Finding sum of complex numbers

**SyNtHeSiS** · October 18th 2010, 01:53 PM

Find the sum of $1 + i + i^2 + i^3+...+i^{16}$

Attempt:

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

but the correct answer was 1

**Plato** · October 18th 2010, 01:59 PM

I get 1. Note that $i^{17}=i$ & $\dfrac{1-i}{1-i}=1$

**SyNtHeSiS** · October 18th 2010, 02:28 PM

Originally Posted by Plato

I get 1. Note that $i^{17}=i$ & $\dfrac{1-i}{1-i}=1$

How did you get this?

**Plato** · October 18th 2010, 02:41 PM

Originally Posted by SyNtHeSiS

How did you get this?

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

$\dfrac{{\left( {1 - i^{17} } \right)}}{{1 - i}} = \dfrac{{\left( {1 - i} \right)}}{{1 - i}}$

**Archie Meade** · October 18th 2010, 02:50 PM

Originally Posted by SyNtHeSiS

How did you get this?

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

$i^{16}=\left(i^2\right)^8=(-1)^{even}=1$

**Plato** · October 18th 2010, 03:04 PM

I prefer this idea. $\bmod (N,4) = k$ means that $k$ is the remainder when $N$ is divided by $4$ .

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

Example: $i^{17}=i^{\bmod (17,4)}=i^1$

**HallsofIvy** · October 19th 2010, 03:24 AM

We might also point out that $(1+ x+ x^2+ \cdot\cdot\cdot+ x^{n-1}+ x^n)(1- x)= (1- x)^{n+1}$ for any positive integer n.

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