Mock AIME 2 2006-2007 Problems/Problem 5
Revision as of 02:35, 23 April 2008 by I like pie (talk | contribs)
Problem
Given that and
find
.
Solution
Multiplying both sides of the equation by , we get
![$iz^3 = z + 2 + \frac{3}{z} + \frac{4}{z^2} + \cdots$](http://latex.artofproblemsolving.com/7/3/d/73dd020de5a5408e6f52c937e7fa8d3ad55908ad.png)
and subtracting the original equation from this one we get
![$iz^2(z-1)=z+1+\frac{1}{z}+\frac{1}{z^2}+\frac{1}{z^3}+\cdots$](http://latex.artofproblemsolving.com/4/8/0/480103d40eb8a07b9f7e95e6b28f7e18bebacdf5.png)
Using the formula for an infinite geometric series, we find
![$iz^2(z-1)=\frac{z}{1-\frac{1}{z}}=\frac{z^2}{z-1}$](http://latex.artofproblemsolving.com/f/c/d/fcd64829e67ebcc4eab6aa004f4ebf575367775d.png)
Rearranging, we get
![$iz^2(z-1)^2=z^2\iff (z-1)^2=\frac{1}{i}=-i\Rightarrow z=1\pm\sqrt{-i}$](http://latex.artofproblemsolving.com/c/9/9/c990fcfac5ea595298da9e3f9b72af3d76bdd42f.png)
Thus , and the answer is
.