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I'm having a problem finding the residue of $f(z)= dfrac{tan z}{z}= dfrac{h(z)}{z-z_0}$ at $z_0=0$ since $tan z=0$ at $z_0=0$ hence I cannot apply a proposition as it requires $h(z)$ to be analytic and nonzero at $z_0.$ I'm not also sure if I can use the Laurent Series for this.

Hints will suffice. Thank you so much.

Another way to tackle problem which contains function with known taylor series ...(Or atleast we can calculate there taylor series expansion)

$tan z=z+z^3/3....$

Residue is the coefficent of $1/z$

SO $dfrac{tan z}{z}=1+z^2/3$

SO it does not have residue

Note : If don't know taylor series expansion then also you can construct then easily.

I suppose you know taylor series exapansion of $sin z $ and $cos z$

$dfrac{sin z}{cos z}=dfrac{z-z^3/6}{1-z^2/2}$

Use binomial expansion for denomiantor and do some small calculation you get required

Here's a hint: what's the residue of $f(z)=z^n$ at $z_0=0$ for different integers $n$? Directly integrating one circle is easy.

Now $tan(x)$ is meromorphic and so has a Taylor series around $z_0=0$, so bring it in and go on from there.

In particular, what does $h(z_0)=0$ meromorphic around $z_0$ imply about the residue of $f(z)=h(z)/z$ at $z_0$?