domain which is connected but not simply connected), but it should drive home the lesson that in these types of theorems computations at the boundary certainly do require that all is well on the inside!
If you have taken multivariable calculus and/or physics I encourage you to think about this phenomenon in these terms.  For instance, I give a treatment of line integrals, conservative vector fields and Green's theorem including some physical reasoning and a proof of Cauchy's Theorem in terms of Green's theorem in these three handouts.  (It would take me more time than I have at the moment to rummage through these notes and pick out the most relevant examples, and this answer suffers because of that: my apologies.)
Green's Theorem stated physically (in circulation form) goes something like this: for a region bounded by a smooth curve, integrating the circulation density (aka the curl) over the region gives the "net" or "total" circulation, a line integral around the boundary.  But if the vector field has a singularity in the middle of the region, we no longer have a closed system and there is no reason to expect this to hold.  Green's Theorem also has a mathematically equivalent version in terms of flux: if you integrate the flux density (aka the divergence) over the region, you get the "net" or "total flux", a normal (i.e., as opposed to tangential) line integral through the boundary.  (I do not have a very highly developed physical intuition, but even for me this statement about fluid flow is very natural.)  Now imagine that we have a hole in our region where there is no well-defined flux density.  This is obviously going to be a problem in integrating the flux density to get the total flux!
Let me end with one more explanation: later on in complex analysis one learns a version of Cauchy's Theorem which says that the line integrals of the same holomorphic function over two homotopic curves are equal.  (This can also be described in terms of multvariable calculus: by the Cauchy-Riemann equations the corresponding vector field is irrotational, but let me not get into this for now.)  If the region is simply connected, then every curve in that region is homotopic to a point, so every line integral is equal to zero.  However, if the region has exactly one hole, like $\mathbb{C} \setminus \{0\}$, then there will be curves which we cannot contract to a point.  We still have a lot of invariance: the value of the integral depends only on the number of times the curve winds around the singularity.  Moreover one goes on to study the common value of all integrals of the function $f$ winding once around the hole in terms of $f$: this leads to the important notions of Laurent series expansion and residue.
A: By "multiple connected" you probably mean "not simply connected", and of course you cannot conclude that those integrals all vanish. A function with a simple pole at the origin is analytic in an annulus around the origin, and the integral over any simple closed cycle within the annulus that winds once around the origin will be nonzero (indeed, it will have the value $2 \pi i$ times the residue at the origin).
A: The question seems to put the burden of proof in the wrong place.  No argument was given to support the conclusion that the integral is 0.  That why we "can't conclude" that.  (The counterexamples given in other answers show that we still won't be able to conclude that later; no argument can lead us to that conclusion because it's false.)
A: If the function is analytic on the contour as well as the points interior to the simple closed contour the surely the closed integral will be zero.
A multiple connected domains can have holes in them. The regions which are not analytic. So when we take the closed integral along the contour we get a non-zero value.
The reason we cannot conclude for every such simple closed contour is because the function may not be analytic for the points interior to the contours.
As others have mentioned you can also look at the function 1/z whose integral along a simple closed contour will be non-zero, i.e. $2\pi i.$
