How find this value of $A$? Question：
Let $z\in C$
Find this value $A$,such 
$$\lim_{k\to +\infty}\left(k-\dfrac{W_{k^2}(z)}{W_{k}(z)}\right)= A\cdot i$$
where $i^2=-1$,and 
$w_{k}(z)$ is Lambert $W$ function:see http://en.wikipedia.org/wiki/Lambert_W_function
and this link give the $W_{k}(x)$ function: 
$$\begin{align} W(x)&=\frac{x}{2\pi}\int_{-\pi}^{\pi}\frac{(1-v\cot v)^2+v^2}{x+v\csc v \cdot e^{-v \cot v}}\\ W(x)&=\int_{-\infty}^{-1/e}-\frac{1}{\pi}\mathfrak{J}\left[\frac{\text{d}}{\text{d}x}W(x)\right]\ln(1-\frac{z}{x})\text{d}x\\ W(x)&=1+(\ln x-1)e^{i/2\pi \int_0^\infty \frac{1}{t+1}\ln\frac{\ln x+t-\ln t-i\pi}{\ln x+t-\ln t+i\pi}\text{d}t}\\ W_k(x)&=1+(\ln x+2k\pi i-1)e^{i/2\pi \int_0^\infty \frac{1}{t+1}\ln\frac{\ln x+t-\ln t+(2k-1)i\pi}{\ln x+t-\ln t+(2k+1)i\pi}\text{d}t} \end{align}$$
My try: let $z=a+bi$,then I Guess $A\to +\infty$ is true? Thank you 
 A: The non-principal branches $W_k(z)$, $k>0$ are always bounded above (and below) by the curve:
$t(y)=-y\cdot\cot(y)+y\cdot i$, where $y\in (2k\pi,2(k+1)\pi)$.
Consequently, considering only the imaginary part, we are interested in the limit:
$$\lim_{k\to\infty}\left( k-\frac{y_2}{y_1}\cdot i\right )$$ with:
$y_2=\theta_2+2k^2\pi$ and $y_1=\theta_1+2k\pi$.
$k$ however is real, so its imaginary part is $-i\cdot k$, therefore it suffices to look at the limit:
$$\lim_{k\to\infty}\left(-i\cdot k-\frac{\theta_2+2k^2\pi}{\theta_1+2k\pi}\cdot i\right)$$
which evaluates (by hand or by Maple) to $-\infty\cdot  i$, in other words negative (signed) imaginary infinity, so $A=-\infty$.
It can also be shown by Maple only, using the asymptotics for the $k$-branch of the Lambert $W_k(z)$, from the Maple help section as:
$$W_k(z)=\log(k,z)-\ln(\log(k,z))+stuff$$
with "stuff" independent of $k$.
Then if we define the complex logarithm log(k,z) as CLog(k,z) in Maple as:

CLog := proc (k, z)
  local rho, theta;
  rho := op(1, polar(z));
  theta := op(2, polar(z));
  ln(rho)+I*(theta+2*k*Pi)
  end;
Wa := (k, z)->CLog(k, z)-ln(CLog(k, z));
expr:=k-Wa(k^2,z)/Wa(k,z);

then Maple evaluates the limit, as:

limit(%, k = infinity);

$\infty\cdot I$
