Taking the negative of a continued fraction If I have a continued fraction for an irrational number $z= \langle a_0;a_1,a_2,a_3,\ldots\rangle$ it seems that $(-1)*z = \langle-a_0;-a_1,-a_2,-a_3,\ldots\rangle$. Is this true?
In general, if you have the continued fraction representation for $y$ and $z$ can you say something about the continued fraction representation of $y*z$?
 A: While evaluating the continued fraction of the negative coefficients of the continued fraction expansion of $z$ does indeed evaluate to $-z$, your formula $[-a_0; -a_1, \dots, ]$ is not regarded as "the continued fraction" of $-z$, which is usually defined using the Euclidean algorithm resulting in non-negative coefficients after the first. In general, for $z \in \mathbb{R}$ and $z= [a_1; a_2, a_3 \dots]$, then 
\begin{align}
-z= [-a_1-1;1,a_2-1, a_3, \dots],
\end{align}
where the terms in the ellipses are identical. For example, $\frac{4}{3} = [1,3]$, while $-\frac{4}{3} = [-2,1,2]$. To address your second question, there are formulas to compute the continued fraction expansion of $\frac{az+b}{cz+d}$, where $a, b, c, d \in \mathbb{Z}$, relying only on the continued fraction expansion of $z$ and certain $2 \times 2$ matrices defined using the coefficients. See An Introduction to Continued Fractions by van der Poorten. 
A: If continued fraction
$$
a_0+\frac{1}{\displaystyle a_1+\frac{1}{a_2+\ddots}}
$$
converges to $z$, where all $a_k$ and $z$ are complex numbers, then continued fraction
$$
-a_0+\frac{1}{\displaystyle -a_1+\frac{1}{-a_2+\ddots}}
$$
converges to $-z$.  
Is that what you mean?  You should be able to prove it!
