- #1
Theman123
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Homework Statement
This problem is probably quite simple for people who have covered this topic but the answer for part A is n=4k where k is any integer.
Part B answer is n= 2+4k
Look below for the working out and i will tell you what i don't understand
Homework Equations
de moivres theorem
The Attempt at a Solution
polar form is ([itex]\sqrt{2}[/itex] cis [itex]\pi[/itex] /4)^n
so [itex]\sqrt{2}[/itex]^n [cos(n[itex]\pi[/itex] /4)] +isin(n[itex]\pi[/itex] /4)] (demoivres theorem)
When Sin (n[itex]\pi[/itex] /4) = 0 the equation is real
n[itex]\pi[/itex] /4 = 0 +k2[itex]\pi[/itex] (For some reason this should be +k[itex]\pi[/itex] which is why i don't get the answer.
n=8k (solving for n)
Part b is prety much the same thing but cos (n[itex]\pi[/itex] /4) The question that i am asking is why is it +k[itex]\pi[/itex] and not +k2[itex]\pi[/itex] as that is usually the period for sin and cosine but [itex]\pi[/itex] is the period for tan i know that.