Question: For each $z \in \mathbb{C}$, evaluate $$\int_0^1 \int_0^{2\pi} \frac{1}{r e^{i\theta} + z} \, d \theta\, dr.$$
My attempt: Let $z \in\mathbb{C}$ be given and if $\phi \colon \mathbb{C} \to \mathbb{C}$ is holomorphic in $D(z,1)$ and continuous in $\overline{D}(z,1)$ then by mean-value property,
$$\phi(z) = \frac{1}{2\pi} \int_0^{2\pi} \phi(z + r e^{i\theta})\, d \theta$$
for any $0 < r\leq 1$ and also
$$\phi(z) = \int_0^1 \phi(z) \, dr = \frac{1}{2\pi} \int_0^1 \int_0^{2\pi} \phi(z + r e^{i\theta}) \,d \theta \, d r$$
So, allowing
$$ \phi(z + r e^{i\theta}) = \frac{1}{r e^{i\theta} + z}\iff \phi(z) = \frac{1}{z}$$
the function $\phi \in H(\mathbb{C} - \{0\})$ so assuming $|z| > 1$ we have
$$\int_0^1 \int_0^{2\pi} \frac{1}{r e^{i\theta} + z} \, d \theta\, dr = \frac{2\pi}{z}$$ if $|z|\leq 1$ then the write $z = s e^{i\varphi}$ where $0 \leq s \leq 1$ and $0 \leq \varphi \leq 2\pi$ the integrand
$$\frac{1}{re^{i\theta} + z} = \frac{1}{re^{i\theta}+ se^{i\varphi}}$$
is undefined when $r = s$ and $\theta = \varphi + \pi \pmod{2\pi}$. So, the integral
$$\int_0^1 \int_0^{2\pi} \frac{1}{re^{i\theta} + z}\, d\theta \, dr = \begin{cases} \frac{2\pi}{z} &\ \text{if $|z| > 1$}\\ \text{undefined} &\ \text{if } |z| \leq 1
\end{cases}$$
May I know if this is valid answer? Or are there any gaps in my proof? Any hints are appreciated.
For any $r>0$, let $\gamma_r(\theta)=re^{i\theta}$, $0\leq\theta\leq2\pi$.
Let $f(z)=\frac{1}{z}$. $f\in H(\mathbb{C}\setminus\{0\})$ and for all $0<r<1$, the chain $\gamma_1-\gamma_r\sim0$ in $\mathbb{C}\setminus\{0\}$. Consequently, for $|z|>1$, \begin{align} 0=f(-z)\operatorname{Ind}_{\gamma_1-\gamma_r}(-z)= \frac{1}{2\pi i} \int_{\gamma_1-\gamma_r}\frac{f(\xi)}{\xi+z}\,d\xi=\frac{1}{2\pi}\big(\int^{2\pi}_0\frac{1}{e^{i\theta}+z}-\frac{1}{re^{i\theta}+z}\Big)\,d\theta \end{align}
Hence, $$\int^1_0\int^{2\pi}_0\frac{1}{re^{i\theta}+z}d\theta\,dr=\int^{2\pi}_0\frac{1}{e^{i\theta}+z}\,d\theta,\qquad |z|>1$$ From $$\frac{1}{z+e^{i\theta}}=\frac{1}{z}\frac{1}{1+z^{-1}e^{i\theta}}=\frac{1}{z}\sum^\infty_{n=0}(-1)^nz^{-n}e^{i\theta n},\qquad |z|>1,$$ it follows that $$\int^{2\pi}_0\frac{1}{e^{i\theta}+z}\,d\theta=\frac{2\pi}{z},\qquad |z|>1$$
More generally, define $$F(r,\theta; z)=\frac{1}{re^{i\theta}+z}$$ where $r>0$, $0\leq \theta\leq2\pi$ and $z\in\mathbb{C}\setminus\{0\}$. Then, for $r<|z|$ $$F(r,\theta;z)=\frac{1}{z}\sum^\infty_{n=0}(-1)^nz^{-n}r^ne^{i\theta n}$$ and so, $$\int^{2\pi}_0F(r,\theta;z)\,d\theta=\frac{2\pi}{z}$$ On the other had, if $|z|<r$, $$F(r,\theta;z)=\frac{1}{re^{i\theta}}\sum^\infty_{n=0}(-1)^nz^nr^{-n}e^{-i\theta n}=\sum^\infty_{n=0}(-1)^nz^nr^{-n-1}e^{-i\theta(n+1)}$$ and so, $$\int^{2\pi}_0F(r,\theta;z)\,d\theta=0$$
Putting things together, $$\int^{2\pi}_0 F(r,\theta;z)\,d\theta=\left\{\begin{array}{lcr} \frac{2\pi}{z} &\text{if} & |z|>r\\ 0 &\text{if} & |z|>r \end{array}\right. $$ Hence, $$\int^1_0\int^{2\pi}_0 F(r,\theta;z)\,d\theta\,dr=2\pi\frac{\min(|z|,1)}{z}$$
(When $r=|z|<1$ and, $F(r,\theta;z)$ has a pole of order $1$ at every point of the form $-re^{i\theta}$, and the integral is not over $\int^{2\pi}_0 F(r,\theta;z)\,d\theta$ is not defined. The singleton $\{r\}$ is however a set of measure zero and $\int^1_0\Big(\int^{2\pi}_0F(r,\theta;z)\,d\theta\Big)\,dr$ is defined.)