1. ## complex analysis

1. Evaluate $\displaystyle \int_{C(0,2)}\frac{dz}{1+z^2}$.

I know $\displaystyle z(t)=2e^{it}, t\in[0,2\pi]$, $\displaystyle dz=i2e^{it}dt$. So $\displaystyle \int_{C(0,2)}\frac{dz}{1+z^2}=\int_0^{2\pi}\frac{i 2e^{it}}{1+(2e^{it})^2}dt$. I use substitution, letting $\displaystyle u=2e^{it}, du=i2e^{it}dt$, then I get $\displaystyle \int_0^{2\pi}\frac{i2e^{it}}{1+(2e^{it})^2}dt=\int _2^2\frac{du}{1+u^2}=0$. Is this okay? I don't know how else to do this...

2. Let $\displaystyle 0<|a|<|b|$. Evaluate $\displaystyle \int_{C(0,r)}\frac{dz}{(z-a)(z-b)}$, where (i) $\displaystyle 0<r<|a|$, (ii) [tex] $\displaystyle |a|<r<|b|$, (3) $\displaystyle |b|<r$.

Can I get some help with this?

2. Originally Posted by dori1123
1. Evaluate $\displaystyle \int_{C(0,2)}\frac{dz}{1+z^2}$.

I know $\displaystyle z(t)=2e^{it}, t\in[0,2\pi]$, $\displaystyle dz=i2e^{it}dt$. So $\displaystyle \int_{C(0,2)}\frac{dz}{1+z^2}=\int_0^{2\pi}\frac{i 2e^{it}}{1+(2e^{it})^2}dt$. I use substitution, letting $\displaystyle u=2e^{it}, du=i2e^{it}dt$, then I get $\displaystyle \int_0^{2\pi}\frac{i2e^{it}}{1+(2e^{it})^2}dt=\int _2^2\frac{du}{1+u^2}=0$. Is this okay? I don't know how else to do this...

2. Let $\displaystyle 0<|a|<|b|$. Evaluate $\displaystyle \int_{C(0,r)}\frac{dz}{(z-a)(z-b)}$, where (i) $\displaystyle 0<r<|a|$, (ii) [tex] $\displaystyle |a|<r<|b|$, (3) $\displaystyle |b|<r$.

Can I get some help with this?
Are you familiar with the residue theorem?

3. Originally Posted by chiph588@
Are you familiar with the residue theorem?
No. Do I have to use that to do my work?

4. Originally Posted by dori1123
No. Do I have to use that to do my work?
Residue theorem - Wikipedia, the free encyclopedia

You don't need it for this problem, but it's an extremely useful theorem.