# Confusion while evaluating the complex continued fraction $\frac{1}{i+\frac{1}\ddots}$

I faced some problems when I tried to evaluate the following continued fraction $$\cfrac{1}{i+\cfrac{1}{i+\cfrac{1}{i+\cfrac{1}{i+\cfrac{1}{i+\cfrac{1}{i+\cdots}}}}}}\$$ The common trick as usual is to set $$x= \cfrac{1}{i+\cfrac{1}{i+\cfrac{1}{i+\cfrac{1}{i+\cfrac{1}{i+\cfrac{1}{i+\cdots}}}}}}\$$ Then, I have $$x=\frac{1}{i+x}$$ Using the quadratic formula, I get $$x=\frac{-i\pm\sqrt{3}}{2}$$ There are two solutions for $$x$$, but I don't know which one is correct. For any real continued fraction, for example $$\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cdots}}}}}}\$$ it is clear to see that the value is $$\frac{-1+\sqrt{5}}{2}$$ instead of $$\frac{-1-\sqrt{5}}{2}$$.

But for the complex continued fraction, I have difficulty distinguishing the correct solution.

Well, Vasily has put his finger on one problem, but I would like to point out a much more serious one.

We write a continued fraction to get a number that is the limit of the convergents, that is, of the expressions that you get when you cut your continued fraction off, to be a finite c.f.

The first convergent is $$\frac1i$$, no problem, but the second is $$\frac1{i+\frac1i}=\frac10\,,$$ a most unfortunate development. My recommendation would be to go on to other complex continued fractions, where the partial denominators are rather larger than $$i$$.

• [+1] very thorough remark. Mar 10, 2019 at 15:02
• I'm not sure what you mean by your last sentence. What about $[x;x,x,\cdots]$ where $x = \sqrt{2}i$? Mar 11, 2019 at 4:47
• I didn’t mean much. But $\sqrt2i$ is not what I had in mind by “rather larger”. Indeed, even $x=2i$ seems to give an uninteresting limiting value. Mar 11, 2019 at 4:54
• Indeed, see my answer, which shows that the continued fraction $[x;x,x,\cdots]$ converges for every complex $x ∉ i(-2,2)$. And the same analysis will reveal exactly which $x$ will result in division by zero. Mar 11, 2019 at 13:26

The problem with this expression is that function $$f(x)=(i+x)^{-1}$$ is a tricky one:

$$f(f(x)) = \frac{1}{i+\frac{1}{i+x}}=\frac{i+x}{-1+ix+1}=-i+\frac1x,\\ f(f(f(x))) = \frac{1}{i+(-i+1/x)}=x.$$

So every number generates the orbit of length 3, except for two numbers you have found, which are fixed points. Thus, the series of nesting functions $$f$$ does not converge if you haven't already started with a fixed point. So the question as it is stated has no sense: you cannot assign any number to this expression.

I shall answer the general question completely. The continued fraction $$[x;x,x,...]$$ converges for any $$x ∈ \mathbb{C} ∖ i(-2,2)$$ $$= \{ z : z ∈ \mathbb{C} ∧ z ∉ \{ ir : r ∈ (-2,2) \} \}$$, and here is a proof sketch.

If $$[x;x,x,...]$$ converges to $$c$$, then $$c = x + 1/c$$ and hence $$c$$ is a root of the quadratic $$( t ↦ t^2 - x t - 1 )$$.

Let $$r,s$$ be the roots of the quadratic $$( t ↦ t^2 - x t - 1 )$$ and so $$r + s = x$$ and $$r s = -1$$.

Let the sequence of approximants be $$(a_n)$$ where $$a_1 = x$$ (and the sequence stops if it becomes $$0$$).

Let $$b_0 = 1$$ and $$b_n = a_n b_{n-1}$$ for each $$n$$ such that $$a_n$$ is defined, giving $$a_n = \frac{b_n}{b_{n-1}}$$.

Given any $$n ∈ \mathbb{N}^+$$ such that $$a_n ≠ 0$$:

$$b_{n+1} = ( x + \frac1{a_n} ) b_n = x b_n + b_{n-1}$$.

Thus $$b_{n+1} - r b_n = s ( b_n - r b_{n-1} ) = s^n ( b_1 - r b_0 ) = (x-r) s^n = s^{n+1}$$.

Thus $$b_n - r^n b_0 = \sum_{k=1}^n r^{n-k} s^k$$ and hence $$b_n = \sum_{k=0}^n r^{n-k} s^k$$.

If $$r ≠ s$$, then $$b_n = {\large \frac{ r^{n+1} - s^{n+1} }{r-s} }$$ and hence $$a_n = {\large \frac{ r^{n+1} - s^{n+1} }{ r^n - s^n } }$$.

If $$r = s$$, then $$b_n = (n+1) r^n$$ and hence $$a_n = \frac{n+1}{n} r$$.

If $$x ∉ i[-2,2]$$,

$$|r| ≠ 1$$ otherwise $$x = r + s = r - \frac{1}{r} = r - r^* = 2i·Im(r) ∈ i[-2,2]$$.

Permute $$r$$,$$s$$ such that $$|r| > 1 > |s|$$, possible since $$|r| · |s| = |rs| = 1$$.

Then by induction we get $$a_n = {\large \frac{ r^{n+1} - s^{n+1} }{ r^n - s^n } } ≠ 0$$ for every $$n ∈ \mathbb{N}^+$$.

Thus $$a_n = r + {\large \frac{ (r-s) s^n } { r^n - s^n } } = r + {\large \frac{r-s}{ (\frac{r}{s})^n - 1 } } \to r$$ as $$n \overset{∈\mathbb{N}}\to \infty$$.

If $$x ∈ \{2i,-2i\}$$,

$$r = s ∈ \{i,-i\}$$ because $$(r-s)^2 = (r+s)^2 - 4rs = 0$$.

Then by induction we get $$a_n = \frac{n+1}{n} r ≠ 0$$ for every $$n ∈ \mathbb{N}^+$$.

Thus $$a_n \to r$$ as $$n \overset{∈\mathbb{N}}\to \infty$$.

If $$x ∈ i(-2,2)$$,

By induction we get $$a_n ∈ i\mathbb{R}$$ whenever $$a_n$$ is defined.

If $$a_n \to c$$ as $$n \to \infty$$:

$$c ∈ i\mathbb{R}$$ since $$i\mathbb{R}$$ is closed.

But $$c ∈ \{r,s\} = {\large \frac{ x \pm \sqrt{x^2+4} }{2} }$$ and hence $$c ∉ i\mathbb{R}$$ since $$x^2+4 > 0$$.

Therefore $$( a_n )$$ either terminates in a $$0$$ or does not converge.
It is worth emphasizing that one cannot assume that the continued fraction converges. If it converges then its limit must be one of the roots of the quadratic (and the above proof shows us explicitly which one). But it may be that it does not converge in the first place, such as for $$x = i$$.
• By the way, from this it is trivial to obtain $[0;x,x,x,\cdots]$, or any other continued fraction with eventually repeating $x$. Mar 11, 2019 at 13:34
For a continued fraction to converge, the sequence of convergents (values finite initial segments of the partial fraction expression) must converge to a particular complex number, that is, to a specific argument and magnitude. For the real continued fraction you mention, the algebraic expression allows the argument to be either $$0$$ (containing "$$+\sqrt{5}$$") or $$\pi$$ (containing "$$-\sqrt{5}$$"). The same thing can be done with the complex version. Your two expressions have arguments $$-\pi/6$$ and $$7\pi/6$$. If your continued fraction had any hope of converging (which, as others have shown, it does not) it converges to something with a specific argument. You generically hope that one of the algebraic solutions you have has an argument matching the limit argument of the convergents. Finding that limit argument usually requires another computation, typically solving another recurrence.