An estimation from Iwaniec-Kowalski book I am reading the book 'Analytic Number Theory' by Iwaniec and Kowalski and I am confused with an estimation from Chapter 19. 
On page 446, they say

If $cd \leq z^{1/2}$ we derive from the Prime Number Theorem that 
  \begin{equation}
-\sum_{\substack{m \leq z/cd \\ (m,cd)=1}} \frac{\mu(m)}{m} \log cdm = \frac{cd}{\varphi(cd)} \{1+O(\tau(cd)(\log z)^{-A})\}.
\end{equation}

The notations in the equation are explained as follows:
Here $c,d$ are positive square-free integers, $\mu$ is the Möbius function, $\varphi$ is Euler's totient function and $\tau$ is the divisor function. Also, $z$ is a large parameter and $A>0$ is arbitrarily large.
I cannot understand how to use PNT to derive this estimation.
Thanks for any help. 
I am sorry for mistakenly posting this question on MathOverflow before.
 A: All the $O$-constants depend only on $A$.


*

*The PNT says that for any fixed $A$ as $x \to \infty$ $$\sum_{n \le x} \frac{\mu(n)}{n} =O( (\log x)^{-A})$$

*To estimate the $\sum_{n\le x,\gcd(n,k)=1}$ sum we'll show by induction, starting with $k=1$ to obtain it for $k=p,pq,pqr,\ldots$,
that if 
$$\sum_{n \le x,\gcd(n,k)=1} \frac{\mu(n)}{n}=O(C_{A,k} (\log x)^{-A}),\qquad C_{A,k}=\prod_{q | k}\frac{A+1}{(1-q^{-1})^2}$$ Then for $p \nmid k $ and $x \ge p$ $\qquad\qquad\scriptstyle( \text{since }\sum_{n,\gcd(n,kp)=1} \mu(n)n^{-s}= \frac1{1-p^{-s}}\sum_{n,\gcd(n,k)=1} \mu(n)n^{-s})$
$$\sum_{n \le x,\gcd(n,kp)=1} \frac{\mu(n)}{n}=
\sum_{l=0}^{\log_p(x)}\frac1{p^l}\sum_{n \le x/p^l,\gcd(n,k)=1} \frac{\mu(n)}{n} $$ $$= O(C_{A,k}\sum_{l= 0}^{\log_p(x)}\frac1{p^l} (\log x/p^l)^{-A})
= O(C_{A,k}\sum_{l= 0}^{\log_p(x)}\frac1{p^l} (\log x)^{-A} (1-\frac{\log p^l}{\log x})^{-A})$$ $$
= O(C_{A,k}\sum_{l\ge  0}\frac1{p^l} (\log x)^{-A} (1+A\frac{l\log p}{\log x}))
=O(C_{A,kp} (\log x)^{-A})
$$

*From partial summation of $\sum_{n > x, \gcd(n,k)=1}\frac{\mu(n)}{n}\log n$ we obtain that 

$$ -\sum_{n \le x, \gcd(n,k)=1}\frac{\mu(n)}{n}\log n = B_k+O(C_{A,k}(\log x)^{1-A})$$
  Note $C_{A,k} \approx (A+1)^{\omega(k)}$ while Iwaniec expects $\tau(k) \approx 2^{\omega(k)}$


*To find $B_k$ let $$F_k(s)=\sum_{n,\gcd(n,k)=1}\!\! \mu(n) n^{-s} = \frac{1}{\zeta(s)\prod_{p | k} (1-p^{-s})}, \  F_k'(s)=-\sum_{n,\gcd(n,k)=1}\!\!\! \mu(n) n^{-s}\log n$$
Then $$\lim_{s \to 1^+} F_k'(s)= \frac1{\prod_{p | k} (1-p^{-1})} = \frac{k}{\phi(k)}$$ so that (from partial summation of the Dirichlet series) $$B_k =\frac{k}{\phi(k)}$$
