Question

Simplify: i137 + i1003

Answer 1

Is that i^137 + i^1003?

Answer 2

offline.
I would guess so though.
What's your answer? :)

Answer 3

0

Answer 4

It's correct, but show your process, in case the user returns...

Answer 5

When multiplying i, there is a pattern.

i*i= -1

-1*i=-i

-i*i=1

Every fourth number is a 1

So if we section the power it's raised to into groups of four we can get (i*i*i*i)=1 * (i*i*i*i) = 1...

for i^137
137/4 = 34 with a remainder of 1

This is (i*i*i*i)*(i*i*i*i)*(i*i*i*i)... 34 times which is 1 times itself 34 times.
multiplied by that remainder of 1

1*i = i

So now we have i + i^1003

1003/4 = 250 with a remainder of 3

This is 1 times itself 250 times multiplied by the three remaining i's which is equal to
1*i*i*i= -i

So we have

i + (-i) = 0

Answer 6

:)
Thanks a bunch :D

Answer 7

Yup :D