simplify: 17i^19?

Notice this pattern:
i = i
i² = -1
i^3 = i * i² = -i
i^4 = i² * i² = -1 * -1 = 1

If you continue this, you'll see it repeats.
i^5 = i^4 * i = 1 * i = i
i^6 = i^4 * i² = -1
i^7 = -i
i^8 = 1
etc.

If you take the exponent and subtract a multiple of 4, you get the same answer.
i^19 will therefore be the same as i^3 = -i

And then that is multiplied by 17.
-17i

Another way to see this is to expand it out as follows:
17i^19 = 17(i^4)(i^4)(i^4)(i^4)(i^3)
= 17 * 1 * 1 * 1 * 1 * -i
= 17 * -i
= -17i

Answer:
-17i

Remember these:
i = squareroot of -1
i^2 = -1
i^3 = -i
i^4 = 1

i^18 is 6 of i^3 which would make it -i, but since it is 17i^19 and not 17i^18, -i should be multiplied by i making it i^4
Which is 1

Therefore, 17 x 1 = 17

 
   17 i¹⁹            ← Note: i⁴ = 1 ... So, use the laws of exponents to rewrite  i¹⁹
= 17 (i⁴)⁴  i² i
= 17(1)⁴ (-1)i
= -17i               ← ANSWER
 

17i^19
= 17i^16 X i^2 X i
= 17(1) X (-1) X i
= -17i <<< answer

hint:
i^2=-1