Complex Recursion Consider the recursive function $C_n = C_{n-1} + iC_{n-2}$, where $C_1 = 1, C_2 = 1.$  If $C_{10}$ is written in the form $a+bi,$ find $b$. 
I solved this problem through brute force with a calculator.  Is there an elegant solution method?  I wasn't able to find any pattern in the terms, nor was I able to find a nice geometric solution. 
 A: The comment of @lulu is most apposite. The equations are a mess, but once I caught my mistakes, the symbolic calculator worked like a charm. It’s perfectly standard recurrence-fu, and here it is:
You have a function $n\mapsto C_n$ that satisfies the stated recursion, and you make the educated guess that it’ll be a linear combination of exponentials, of which a basic one looks like $n\mapsto T^n$ for some value of $T$. The recurrence says $T^{n+2}=T^{n+1}+iT$, from which you immediately factor out $T^n$ to get the quadratic $T^2-T-i=0$. Roots of course are
$$
t=\frac{1+\sqrt{1+4i}}2\,,\qquad t'=1-t_1\,,
$$
and  you must absolutely not try to go numerical at this stage.
You expect that $C_n=At^n+B(1-t)^n$, and by trying this with $C_0=1$ and $C_1=1$, you get $A=t/(2t-1)$ and $B=(t-1)/(2t-1)$. This gives the closed-form description
$$
C_n=\frac{t^{n+1}}{2t-1}\>-\> \frac{(1-t)^{n+1}}{2t-1}\,.
$$
Only now do you call in your symbolic calculator, and find that when $n=10$, you get, lo and behold, $C_{10}=-12-25i$.
