we are dealing with imaginary/complex numbers.
how would you simplify this.

Question

we are dealing with imaginary/complex numbers. 
how would you simplify this.

eyust707 · Answer

this?
$$i^{80}$$

anonymous · Answer

yes that! sorry cant figure out how to work this..

anonymous · Answer

@eyust707  how did you do that?

eyust707 · Answer

$$i^1 = i$$
$$i^2 = -1$$
$$i^3 = i^2*i=-i$$
$$i^4 = -1 *-1 = 1$$
$$i^5 = -1 * -1 * i = i$$
...

anonymous · Answer

$$i ^{80} =  (i^4)^{20}$$

anonymous · Answer

$$i^4 = 1$$

anonymous · Answer

what language is this O_o

anonymous · Answer

can u explain what u did and why you did this @Yahoo! please...
 @xxfreshboy its algebra 2 .-.

anonymous · Answer

Im taking the same thing so whats good with the -_-? it was just a question nothing serious....

eyust707 · Answer

marird do you understand my post?

anonymous · Answer

lol no, it wasnt meant for u. it was meant for my feelings towards the subject.. so far, not liking algebra so much.

anonymous · Answer

umm yes i understand but would i have to do that process all the way till i get to 80?

anonymous · Answer

ooo iight lol.. my fault then if My tone sounded alil off no hard feelings beautiful :)

eyust707 · Answer

no well check it out do you agree that?
$$i^{80}=(i^{4})^{20}$$

anonymous · Answer

lol no its kool (:

anonymous · Answer

yes i agree

eyust707 · Answer

sweet so and since we know that $i^{4} = 1$ we can replace it with 1

we get:
$$(1)^{20}$$
Which is just 1

eyust707 · Answer

We can try another one if you would like.. after you do a few it will make more sense

anonymous · Answer

okay yeah i thnk that would help.

anonymous · Answer

take the integer remainder when you divide the exponent by 4
pattern is 
$$i^0=1, i^1=i, i^2=-1,i^3=-i, i^4=1$$
so for example 
$$i^{51}=i^3=-i$$ because when you divide 51 by 4 the remainder is 3