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# Can someone help? (Complex numbers)

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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) n> 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 :)

Jun 26, 2021

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As follows:

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

Jun 26, 2021

#1
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As follows:

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

Alan Jun 26, 2021
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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!

amygdaleon305  Jun 26, 2021
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No, I didn't assume n1 = n2.  The conclusion holds for each pair of terms separately.

Alan  Jun 26, 2021
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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

amygdaleon305  Jun 27, 2021
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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
edited by Alan  Jun 27, 2021
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Oh I see it now okay! Thanks a lot :)

amygdaleon305  Jun 27, 2021