1. Simplify i

$\displaystyle = (i^3)^7$
$\displaystyle =-i$

Also can someone help me fill in my i chart that my teacher started...

$\displaystyle i^1 = ?$
$\displaystyle i^2 = -1$
$\displaystyle i^3 = i^2 * i = -i$
$\displaystyle i^4 = i^2 * i^2 = 1$
$\displaystyle i^5 = i^4 * i = i$
$\displaystyle i^6 = i^3 * i^3= ?$
$\displaystyle i^7 = i^4 * i^3= ?$
$\displaystyle i^8 = i^4 * i^4= ?$
$\displaystyle i^9 = i^5 * i^4= ?$
$\displaystyle i^10 = i^5 * i^5= ?$
$\displaystyle i^11 = i^5 * i^6= ?$
$\displaystyle i^12 = i^5 * i^7= ?$
$\displaystyle i^13 = i^5 * i^8= ?$

2. Originally Posted by fluffy_penguin

$\displaystyle = (i^3)^7$
$\displaystyle =-i$
that's incorrect $\displaystyle i^{21} = i \cdot i^{20} = i \cdot \left( i^2 \right)^{10} = i$

3. Also can someone help me fill in my i chart that my teacher started...

$\displaystyle i^1 = ?$
$\displaystyle i^2 = -1$
$\displaystyle i^3 = i^2 * i = -i$
$\displaystyle i^4 = i^2 * i^2 = 1$
$\displaystyle i^5 = i^4 * i = i$
$\displaystyle i^6 = i^3 * i^3= ?$
$\displaystyle i^7 = i^4 * i^3= ?$
$\displaystyle i^8 = i^4 * i^4= ?$
$\displaystyle i^9 = i^5 * i^4= ?$
$\displaystyle i^10 = i^5 * i^5= ?$
$\displaystyle i^11 = i^5 * i^6= ?$
$\displaystyle i^12 = i^5 * i^7= ?$
$\displaystyle i^13 = i^5 * i^8= ?$
just reuse the previous values.

for instance, $\displaystyle i^6 = i^3 \cdot i^3$

so look up in your chart for $\displaystyle i^3$, we see that $\displaystyle i^3 = -i$

thus, $\displaystyle i^6 = i^3 \cdot i^3 = -i \cdot -i = i^2 = -1$

4. Thanks.

5. Hello, fluffy_penguin!

Can someone help me fill in my chart that my teacher started...

$\displaystyle \begin{array}{ccccccc}i^1 & = & & & & &{\color{blue}i} \\i^2 & = & & & & & {\color{blue}\text{-}1} \\i^3 & =& \;\,i^2\cdot i \;&= & (\text{-}1)i &=& {\color{blue}\text{-}i} \\i^4 &=& \;\,i^3\cdot i \;& = & (\text{-}i)(i) &=& {\color{blue}1}\end{array}$
$\displaystyle \begin{array}{ccccccc}i^5 &=& i^4\cdot i &=& (1)(i)& = & {\color{red}i} \\i^6 &=& i^5\cdot i &=& (i)(i ) &=& {\color{red}\text{-}1} \\i^7 &=& (i^6)(i) &=&(\text{-}1)(i) &=& {\color{red}\text{-}i} \\i^8 &=& (i^7)(i) &=& (\text{-}i)(i) &=& {\color{red}1}\end{array}$
Don't you see the pattern?

The successive powers are: .$\displaystyle i,\;-1,\;-i,\;1$ . . . . . over and over.

Simplify: .$\displaystyle i^{21}$
We know that: .$\displaystyle i^4 \,=\,1$

So we have: .$\displaystyle i^{21} \:=\:i^{20}\cdot i \;=\;(i^4)^5\cdot i \;=\;(1)^5\cdot i \;=\;i$