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For some reason I can't seem to see why
$$\frac{az+b}{cz+d} = \frac{bc-ad}{c^2\left(z+\frac{d}{c}\right)} + \frac{a}{c}.$$ Its been so many years since I've taken Algebra I, so when this was stated in class, without even the motions of manipulation, I felt lost. Sorry if this question is too dumb to belong on the site.

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2 Answers 2

up vote 11 down vote accepted

Working on the right hand side, first multiply through by one $c$ to get rid of the compound fraction; then get the common denominator $c(zc+d)$, and do the addition: \begin{align*} \frac{bc-ad}{c^2\left(z+\frac{d}{c}\right)}+\frac{a}{c} &= \frac{bc-ad}{c(zc+d)} + \frac{a}{c}\\ &= \frac{bc-ad}{c(zc+d)} + \frac{a(zc+d)}{c(zc+d)}\\ &= \frac{bc-ad+a(zc+d)}{c(zc+d)}\\ &= \frac{bc-ad + azc + ad}{c(zc+d)}\\ &= \frac{bc+azc}{c(zc+d)}\\ &= \frac{c(b+az)}{c(zc+d)}\\ &= \frac{b+az}{zc+d} = \frac{az+b}{cz+d}. \end{align*} Of course, this doesn't really address how one can go from one to the other "mentally", or how one can "spot" the equality. You are probably thinking, "And how on Earth would anybody ever come up with that on the fly?!"

There is a piece of information you may not have that is being used "in the background": you are dealing with a Moebius transformation, a function from $\mathbb{C}$ to $\mathbb{C}$ that is given by $$z\longmapsto \frac{az+b}{cz+d}\qquad\mbox{with $ad-bc\neq 0$.}$$ Any Moebius transformation can be decomposed as a composition of a translation, an inversion, a dilation, and a translation in a standard way (see the Decomposition section in the Wikipedia page). First you take $z$ and you translate it by $\frac{d}{c}$, which gives you the expression $z+\frac{d}{c}$. Then you invert, giving you $$\frac{1}{z+\frac{d}{c}}.$$ Then you dilate by multiplying by $\frac{bc-ad}{c^2}$, to get $$\frac{bc-ad}{c^2\left(z+\frac{d}{c}\right)}$$ and finally you translate by $\frac{a}{c}$ by adding it to the running total; this is exactly the expression you have. If you are accustomed to playing with Moebius transformations, this is a very common decomposition, so it is an identity that you would have "at your fingertips", as it were. The reason for this particular decomposition is that the pieces (translations, inversions, and dilations) are very easy to understand, so they make particularly good pieces to break the Moebius transformation into.

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Thanks so much. This clarifies a lot. And thanks for the additional information on Moebius transformations. –  I Love Cake Feb 15 '11 at 5:18
    
By the way, this equation was indeed introduced in class in the section on Möbius transformations. –  I Love Cake Feb 15 '11 at 5:24

Multiplying through by the common denominator $\rm\ c\ (c\:z + d)\ $ reduces the "fractional" identity to the equivalent "integral" identity $\rm\ c\ (a\:z+b)\ =\ bc-ad + a\ (c\:z + d)\:,\ $ which is easily verified.

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