# Was Wronski Wrong?

I was reminded of a problem from a text book from school. The book was MEI Pure Mathematics 5, First Edition. It had the following question:

The Polish mathematician Hoëné Wronski (1778-1853) once wrote that:

$\pi = \frac{2\infty}{\sqrt{-1}}\left\{ (1+\sqrt{-1})^{1/\infty} - (1-\sqrt{-1})^{1/\infty}\right\}$

Was Wronski wrong?

At the time I found this problem very hard and I was very proud to have solved it. Now I am older, I don’t understand how I could have ever found it difficult, but I suppose that is a good thing.

Let us first turn it into modern notation. We wish to find q, where:

$q = \lim_{x \rightarrow \infty} \frac{2x}{i}\left\{ (1+i)^{1/x} - (1-i)^{1/x}\right\}$

Note that $1+i = \sqrt{2} e^{i \pi/4}$ and $1-i = \sqrt{2} e^{-i \pi/4}$. Using these representations,

$q = \lim_{x \rightarrow \infty} -2^{1+\frac{1}{2x}} xi \left\{ e^{i\pi/4x} - e^{-i\pi/4x}\right\}$

And so

$q = \lim_{x \rightarrow \infty} -2ix2^{1/2x} (2i \sin (\pi/4x))$

Now, if we use the fact that $\sin {x} \approx x$ for small x, we have:

$q = \lim_{x \rightarrow \infty} 2^{\frac{1}{2x}} \pi$

and hence $q =\pi$ as required.

So, yes, Wronski was right.