# imaginary number

• Aug 19th 2009, 08:22 PM
netbook
imaginary 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 PM
JG89
I'm sure it was just a typo, but it should be $i^2 = -1$. To compute $i^5 + i^6 + i^7$ just calculate what $i^5$ is, what $i^6$ is and what $i^7$ is, then add them together.

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

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

Hence using this result,

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

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

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

and we know, $i^4=1$

Hence the required answer is $-1$
• Aug 19th 2009, 10:18 PM
netbook
so how do we know that i^4 = 1 and why then would -i^8 = -1
• Aug 19th 2009, 10:25 PM
bandedkrait
since $i^2= -1$

Square both sides,

$i^4=1$

Now, square both sides again,

$i^8=1$

so clearly $-i^8=-1$
• Aug 19th 2009, 10:33 PM
netbook
ok thanks again for your patience