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

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- Mar 11th 2010, 05:41 AM #1

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- Mar 11th 2010, 05:51 AM #2

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- Mar 11th 2010, 06:07 AM #3

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- Mar 11th 2010, 07:18 AM #4

- Mar 11th 2010, 07:51 AM #5

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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 $\displaystyle i,\ i^2,\ i^3,\ i^4,\ i^5=i,\ i^6=i^2.\ i^7=i^3,\ i^8=i^4,\ i^9=i....$

- Mar 11th 2010, 10:20 AM #6

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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 $\displaystyle 111 \div 4 = 27 \: {\text{Remainder 3}}$