= Solution
The first arrival time $A$ has an <exponential distribution> of rate $\lambda$ and is independent of the following <busy period>. Thus
$$
\phi_{T_k}(\theta)=\frac{\lambda}{\lambda-\theta}\phi_{B_k}(\theta).
$$
Substitute $\phi_{B_k}=(\lambda-\theta)\phi_{T_k}/\lambda$ into part (c). Its service-transform argument becomes $\theta+\lambda(\phi_{B_k}-1)=(\lambda-\theta)(\phi_{T_k}-1)$. Therefore
$$
\boxed{\frac{\lambda-\theta}{\lambda}\phi_{T_k}(\theta)
=\phi_{S_k}((\lambda-\theta)(\phi_{T_k}(\theta)-1))}.
$$
As before, positive arguments must lie in the domain of finite moment-generating functions.
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