# Finding sum of complex numbers

• Oct 18th 2010, 01:53 PM
SyNtHeSiS
Finding sum of complex numbers
Find the sum of $\displaystyle 1 + i + i^2 + i^3+...+i^{16}$

Attempt:

$\displaystyle \sum_{k=0}^{n}ar^k$
$\displaystyle =\frac{a(1-r^{n+1})}{1-r}$
$\displaystyle =\frac{(1-i^{16+1})}{1-r}$
$\displaystyle =\frac{(1-i^{17})}{1-r}$

but the correct answer was 1
• Oct 18th 2010, 01:59 PM
Plato
I get 1. Note that $\displaystyle i^{17}=i$ & $\displaystyle \dfrac{1-i}{1-i}=1$
• Oct 18th 2010, 02:28 PM
SyNtHeSiS
Quote:

Originally Posted by Plato
I get 1. Note that $\displaystyle i^{17}=i$ & $\displaystyle \dfrac{1-i}{1-i}=1$

How did you get this?
• Oct 18th 2010, 02:41 PM
Plato
Quote:

Originally Posted by SyNtHeSiS
How did you get this?

$\displaystyle \dfrac{{a\left( {1 - r^{n + 1} } \right)}} {{1 - r}},\;a = 1,\;r = i\;\& \;n = 16$

$\displaystyle \dfrac{{\left( {1 - i^{17} } \right)}}{{1 - i}} = \dfrac{{\left( {1 - i} \right)}}{{1 - i}}$
• Oct 18th 2010, 02:50 PM
Quote:

Originally Posted by SyNtHeSiS
How did you get this?

If you are wondering about $\displaystyle i^{17}$

$\displaystyle i^{16}=\left(i^2\right)^8=(-1)^{even}=1$
• Oct 18th 2010, 03:04 PM
Plato
I prefer this idea. $\displaystyle \bmod (N,4) = k$ means that $\displaystyle k$ is the remainder when $\displaystyle N$ is divided by $\displaystyle 4$.

So $\displaystyle i^N=i^{\bmod (N,4)}$

Example: $\displaystyle i^{17}=i^{\bmod (17,4)}=i^1$
• Oct 19th 2010, 03:24 AM
HallsofIvy
We might also point out that $\displaystyle (1+ x+ x^2+ \cdot\cdot\cdot+ x^{n-1}+ x^n)(1- x)= (1- x)^{n+1}$ for any positive integer n.