# Question Video: Simplifying Imaginary Numbers with Powers Mathematics

Given that 𝑛 is an integer, simplify 𝑖^(16𝑛 − 35).

02:14

### Video Transcript

Given that 𝑛 is an integer, simplify 𝑖 to the power of 16𝑛 minus 35.

Remember the cycle that helps us remember the identities for various powers of 𝑖 is as shown. So we can do one of two things. Our first method is to use the laws of exponents to essentially unsimplify our expression a little. We know that 𝑥 to the power of 𝑎 times 𝑥 to the power of 𝑏 is the same as 𝑥 to the power of 𝑎 plus 𝑏. So we can reverse this and say that 𝑖 to the power of 16𝑛 minus 35 is equal to 𝑖 to the power of 16𝑛 times 𝑖 to the power of negative 35.

𝑖 to the power of 16𝑛 can actually be further written as 𝑖 to the power of four to the power of four 𝑛. This corresponds to the part of our cycle 𝑖 to the power of four 𝑛. So we can see that 𝑖 to the power of 16𝑛 can be written as one. And what about 𝑖 to the power of negative 35? This one is a little more complicated. We’re going to write negative 35 in the form four 𝑎 plus 𝑏, where 𝑏 can take the values zero, one, two, or three to correspond with the values in our cycle. It’s the same as four times negative nine plus one.

Remember four times negative nine is negative 36 and adding one gets us negative 35. And we chose negative nine instead of negative eight as we needed 𝑏 to be zero, one, two, or three. And we certainly don’t want it to be a negative value. So 𝑖 to the power of negative 35 will have the same result as 𝑖 to the power of four 𝑛 plus one in our cycle; that’s 𝑖. So 𝑖 to the power of 16𝑛 minus 35 is one multiplied by 𝑖 which is 𝑖.

Let’s have a look at the alternative method. Here we would have jumped straight into writing the power — that’s 16 𝑛 minus 35 — in the form four 𝑎 plus 𝑏, where 𝑏 again is zero, one, two, or three. We can write 16𝑛 as four times four 𝑛 and negative 35 as four times negative nine plus one. We can factor this expression and we see that 16 𝑛 minus 35 is the same as four multiplied by four 𝑛 minus nine plus one. So we can see that once again 𝑖 to the power of 16𝑛 minus 35 will have the same result as 𝑖 to the power of four 𝑛 plus one in our cycle; that’s 𝑖.