+526 For positive integers n1 and n2 the value of expression \((1 + i)^{{n}_{1}}+(1 + i^3)^{{n}_{1}}+(1 + i^5)^{{n}_{2}}+(1 + i^7)^{{n}_{2}}\), where \(i = \sqrt{-1}\) is a real number iff
a) n1 = n2 + 1
b) n1 = n2 - 1
c) n1 = n2
d) n1 > 0, n2 > 0
I managed to simplify this as
\((1 + i)^{{n}_{1}}+(1 + i^3)^{{n}_{1}}+(1 + i^5)^{{n}_{2}}+(1 + i^7)^{{n}_{2}} \)
\(= (1 + i)^{{n}_{1}}+(1 - i)^{{n}_{1}}+(1 + i)^{{n}_{2}}+(1 - i)^{{n}_{2}}\)
\(=(cos{\pi\over 4}+isin{\pi\over 4})^{{n}_{1}}+(cos{\pi\over 4}-isin{\pi\over 4})^{{n}_{1}}+(cos{\pi\over 4}+isin{\pi\over 4})^{{n}_{2}} +(cos{\pi\over 4}-isin{\pi\over 4})^{{n}_{2}}\)
Can anyone help me after this step?
Thank you :)
+33659 As follows:
The quoted answer d) is rather strange, since the question specifies that the n's are positive integers!.
+526 Yes indeed option d is too obvious, option c would be the right one since you arrived at the answer by only taking n1 = n2 (= n) commonly right?
Thank you so much!
+33659 No, I didn't assume n1 = n2. The conclusion holds for each pair of terms separately.
+526 But as per the solution, at one point you've seem to take n instead of n1 and n2 what about that? That's why I presumed the solution might be holding for n1 = n2
+33659 I used n as a general value. You can replace it with n1 and the first two terms produce a real number; replace it with n2 and the second two terms produce a real value. So your original expression is true for all positive integer values of n1 and n2.
The question asks for a proof "if and only if" (iff). The expression is a real number for all the possibilities listed, however, it is also true for other possibilities not listed (such as n1 = n2+3, for example), which means none of a) to d) provide the correct answer (they all conform to the "if" part, but not the "only if" part).
