For anyone who wants a little puzzle to start the week: if I set z_0 = i, and z_(n+1) = 0.5 (z_n + |z_n|), what is the limit of z_n as n tends to infinity?

(Spoiler below) . . . . . . . . . . . . . . . . . . . . . . . . . . Cute. The limit in question is 2 / pi . You can do this for any initial value of z_0 . For real values of z_0 , it's easy, so suppose instead that z_0 is non-real. Write z_0 = r (cotan(alpha) + i) , where r = Im(z_0) and alpha = Arg(z_0) is in the interval (-pi , pi) . With this choice of argument, r and alpha have the same sign. We claim that z_n = 2^(-n) r (cotan(alpha / 2^n) + i) , which is immediate for n = 0 . Then we have |z_n| = 2^(-n) |r| |cosec(alpha / 2^n)| = 2^(-n) r cosec(alpha / 2^n) because r and alpha / 2^n and cosec(alpha / 2^n) all have the same sign. Now we have z_(n+1) = 2^(-(n+1)) r (cotan(alpha / 2^n) + cosec(alpha / 2^n) + i) = 2^(-(n+1)) r (cotan(alpha / 2^(n+1)) + i) , by using the "half angle" formula cotan(theta) + cosec(theta) = cotan(theta / 2) . This establishes the claim, and now the limit is easily evaluated as r / alpha = Im(z_0) / Arg(z_0) . Michael Reid