complex arithmeticgeometric mean
It is also possible to define the arithmeticgeometric mean^{} for complex numbers^{}. To do this, we first must make the geometric mean^{} unambiguous by choosing a branch of the square root^{}. We may do this as follows: Let $a$ and $b$ br two nonzero complex numbers such that $a\ne sb$ for any real number $$. Then we will say that $c$ is the geometric mean of $a$ and $b$ if ${c}^{2}=ab$ and $c$ is a convex combination of $a$ and $b$ (i.e. $c=sa+tb$ for positive real numbers $s$ and $t$).
Geometrically, this may be understood as follows: The condition $a\ne sb$ means that the angle between $0a$ and $0b$ differs from $\pi $. The square root of $ab$ will lie on a line bisecting this angle, at a distance $\sqrt{ab}$ from $0$. Our condition states that we should choose $c$ such that $0c$ bisects the angle smaller than $\pi $, as in the figure below:
$$\text{{xy}},(2,1)*0,(0,0);(50,50)**\mathrm{@};(52,52)*b,(0,0);(16,16)**\mathrm{@},(18,18)*a,(0,0);(0,40)**\mathrm{@},(0,42)*c,(0,0);(0,40)**\mathrm{@},(0,42)*c$$ 
Analytically, if we pick a polar representation $a=a{e}^{i\alpha}$, $b=b{e}^{i\beta}$ with $$, then $c=\sqrt{ab}{e}^{i\frac{\alpha +\beta}{2}}$. Having clarified this preliminary item, we now proceed to the main definition.
As in the real case, we will define sequences of geometric and arithmetic means^{} recursively and show that they converge to the same limit. With our convention, these are defined as follows:
${g}_{0}$  $=a$  
${a}_{0}$  $=b$  
${g}_{n+1}$  $=\sqrt{{a}_{n}{g}_{n}}$  
${a}_{n+1}$  $={\displaystyle \frac{{a}_{n}+{g}_{n}}{2}}$ 
We shall first show that the phases of these sequences converge. As above, let us define $\alpha $ and $\beta $ by the conditions $a=a{e}^{i\alpha}$, $b=b{e}^{i\beta}$, and $$. Suppose that $z$ and $w$ are any two complex numbers such that $z=z{e}^{i\theta}$ and $w=w{e}^{i\varphi}$ with $$. Then we have the following:

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The phase of the geometric mean of $z$ and $w$ can be chosen to lie between $\theta $ and $\varphi $. This is because, as described earlier, this phase can be chosen as $(\theta +\varphi )/2$.

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The phase of the arithmetic mean of $z$ and $w$ can be chosen to lie between $\theta $ and $\varphi $.
By a simple induction argument^{}, these two facts imply that we can introduce polar representations ${a}_{n}={a}_{n}{e}^{i{\theta}_{n}}$ and ${g}_{n}={g}_{n}{e}^{i{\varphi}_{n}}$ where, for every $n$, we find that ${\theta}_{n}$ lies between $\alpha $ and $\beta $ and likewise ${\varphi}_{n}$ lies between $\alpha $ and $\beta $. Furthermore, since ${\varphi}_{n+1}=({\varphi}_{n}+{\theta}_{n})/2$ and ${\theta}_{n+1}$ lies between ${\varphi}_{n}$ and ${\theta}_{n}$, it follows that
$${\varphi}_{n+1}{\theta}_{n+1}\le \frac{1}{2}{\varphi}_{n}{\theta}_{n}.$$ 
Hence, we conclude that ${\varphi}_{n}{\theta}_{n}\to 0$ as $n\to \mathrm{\infty}$. By the principle of nested intervals, we further conclude that the sequences ${\{{\theta}_{n}\}}_{n=0}^{\mathrm{\infty}}$ and ${\{{\varphi}_{n}\}}_{n=0}^{\mathrm{\infty}}$ are both convergent^{} and converge to the same limit.
Having shown that the phases converge, we now turn our attention to the moduli. Define ${m}_{n}=\mathrm{max}({a}_{n},{g}_{n})$. Given any two complex numbers $z,w$, we have
$$\sqrt{zw}\le \mathrm{max}(z,w)$$ 
and
$$\left\frac{z+w}{2}\right\le \mathrm{max}(z,w),$$ 
so this sequence ${\{{m}_{n}\}}_{n=0}^{\mathrm{\infty}}$ is decreasing. Since it bounded from below by $0$, it converges.
Finally, we consider the ratios of the moduli of the arithmetic and geometric means. Define ${x}_{n}={a}_{n}/{g}_{n}$. As in the real case, we shall derive a recursion relation for this quantity:
${x}_{n+1}$  $={\displaystyle \frac{{a}_{n+1}}{{g}_{n+1}}}$  
$={\displaystyle \frac{{a}_{n}+{g}_{n}}{2\sqrt{{a}_{n}{g}_{n}}}}$  
$={\displaystyle \frac{\sqrt{{a}_{n}^{2}+2{a}_{n}{g}_{n}\mathrm{cos}({\theta}_{n}{\varphi}_{n})+{{g}_{n}}^{2}}}{2\sqrt{{a}_{n}{g}_{n}}}}$  
$={\displaystyle \frac{1}{2}}\sqrt{{\displaystyle \frac{{a}_{n}}{{g}_{n}}}+2\mathrm{cos}({\theta}_{n}{\varphi}_{n})+{\displaystyle \frac{{g}_{n}}{{a}_{n}}}}$  
$={\displaystyle \frac{1}{2}}\sqrt{{x}_{n}+2\mathrm{cos}({\theta}_{n}{\varphi}_{n})+{\displaystyle \frac{1}{{x}_{n}}}}$ 
For any real number $x\ge 1$, we have the following:
$x1$  $\ge 0$  
${(x1)}^{2}$  $\ge 0$  
${x}^{2}2x+1$  $\ge 0$  
${x}^{2}+1$  $\ge 2x$  
$x+{\displaystyle \frac{1}{x}}$  $\ge 2$ 
If $$, then $1/x>1$, so we can swithch the roles of $x$ and $1/x$ and conclude that, for all real $x>0$, we have
$$x+\frac{1}{x}\ge 2.$$ 
Applying this to the recursion we just derived and making use of the halfangle identity for the cosine, we see that
$${x}_{n+1}\ge \frac{1}{2}\sqrt{2+2\mathrm{cos}({\theta}_{n}{\varphi}_{n})}=\mathrm{cos}\left(\frac{{\theta}_{n}{\varphi}_{n}}{2}\right).$$ 
Title  complex arithmeticgeometric mean 

Canonical name  ComplexArithmeticgeometricMean 
Date of creation  20130322 17:10:05 
Last modified on  20130322 17:10:05 
Owner  rspuzio (6075) 
Last modified by  rspuzio (6075) 
Numerical id  15 
Author  rspuzio (6075) 
Entry type  Result 
Classification  msc 33E05 
Classification  msc 26E60 