lol how when that will give you a real value hahaha
umad?
? if you have n=2 then it becomes e^(-pi) which I'm pretty sure is not equal to i^i (correct me if I am wrong)Umm....
where n is a positive odd integer
Essentially, you need to know Euler's Identity to solve this.lol how when that will give you a real value hahaha
well Mirakon what you reckon??
buddy learn to read, n has a restriction for odd integers only!? if you have n=2 then it becomes e^(-pi) which I'm pretty sure is not equal to i^i (correct me if I am wrong)
aiight but in the last line how did you get thatEssentially, you need to know Euler's Identity to solve this.
e^(i.pi) + 1 = 0
e^(i.pi) = -1
e^(i.pi/2) = i (square root)
e^(-pi/2) = i^i (raise to the i; i*i = -1)
This guy knows his complex numbers.Umm....
where n is a positive odd integer
Do i detect sarcasm?This guy knows his complex numbers.
aiight cool but where does the e come from or is this a rule??
It is basically just the usual index laws combined with the fact that:
No, I was giving you a compliment.Do i detect sarcasm?
You'd cover it in most first year algebra courses at uni. Don't think it is 4U though.i do not think Euler's Identity is syllabus, but if you do a bit of extra-curricular maths you may know it