Compute, for all integers $m,n$, $\int_{|z|=2}z^n(1-z^m)dz$

Question

Compute, for all integers $m,n$, $\int_{|z|=2}z^n(1-z^m)dz$.

Isn't $z^n-z^{mn}$ analytic, so the value is 0? And if it isn't analytic, letting $z(t)=2e^{it},t\in[0,2\pi]$ gives $$\int_{0}^{2\pi}(2^ne^{nit}-2^{nm}e^{nmit})2ie^{it}dt$$ $$i\int_0^{2\pi}2^{n+1}e^{it(n+1)}-2^{nm+1}e^{it(nm+1)}dt$$ $$=i\bigg[\frac{2^{n+1}}{i(n+1)}e^{it(n+1)}-\frac{2^{nm+1}}{i(nm+1)}e^{it(nm+1)}\bigg]^{2\pi}_0$$ $$=0$$ Is this correct though? This was mean't to be a challenging problem so I doubt the answer would be so trivial.

Answer 1

Hint:

You need to spilt the cases:

$(1)$ $n<0$ and $m>0$ so you have pole at $z=0$

$(2)$ $n>0$ and $m>0$ so $I=0$

$(3)$ $n<0$ and $m<0$

$(4)$ $n>0$ and $m<0$

I will solve the first and will let you solve $(3)$ and $(4)$ the answer for $(2)$ is zero because the function is analytic

$(1)$ For the first: $n<0$ and $m>0$

$$\int_{|z|=2}z^n(1-z^m)dz$$

Pole at $z=0$

Use Cauchy's integral formula:

Let's define $g(z):=1-z^m$

so $$I=\frac{2\pi i}{(n-1)!}g^{(n-1)}(z=0)$$