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I tried to calculate $$\sqrt{ ... {\sqrt{\sqrt i}}}$$ by saying $k =\sqrt{ ... {\sqrt{\sqrt i}}}$ , so the equation $$ \sqrt k = k$$ must be true. When we square each side of the equation, we get$$ k^2 - k = 0$$ so we get $\sqrt{ ... {\sqrt{\sqrt i}}} = 0,1$.

However, I learned that $ai \not = b $ when $a,b \subset \mathbb {R}$, so the equation can't be correct. Where is the flaw in my logic?

Jay Lee
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    Are your terms well defined? What rule do you follow in selecting each square root? – lulu Apr 17 '18 at 13:13
  • Just to remind you that you have to be careful when you take square roots in the complex plane, what is your definition of $\sqrt{}$ ? – Delta-u Apr 17 '18 at 13:14
  • you cannot use infinity this way. $a+\infty=b+\infty=\infty$ for real $a\ne b$ – Vasili Apr 17 '18 at 13:15
  • If, say, you write $i=\exp(i\frac {\pi}2)$ and look at the sequence $a_n=\exp(i\frac {\pi}{2^n})$ then you do indeed get $a_n\to 1$. But is this what you had in mind? – lulu Apr 17 '18 at 13:16
  • using the principal value of the logarithm your question is equivalent to evaluate $\lim_{n\to\infty}e^{i\frac{\pi}{2^n}}$ – Masacroso Apr 17 '18 at 13:16

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If for $z=r e^{i \theta}$, $\theta \in (-\pi,\pi)$ you define $\sqrt{z}$ by $\sqrt{r} e^{i \frac{\theta}{2}}$ then with: $$u_0=i=e^{i \frac{\pi}{2}}$$ $$u_{k+1}=\sqrt{u_k}$$ then your recurrence is well defined and:

$$u_k=e^{i \frac{\pi}{2^{k+1}}}$$

so indeed $u_k \to e^0=1$.

Delta-u
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If you always use the principal square root, the answer is one. $i$ has modulus 1 and argument $\pi/2$. Each time you take the square root, the modulus stays the same and the argument is cut in half. Thus, in the limit you get the complex number with modulus 1 and argument 0, which is otherwise known as 1.

C Monsour
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