if 1^2 = -1 then what does i^5+i^6+i^7= anyone give me somewhere to start on this one

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- Aug 19th 2009, 08:22 PMnetbookimaginary number
if 1^2 = -1 then what does i^5+i^6+i^7= anyone give me somewhere to start on this one

- Aug 19th 2009, 08:57 PMJG89
I'm sure it was just a typo, but it should be $\displaystyle i^2 = -1 $. To compute $\displaystyle i^5 + i^6 + i^7 $ just calculate what $\displaystyle i^5 $ is, what $\displaystyle i^6 $ is and what $\displaystyle i^7 $ is, then add them together.

For example, you know that $\displaystyle i^2 = -1 $, so $\displaystyle i^3 = i^2*i = -i $. Continuing on, $\displaystyle i^4 = i^3 * i = -i*i = -i^2 = 1 $ and continue in this process. - Aug 19th 2009, 10:10 PMbandedkrait
A simple but interesting result which we obtain from above is that :

**The sum of any 4 consecutive powers of $\displaystyle i$ is equal to zero.**

Hence using this result,

$\displaystyle i^5+i^6+i^7+i^8=0$

Hence the value of the given expression is merely $\displaystyle -i^8$

i.e. http://www.mathhelpforum.com/math-he...acaf41ab-1.gif= $\displaystyle -i^8$

and we know, $\displaystyle i^4=1$

Hence the required answer is $\displaystyle -1$ - Aug 19th 2009, 10:18 PMnetbook
so how do we know that i^4 = 1 and why then would -i^8 = -1

- Aug 19th 2009, 10:25 PMbandedkrait
since $\displaystyle i^2= -1$

Square both sides,

$\displaystyle i^4=1$

Now, square both sides again,

$\displaystyle i^8=1$

so clearly $\displaystyle -i^8=-1$ - Aug 19th 2009, 10:33 PMnetbook
ok thanks again for your patience