Question

Prove that i^98+w^99=0
.
.
where i=√(-1) and w is the complex cube root of unity.

Answer 1

w^99-1=0
w^99=1
1=1

Answer 2

w^99-1=0 how??

Answer 3

what happened to i

Answer 4

i^0 = 1
i^1 = i
i^2 = -1
I^3 = -i (from i*i^2= i*-1)
i^4 = 1 (i*i^3= i*-i= -(i*i)= -(-1) = +1. or i^2*i^2= -1*-1= 1)
i^5 = i
there is a pattern here...

notice we can rewrite as
i^98= i^96 * i^2
i^96= (i^4)^24 * i^2
i^4= 1 and 1^24 is 1
i^2= -1
1*-1= -1

for W^99, we are told W^3= 1
rewrite W^99= (W^3)^33 = 1^33= 1

Answer 5

I used rules for exponents
\[ (x^a)^b = x^{ab}\]
and
\[ x^{a+b}=x^a\cdot x^b \]

Answer 6

ok

Answer 7

how it is coming -1

Answer 8

If you don't like using exponents, you could write out
i^98 = i*i*i*i....*i 98 times
put parens around each group of 4 i's:
i^98= (i*i*i*i) * (i*i*i*i) * ... (i*i*i*i) * (i*i*i*i) * i*i
you will get 24 groups of i*i*i*i each = 1
and i*i left over: 1*1*1... *1 * -1= -1

Answer 9

(i^4)^24=-1 how

Answer 10

(i^4)^24 = 1

notice (i^4)^24 = i^96
we want i^98

Answer 11

maybe the big numbers are confusing...
if you had
i^6= i*i*i*i * i*i (that is what i to the 6th power means)
but we know i*i*i*i= 1 (see above)
so i^6= 1* i*i
but i*i= -1 (by definition)
so i^6= 1* -1 = -1

now if we had i^10 it would be
i^10 = i*i*i*i * i*i*i*i * i*i
but this is
i^10 = 1 * 1 * -1
i^10= -1
each time we multiply by i^4, we are multiplying by 1

Answer 12

pretty confusing

Answer 13

but got it

Answer 14

think about it, it should make sense.

Answer 15

w^99=1 how

Answer 16

Hint : i^4 =1 w^3 =1

Answer 17

i know this

Answer 18

just need to ask w^99 =1 how?

Answer 19

you know w^3= 1 (w is the cube root of 1)

so think like this
w*w*w * w*w*w * w*w*w * ....
in groups of 3
you will have exactly 33 groups
each w*w*w = 1
and 1 times itself 33 times is still 1

Answer 20

i^98 = (i^4)^24 * i^2

Answer 21

w^99 = (w^3)^33 = 1

1^2 = -1


so -1+1 =0

Answer 22

thanks alot