imaginary number

**netbook** · August 19th 2009, 08:22 PM

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

**JG89** · August 19th 2009, 08:57 PM

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.

**bandedkrait** · August 19th 2009, 10:10 PM

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.