1. ## what is i^3?

if i^2=-1, does that make i^3=1, im not really sure?

2. $\displaystyle i^3 = i^{2+1} = i^2\cdot i = (-1)\cdot i = -i$

3. @josipive
why are you allowed to do the 3rd step (-1).i,
while I am writing I might know the answer, is it because the root of i is -1?
sorry got confused becasue you wrote (-1).i instead of i.(-1)

4. Originally Posted by MarNie
@josipive
why are you allowed to do the 3rd step (-1).i,
while I am writing I might know the answer, is it because the root of i is -1?
sorry got confused becasue you wrote (-1).i instead of i.(-1)
1- $\displaystyle i^2=-1$.
2- $\displaystyle -i=i-$.

5. Originally Posted by hyzlemon
if i^2=-1, does that make i^3=1, im not really sure?
Hi hyzlemon,

$\displaystyle i^2=-1$ makes $\displaystyle i^2(i^2)=-1(-1)=1$

All odd powers of $\displaystyle i$ are $\displaystyle {\pm}\ i$
All even powers of $\displaystyle i$ are $\displaystyle {\pm}1$

$\displaystyle i=\sqrt{-1}$

$\displaystyle i^2=\sqrt{-1}\sqrt{-1}=-1$

$\displaystyle i^3=i^2i=ii^2=(-1)i=i(-1)=-i$

$\displaystyle i^4=i^2i^2=(-1)(-1)=1\ or\ i^3i=ii^3=i(-i)=-i^2=-(-1)=1$

$\displaystyle i^5=i^4i=(1)i=i$

Hence all higher powers continue on in this cycle $\displaystyle i,\ i^2,\ i^3,\ i^4,\ i^5=i,\ i^6=i^2.\ i^7=i^3,\ i^8=i^4,\ i^9=i....$

6. I have found it useful when teaching algebra students about the powers of i, is just jot down the following:

$\displaystyle i^0=1$

$\displaystyle i^1=i$

$\displaystyle i^2=-1$

$\displaystyle i^3=-i$

Now when encountering any power of i, divide the exponent by 4. Using a little modular arithmetic, we arrive at:

[1] If the remainder is 0, then the answer is 1.

[2] If the remainder is 1, then the answer is i.

[3] If the remainder is 2, then the answer is -1.

[4] If the remainder is 3, then the answer is -i.

Example: $\displaystyle i^{111}=-i$ because $\displaystyle 111 \div 4 = 27 \: {\text{Remainder 3}}$