I have a question, for example if $f(z)$ is a function with a removable singularity at one point say $z = z_0$, then can i conclude that $\int_{C}f(z)dz = 0$?
For example the function $\int_{C}\dfrac{e^{iz}-1}{z}dz$
Does it hold for any function with removable singularity.
Yes. Removable singularities aren't really 'singularities'. I like to think of them more as a poor choice of domain. Your function $$f(x)=\frac{e^{iz}-1}{z}$$ is a great example. Let's consider $f$ as a function from $\mathbb{C} \backslash \{0\}$ to $\mathbb{C}$ - This function is clearly analytic/holomorphic on $\mathbb{C} \backslash \{0\}$ and when we write $f$ as a Laurent expansion on this domain, we notice that what we get is in fact a Taylor expansion. We can extend $f$ to the point $0$ by using this Taylor expansion. Let's call this function $\bar{f}:\mathbb{C} \to \mathbb{C}$. We have that $\bar{f}$ is holomorphic on $\mathbb{C}$ because it is given by its Taylor expansion.
Now, take a curve $C$ in $\mathbb{C} \backslash \{0\}$. We have that (since $f$ and $\bar{f}$ take the same values away from $0$), $$\int_C f(z) \ dz = \int_C \bar{f}(z) \ dz = 0.$$
The moral is, that although we avoided $0$ in the domain of $f$, it is actually possible to think of $f$ as entire function. This is what I meant about the 'poor choice of domain'. I hope this helps.
It is not really clear what is $C$ in your question, but suppose that $C$ is a contour such that $f$ has a removable singularity $z_0$ inside $C$. Then if it is the only singularity we have with Residue theorem : $$\oint_C f(z)~\mathrm dz= 2\pi i \operatorname{Res}( f, z_0 )\,\mathrm{Ind}_C(z_0)$$ But since $z_0$ is a removable singularity we have $\operatorname{Res}( f, z_0 )=0$. So finally : $\oint_C f(z)~\mathrm dz=0$.
I had the same problem when I encountered this material. The terminology is quite confusing in complex analysis, but the naming scheme has significance. Think of a removable singularity as a function that can be turned into an analytic function by removing one point. How can you remove such a point? Well, if you write out the Laurent series you can express the function so that the numerator can be multiplied by powers of $z$ which will cancel the singularity in the denominator. The function that you listed has a "simple", order 1, pole at $z$=0. So, it has a removable singularity.
These are the easy problems. They can be solved with the Cauchy Integral Formula. Your integral follows the form $\oint f(z) dz$= 2$\pi$i $a_{-1}$. Where $a_{-1}$ is a Laurent series coefficient.
