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

1 Answer
May 23, 2018

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

Explanation:

Let S = \sqrt{-2+2\sqrt{-2+2\sqrt{-2+2\sqrt{-2+2\sqrt{-2+...}}}}}

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

S^2 = -2+2\sqrt{-2+2\sqrt{-2+2\sqrt{-2+2\sqrt{-2+...}}}}

S^2 + 2 = 2\sqrt{-2+2\sqrt{-2+2\sqrt{-2+2\sqrt{-2+...}}}}

\frac{S^2 + 2}{2} = \sqrt{-2+2\sqrt{-2+2\sqrt{-2+2\sqrt{-2+...}}}}

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

And if you solve for S:

S^2 + 2 = 2S, S^2 - 2S + 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 \sqrt{-2} into a calculator that can handle imaginary numbers and use the the recurrence relation :

f(1) = \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+1.001394137i
pretty good approximation