# 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, 10 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, $$\LARGE n = \boxed{35}, \boxed{36}$$

- 4 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.

- 4 years, 10 months ago

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

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

- 4 years, 10 months ago

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

- 4 years, 10 months ago

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

- 4 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...

- 4 years, 10 months ago

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.

- 4 years, 10 months ago

Oh! You're welcome.

- 4 years, 10 months ago