If z_(n+1)=dfrac12(z_n+dfrac1z_n) for n=0,1,2,... and…

If $z_{n+1}=\dfrac12\left(z_n+\dfrac1{z_n}\right)$ for $n=0,1,2,\ldots$ and $-\dfrac\pi2<\arg z_0<\dfrac\pi2$, then $\lim\limits_{n\to\infty}z_n=1$. Prove this.

Step 1 · Rewrite the recursion algebraically

$$ z_{n+1}=\frac{z_n^2+1}{2z_n} $$

Compute $z_{n+1}\pm1$ by combining over the common denominator $2z_n$:

$$ z_{n+1}-1=\frac{z_n^2-2z_n+1}{2z_n}=\frac{(z_n-1)^2}{2z_n}, \qquad z_{n+1}+1=\frac{z_n^2+2z_n+1}{2z_n}=\frac{(z_n+1)^2}{2z_n} $$

Step 2 · The substitution linearizes the recursion

Let $w_n=\dfrac{z_n-1}{z_n+1}$. Dividing the two identities above:

$$ w_{n+1}=\frac{z_{n+1}-1}{z_{n+1}+1}=\frac{(z_n-1)^2}{(z_n+1)^2}=w_n^2 $$

Iterating: $\ \boxed{w_n=w_0^{2^n}}$.

Step 3 · $|w_0|<1$ because $\text{Re}(z_0)>0$

For $z=x+iy$ with $x>0$:

$$ |z-1|^2-|z+1|^2=\big[(x-1)^2+y^2\big]-\big[(x+1)^2+y^2\big]=-4x<0 $$

so $|z-1|<|z+1|$, i.e. $|w|=\left|\dfrac{z-1}{z+1}\right|<1$ whenever $\text{Re}(z)>0$. Since $-\frac\pi2<\arg z_0<\frac\pi2$ means exactly $\text{Re}(z_0)>0$, we get $|w_0|<1$ strictly.

Step 4 · $w_n\to0$

Since $|w_0|<1$:

$$ |w_n|=|w_0|^{2^n}\to0 \quad\text{as } n\to\infty $$

— in fact doubly-exponentially fast, the hallmark of Newton's method.

Step 5 · Invert to recover $z_n\to1$

Solving $w=\dfrac{z-1}{z+1}$ for $z$ gives $z=\dfrac{1+w}{1-w}$, a continuous function of $w$ on the open unit disk (no singularity, since $|w_n|<1$ never reaches $w=1$). Therefore:

$$ \boxed{\lim_{n\to\infty}z_n=\lim_{n\to\infty}\frac{1+w_n}{1-w_n}=\frac{1+0}{1-0}=1} $$

Why the strict inequality matters

If $z_0$ were purely imaginary ($\arg z_0=\pm\frac\pi2$, the excluded boundary), then $\text{Re}(z_0)=0$ gives $|w_0|=1$ exactly, and $w_n=w_0^{2^n}$ stays on the unit circle forever rather than shrinking to $0$ — so $z_n$ need not converge to $1$ in that boundary case, which is exactly why the problem requires the strict inequality $-\frac\pi2<\arg z_0<\frac\pi2$.

Verification

Starting from $z_0=1+i$ (so $\arg z_0=\frac\pi4$, within range): direct iteration gives $z_1=0.75+0.25i,\ z_2=0.975-0.075i,\ z_3\approx0.99730+0.00172i,\ z_4\approx1.0000022-0.0000046i,\ z_5\approx1-1.0\times10^{-11}i$, rapidly converging to $1$ — matching $w_0=(z_0-1)/(z_0+1)=0.2+0.4i$, $|w_0|=\sqrt{0.2}\approx0.447<1$, with $w_n=w_0^{2^n}$ predicting each $z_n$ exactly.

(Note: contrary to a superficially plausible claim that $|z_n|\ge1$ for all $n$, this is false in general — here $|z_1|\approx0.79<1$ — so any correct proof must avoid assuming it; the Cayley-transform argument above sidesteps this issue entirely.)