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3psilon
How to tell what i^33 is . I remember doing this in algebra 2 but I forgot how to
i^4 = 1 33 = 4x8+1 => i^33=i^1=i
i^1=i i^2=-1 i^3=-i i^4=1 That's the pattern, so look for anay number smaller than the given or the given and check if it's divisible by 4. In this case as rahul91 wrote 4*8+1=33 so you know that i^32 equals 1 and i^1 equals i so the answer is 1*i
Any exponentnummber divisible by four equals 1 by the pattern above
Thank you @frx very helpful :)