Prove that $S_R^+(b)$ and $S_r^+(a)$ are homotopic in a domain $D$. currently I'm working on the following exercise:

Let $D \subset \mathbb C$ be a domain. Let $a,b \in \mathbb C$ and $r,R > 0$ such that
  $B_r(a) \subset B_R(b)$ and
  \begin{align*}
A := \{z \in \mathbb C: |z-b| \le R, \, |z-a|\ge r\} \subset D.
\end{align*}
  Prove that the circle $S_R^+(b)$ and $S_r^+(a)$ are homotopic in $D$.

I wanted to use the following homotopy:
\begin{align*}
H(t,s) &:= (1-s)(b+Re^{it}) + s(a+re^{it}).
\end{align*}
We have that
\begin{align*}
H(t,0) = b+Re^{it} = S_R^+(b), \quad H(t,1) = a+re^{it} = S_r^+(a), \quad \forall t \in [0,2\pi].
\end{align*}
Furthermore
\begin{align*}
H(0,s) = (1-s)(b+R)+s(a+r) = H(2\pi,s), \quad \forall s \in [0,1].
\end{align*}
We need to show that $H(t,s) \in D$, $\forall (t,s) \in [0,2\pi] \times [0,1]$. We can do that by showing $H(t,s) \in A$. We have that
\begin{align*}
|H(t,s)-b| &= |(1-s)(b+Re^{it}) + s(a+re^{it})-b| \\
&= |(1-s)b-b+sa+(1-s)Re^{it}+sre^{it}| \\
&= |b(1-s-1)+sa+e^{it}((1-s)R+sr)|
= |sa-sb+e^{it}(R-sR+sr)| \\
&= |s(a-b)+e^{it}(R-s(R-r))|
\le |s||a-b|+|R-s(R-r)|.
\end{align*}
We know that $B_r(a) \subset B_R(b)$. That means
\begin{align*}
|z-a| < r \Rightarrow |z-b|<R, \quad \forall z \in B_r(a).
\end{align*}
But I don't know how to use this in my estimation.
Edit: I worked my way through the solution which I got, and it uses the estimation $|a-b| \le R-r$. I don't understand why this is true.
 A: Some comments:


*

*Good work. Nevertheless, I don't know why you need your "Furthermore...".  :-?

*Also, in order to show that $H(s,t) \in A$, you need to prove that $|H(s,t) - a| \geq r$ too, don't you?

*As for your first question, I would try the following estimation instead:


\begin{align*}
|H(t,s)-b| &= |(1-s)(b+Re^{it}) + s(a+re^{it})-b| \\
&= |(1-s)(b+Re^{it}) +sb -sb + s(a+re^{it})-b| \\
&= |(1-s)Re^{it} + s(a-b + re^{it})| \\
&\le (1-s)|Re^{it}| + s|a-b| + s|re^{it}| \\
& = (1-s)R + s|a-b| + sr  \\
\end{align*}
Now you use the hint $|a-b| < R - r$, and get
\begin{align*}
&< (1-s)R + s(R-r) + sr = R
\end{align*}
So, it's ok: your homotopy $H(s,t)$ never goes outside the big disk $B_R(b)$.


*

*As for your second question, namely this last inequality $|a-b| < R - r$ comes from the fact that $a\in B_R(b)$, so $|a-b| < R$, plus the fact you can still go further away $r$ steps in the direction of the vector $a-b$ and you're still inside the big disk $B_R(b)$, because $B_r(a) \subset B_R(b)$. Hence, $|a-b|+r < R$.

