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 $i=\sqrt{-1}$ .

#NumberTheory #MathProblem #Math

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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 - 7 years, 10 months ago

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

Kishan k - 7 years, 10 months ago

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 - 7 years, 10 months ago

@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 - 7 years, 10 months ago

@Krishna Jha – Oh! You're welcome.

Pi Han Goh - 7 years, 10 months ago

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

You wrote $-\frac{1}{2}$ ..

Krishna Jha - 7 years, 10 months ago

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

Pi Han Goh - 7 years, 10 months ago

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 - 7 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$