An algebra problem by Brian Wang

${i}^{1}=i$ , ${i}^{2}=-1$ , ${i}^{3}=-i$ , ${i}^{4}=1$ , ${i}^{5}=i$ , Et cetera. Divide 2015 by 4, and you get a remainder of 3. Since i to the power of any multiple of 4 equals 1, and 1 times i is i, we can ignore everything except the remainder. Therefore, ${i}^{2015}={i}^{3}=$ ${-i}$
$\boxed{-i}$

divide 2015 by 2, you get a remainder of 1.keep in mind that i^2=-1. now you can write (i)^(2015)=[i^2]^(1007) multiplied by i. this is equal to (-1)^(1007) multiplied by i. now (-1) to the power any odd number is equal to (-1). so we have (-1) multiplied by i, hence correct ans is -i.