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Complex number problem

If \(n\) is a natural number such that \(\mid i+2i^{2}+3i^{3}...ni^{n}\mid=18\sqrt{2}\).

Then what is(are) the value of \(n\)?

Where \(i=\sqrt{-1}\).

Note by Krishna Jha
4 years, 2 months ago

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Given \( | i + 2i^2 + 3i^3 + \ldots + n i^{n} | = 18 \sqrt{2} \)

\( i + 2i^2 + 3i^3 + \ldots + n i^{n} \)

\( = (i + i^2 + i^3 + \ldots + i^n) \)

\( \space \space \space + \space \space (i^2 + i^3 + \ldots + i^n) \)

\( \space \space \space + \space \space \space \space \space \space \space \space \space (i^3 + \ldots + i^n) \)

\( \space \space \space \ldots \)

\( \space \space \space + \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space (i^n) \)

\( = \large \frac {i(i^n -1)}{i-1} + \frac {i^2 (i^{n-1} -1)}{i-1} + \frac {i^3 (i^{n-2} -1)}{i-1} + \ldots + \frac {i^n (i^{1} -1)}{i-1} \)

\( = \large \frac {1}{i-1} [ i(i^n -1) + i^2 (i^{n-1} - 1) + i^3 (i^{n-2} - 1) + \ldots + i^{n-1} ( i^2 - 1) + i^n (i - 1) ] \)

\( = \large \frac {1}{i-1} [ n \space i^{n+1} - \frac {i(i^n -1)}{i-1} ]\)

\( = \large ( \frac {1}{i-1} )^2 [ (i-1) n \space i^{n+1} - i(i^n -1) ]\)

\( = \LARGE \frac {1}{2} [-n \space i^{n+1} + i^n (n+1) - 1 ] \)

Case I: \( n \pmod{4} = 0 \Rightarrow i^{n+1} = i, \space i^{n} = 1 \)

\( \large \Rightarrow -n \space i^{n+1} + i^n (n+1) - 1 = n - ni \)

\( \large \Rightarrow | -n \space i^{n+1} + i^n (n+1) - 1 | = \sqrt{ n^2 + n^2 } \)

\( \large \Rightarrow 36 \sqrt 2 = \sqrt{ n^2 + n^2 } \Rightarrow n = 36 \)

Case II: \( n \pmod{4} = 1 \Rightarrow i^{n+1} = -1, \space i^{n} = i \)

Solve this the same way and it yields no integer solution

Case III: \( n \pmod{4} = 2 \Rightarrow i^{n+1} = -i, \space i^{n} = -1 \)

Likewise. No integer solution.

Case IV: \( n \pmod{4} = 3 \Rightarrow i^{n+1} = 1, \space i^{n} = -i \)

Likewise, we get \(n = 35 \)

Hence, \( \LARGE n = \boxed{35}, \boxed{36} \)

Pi Han Goh - 4 years, 2 months ago

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Hey man, good solution!! When I first saw the problem I also thought about the many sums of APs, but I definitely did not know how to work on it.

Leonardo Cidrão - 4 years, 2 months ago

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Hey.. I think \(\frac{1}{(i-1)^{2}}=\frac{i}{2}\)?

You wrote \(-\frac{1}{2}\)..

Krishna Jha - 4 years, 2 months ago

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Oh man I'm horrible in spotting my own mistakes... Thanks!

Pi Han Goh - 4 years, 2 months ago

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Too lengthy , I also tried in the same way, but is there any short trick?.Otherwise your solution is nice....

Kishan K - 4 years, 2 months ago

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Well, I not sure if this is better or not. But....

Let \( \large f(n) = | \displaystyle \sum_{j=1}^n j \space i^j | \)

Then by trial and error, we have

\( f(3) = f(4) = 2 \sqrt2, \space f(7) = f(8) = 4 \sqrt2, \space f(11) = f(12) = 6 \sqrt2 \)

This suggests that \( f(2k-1) = f(2k) = k \sqrt2 \) for some positive integer \(k>1\). I'm not sure I can prove this rigorously. Induction perhaps? Still thinking.

With that, we get \(k = 18 \Rightarrow 2k-1 = 35, 2k = 36 \Rightarrow n=35,36 \)

On the other hand, since we have a scalar multiple of \( \sqrt2 \) on \(RHS\), this suggests that \( | Re(i+2i^2 + 3i^3 + \ldots + n i^n) | = | Im(i+2i^2 + 3i^3 + \ldots + n i^n) | \). Not sure how to continue...

Pi Han Goh - 4 years, 2 months ago

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@Pi Han Goh A shorter way: Let \(S=i+2i^{2}+3i^{3}+...+ni^{n}...(1)\)

\(\hspace{30mm}iS =\hspace{4mm}i^{2}+2i^{3}....+(n-1)i^{n}+ni^{n+1}...(2)\)

Subtracting terms having same power of i in \((1)\) and \((2)\)...

We have \(S(1-i)=i+i^{2}+i^{3}...i^{n}-ni^{n+1}\)

After this applying formula for sum of G.P.

\(S(1-i)=\frac{i(1-i^{n})}{1-i}-ni^{n+1}\)

\(\Rightarrow S=\frac{i(1-i^{n})}{(1-i)^{2}}-\frac{ni^{n+1}}{1-i}\)

Then the cases that you told... thanks for the cases. They just didnt come to my mind.

Krishna Jha - 4 years, 2 months ago

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@Krishna Jha Oh! You're welcome.

Pi Han Goh - 4 years, 2 months ago

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