calculus - How prove this limit $ lim_{ntoinfty}left(n+frac{1}{2}right)left(z_{n}-left(n+frac{1}{2}right)piright)=frac{H}{pi}$

Question:

Let $H\in R$,Prove that the transcendental equation $$z\cot{z}+H=0$$ has a countable number of zeros $z_{n}$ and that

$$\lim_{n\to\infty}\left(n+\dfrac{1}{2}\right)\left(z_{n}-\left(n+\dfrac{1}{2}\right)\pi\right)=\dfrac{H}{\pi}$$

My try: we must only prove this

$$z_{n}=\left(n+\dfrac{1}{2}\right)\pi+\dfrac{H}{\left(n+\dfrac{1}{2}\right)\pi}+o\left(\dfrac{1}{n^2}\right)$$
if this problem don't tell this limit reslut,then we how find this limit? Thank you

can you someone help me,Thank you very much!

Answer

Your equation can be rearranged to:

$$\cot(z) = -\frac{H}{z}$$

Put $z = y + x$ where $y = (n+\frac{1}{2})\pi$ and use $\cot(y + x) = -\tan(x)$:

$$\frac{H}{y+x} = \tan(x)$$

We need to show $\lim_{y\rightarrow\inf}xy = H$. Taylor expand both sides:

$$\frac{H}{y}(1 - \frac{x}{y}) = x + O(x^2)$$

$$\left(\frac{H}{y^2} - 1\right)x = -\frac{H}{y} + O(x^2)$$

$$x = \frac{H/y}{1-H/y^2} + O(x^2)$$

From this we can see that $x \in O(1/y)$, so $O(x^2) = O(1/y^2)$, and $xy = H + O(1/y)$ as required.