# Show that,sqrt(-2+2sqrt(-2+2sqrt(-2+2sqrt(-2+.............))))=1+-i?

May 23, 2018

Converges to $1 + i$ (on my Ti-83 graphing calculator)

#### Explanation:

Let $S = \setminus \sqrt{- 2 + 2 \setminus \sqrt{- 2 + 2 \setminus \sqrt{- 2 + 2 \setminus \sqrt{- 2 + 2 \setminus \sqrt{- 2 + \ldots}}}}}$

First, Assuming that this infinite series converges (i.e. assuming S exists and takes the value of a complex number),

${S}^{2} = - 2 + 2 \setminus \sqrt{- 2 + 2 \setminus \sqrt{- 2 + 2 \setminus \sqrt{- 2 + 2 \setminus \sqrt{- 2 + \ldots}}}}$

${S}^{2} + 2 = 2 \setminus \sqrt{- 2 + 2 \setminus \sqrt{- 2 + 2 \setminus \sqrt{- 2 + 2 \setminus \sqrt{- 2 + \ldots}}}}$

$\setminus \frac{{S}^{2} + 2}{2} = \setminus \sqrt{- 2 + 2 \setminus \sqrt{- 2 + 2 \setminus \sqrt{- 2 + 2 \setminus \sqrt{- 2 + \ldots}}}}$

$\setminus \frac{{S}^{2} + 2}{2} = S$

And if you solve for S:

${S}^{2} + 2 = 2 S , {S}^{2} - 2 S + 2 = 0$
and applying the quadratic formula you get:

 S = \frac{2\pm \sqrt{4-8}}{2} = \frac{2\pm \sqrt{-4}}{2} = \frac{2\pm 2i}{2} = 1 \pm i

Usually the square root function takes the positive value thus $S = 1 + i$

Thus, if it converges then it must converge to $1 + i$

Now all you have to do is prove that it converges or if you are lazy like me then you can plug $\setminus \sqrt{- 2}$ into a calculator that can handle imaginary numbers and use the the recurrence relation :

$f \left(1\right) = \setminus \sqrt{- 2}$

 f(n+1) = \sqrt{-2 + 2\sqrt{f(n)}

I repeated this many times on my Ti - 83 and found that it does get closer for example after I repeated it somewhere like 20 times I got approximately
$1.000694478 + 1.001394137 i$
pretty good approximation