Simplify the complex number i^{57} as much as possible.

Question

Accepted Answer

Step:1 ：The complex number (i) has a cycle of 4. That is, (i^1 = i), (i^2 = -1), (i^3 = -i), and (i^4 = 1). After that, the cycle repeats. Therefore, to simplify (i^{57}), we need to find the remainder when 57 is divided by 4. This will tell us where in the cycle (i^{57}) falls.Step:2 ：Calculate the remainder when 57 is divided by 4: remainder = 1Step:3 ：Since the remainder is 1, this means that (i^{57}) is equivalent to (i^1), which is simply (i).Step:4 ：Final Answer: (boxed{i})

Answer

Step: 1 ：The complex number (i) has a cycle of 4. That is, (i^1 = i), (i^2 = -1), (i^3 = -i), and (i^4 = 1). After that, the cycle repeats. Therefore, to simplify (i^{57}), we need to find the remainder when 57 is divided by 4. This will tell us where in the cycle (i^{57}) falls.Step: 2 ：Calculate the remainder when 57 is divided by 4: remainder = 1Step: 3 ：Since the remainder is 1, this means that (i^{57}) is equivalent to (i^1), which is simply (i).Step: 4 ：Final Answer: (boxed{i})

Simplify the complex number $i^{57}$ as much as possible.

Answer

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