For positive integers n_{1} and n_{2} 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) n_{1} = n_{2} + 1

b) n_{1} = n_{2} - 1

c) n_{1} = n_{2}

d) n_{1 }> 0, n_{2} > 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 :)

amygdaleon305 Jun 26, 2021

#1**+3 **

Best Answer

As follows:

The quoted answer d) is rather strange, since the question specifies that the n's are positive integers!.

Alan Jun 26, 2021

#2**+2 **

Yes indeed option d is too obvious, **option c** would be the right one since you arrived at the answer by only taking n_{1} = n_{2} (= n) commonly right?

Thank you so much!

amygdaleon305
Jun 26, 2021

#3**+3 **

No, I didn't assume n1 = n2. The conclusion holds for each pair of terms separately.

Alan
Jun 26, 2021

#4**+1 **

But as per the solution, at one point you've seem to take n instead of n_{1} and n_{2} what about that? That's why I presumed the solution might be holding for n_{1} = n_{2}

amygdaleon305
Jun 27, 2021

#5**+2 **

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).

Alan
Jun 27, 2021