complex arithmetic-geometric mean

It is also possible to define the arithmetic-geometric mean forcomplex numbers. To do this, we first must make the geometricmean unambiguous by choosing a branch of the square root. Wemay do this as follows: Let $a$ and $b$ br two non-zero complexnumbers such that $a\eq sb$ for any real number $s<0$ . Thenwe 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 $a\eq sb$ means that the angle between $0a$ and $0b$ differsfrom $\\pi$ . The square root of $a b$ will lie on a line bisectingthis angle, at a distance $\\sqrt{|ab|}$ from $0$ . Our conditionstates that we should choose $c$ such that $0c$ bisects the anglesmaller than $\\pi$ , as in the figure below:

\\xy,(2,-1)*{0},(0,0);(50,50)**@{-};(52,52)*{b},(0,0);(-16,16)**@{-},(-18,18)*{%a},(0,0);(0,40)**@{-},(0,42)*{c},(0,0);(0,-40)**@{-},(0,-42)*{-c}

Analytically, if we pick a polar representation $a=|a|e^{i\\alpha}$ , $b=|b|e^{i\\beta}$ with $|\\alpha-\\beta|<\\pi$ , then $c=\\sqrt{|ab|}e^{i{\\alpha+\\beta\\over 2}}$ . Having clarified this preliminary item,we now proceed to the main definition.

As in the real case, we will define sequences of geometric and arithmeticmeans recursively and show that they converge to the same limit. With ourconvention, these are defined as follows:

	$\\displaystyle g_{0}$	$\\displaystyle=a$
	$\\displaystyle a_{0}$	$\\displaystyle=b$
	$\\displaystyle g_{n+1}$	$\\displaystyle=\\sqrt{a_{n}g_{n}}$
	$\\displaystyle a_{n+1}$	$\\displaystyle={a_{n}+g_{n}\\over 2}$

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 $|\\alpha-\\beta|<\\pi$ . Suppose that $z$ and $w$ are any two complex numbers such that $z=|z|e^{i\\theta}$ and $w=|w|e^{i\\phi}$ with $|\\phi-\\theta|<\\pi$ . Then we have the following:

•
The phase of the geometric mean of $z$ and $w$ can be chosen to liebetween $\\theta$ and $\\phi$ . This is because, as described earlier, thisphase can be chosen as $(\\theta+\\phi)/2$ .
•
The phase of the arithmetic mean of $z$ and $w$ can be chosen to liebetween $\\theta$ and $\\phi$ .

By a simple induction argument, these two facts imply that we can introducepolar representations $a_{n}=|a_{n}|e^{i\\theta_{n}}$ and $g_{n}=|g_{n}|e^{i\\phi_{n}}$ where, for every $n$ , we find that $\\theta_{n}$ lies between $\\alpha$ and $\\beta$ and likewise $\\phi_{n}$ lies between $\\alpha$ and $\\beta$ .Furthermore, since $\\phi_{n+1}=(\\phi_{n}+\\theta_{n})/2$ and $\\theta_{n+1}$ lies between $\\phi_{n}$ and $\\theta_{n}$ , it follows that

|\\phi_{n+1}-\\theta_{n+1}|\\leq{1\\over 2}|\\phi_{n}-\\theta_{n}|.

Hence, we conclude that $|\\phi_{n}-\\theta_{n}|\\to 0$ as $n\\to\\infty$ . Bythe principle of nested intervals, we further conclude that the sequences $\\{\\theta_{n}\\}_{n=0}^{\\infty}$ and $\\{\\phi_{n}\\}_{n=0}^{\\infty}$ are both convergentand converge to the same limit.

Having shown that the phases converge, we now turn our attention to themoduli. Define $m_{n}=\\max(|a_{n}|,|g_{n}|)$ . Given any two complexnumbers $z, w$ , we have

|\\sqrt{zw}|\\leq\\max(|z|,|w|)

and

\\left|{z+w\\over 2}\\right|\\leq\\max(|z|,|w|),

so this sequence $\\{m_{n}\\}_{n=0}^{\\infty}$ is decreasing. Since it bounded frombelow by $0$ , it converges.

Finally, we consider the ratios of the moduli of the arithmetic and geometricmeans. Define $x_{n}=|a_{n}|/|g_{n}|$ . As in the real case, we shall derive arecursion relation for this quantity:

	$\\displaystyle x_{n+1}$	$\\displaystyle={\|a_{n+1}\|\\over\|g_{n+1}\|}$
		$\\displaystyle={\|a_{n}+g_{n}\|\\over 2\\sqrt{\|a_{n}g_{n}\|}}$
		$\\displaystyle={\\sqrt{\|a_{n}^{2}\|+2\|a_{n}\|\|g_{n}\|\\cos(\\theta_{n}-\\phi_{n})+\|g_{%n}\|^{2}}\\over 2\\sqrt{\|a_{n}g_{n}\|}}$
		$\\displaystyle={1\\over 2}\\sqrt{{\|a_{n}\|\\over\|g_{n}\|}+2\\cos(\\theta_{n}-\\phi_{n})%+{\|g_{n}\|\\over\|a_{n}\|}}$
		$\\displaystyle={1\\over 2}\\sqrt{x_{n}+2\\cos(\\theta_{n}-\\phi_{n})+{1\\over x_{n}}}$

For any real number $x\\geq 1$ , we have the following:

	$\\displaystyle x-1$	$\\displaystyle\\geq 0$
	$\\displaystyle(x-1)^{2}$	$\\displaystyle\\geq 0$
	$\\displaystyle x^{2}-2x+1$	$\\displaystyle\\geq 0$
	$\\displaystyle x^{2}+1$	$\\displaystyle\\geq 2x$
	$\\displaystyle x+{1\\over x}$	$\\displaystyle\\geq 2$

If $0<x<1$ , then $1/x>1$ , so we can swithch the roles of $x$ and $1/x$ andconclude that, for all real $x>0$ , we have

x+{1\\over x}\\geq 2.

Applying this to the recursion we just derived and making use of the half-angleidentity for the cosine, we see that

x_{n+1}\\geq{1\\over 2}\\sqrt{2+2\\cos(\\theta_{n}-\\phi_{n})}=\\cos\\left({\\theta_{n}%-\\phi_{n}\\over 2}\\right).