# 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 \$\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 PM
bandedkrait
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 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 \$\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 PM
netbook
ok thanks again for your patience