Find $n$ where $\displaystyle \bigg|i+2i^2+3i^3+\cdots \cdots +ni^n\bigg|=18\sqrt{2}$.

Question

Suppose that $n$ is a natural number such that $\displaystyle \bigg|i+2i^2+3i^3+\cdots \cdots +ni^n\bigg|=18\sqrt{2}$. Find the value of $n$.

What I try : Let $\displaystyle S =i+2i^2+3i^3+\cdots +ni^n\cdots (1)$

Then $\displaystyle iS =i^2+2i^3+3i^4+\cdots +ni^{n+1}\cdots (2)$

So $\displaystyle S(1-i)=i+i^2+i^3+\cdots +i^n-ni^{n+1}$

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

So we have $\displaystyle \bigg|i^n-ni^{n+1}(1+i)\bigg|=36\sqrt{2}$

How do i find value of $n$ , please help me, Thanks

From the left hand side, it should be possible to get the real part and imaginary part of that expression (and hence the value of the modulus), first assuming that $n$ is odd and then assuming that $n$ is even. — Sarvesh Ravichandran Iyer, Mar 16 '24 at 07:53
Shouldn’t it be $\bigg|i^n-1-ni^{n+1}(1+i)\bigg|=36\sqrt{2}$? — Aig, Mar 16 '24 at 08:12

score 4 · Answer 1 · answered Mar 16 '24 at 08:06

4

Hint: $$\begin{align} 36\sqrt{2}&=\bigg|i^n-ni^{n+1}(1+i)\bigg|\\&=|i^n|\cdot \left|1-ni(i+1) \right|\\&=\left|1-ni(i+1)\right| \\&=\left|(1+n)-ni\right|\\&=\sqrt{(1+n)^2+n^2} \end{align}$$

answered Mar 16 '24 at 08:06

NN2

15,892

1

square both sides and you will have an even plus an odd giving an even number, so no natural n? – hellofriends Mar 16 '24 at 08:09
This seems to have nothing to do with the question. – Aig Mar 16 '24 at 08:38

score 2 · Accepted Answer · answered Mar 16 '24 at 08:35

So we have $$\bigg|i^n-1-ni^{n+1}(1+i)\bigg|=36\sqrt{2}.$$

If $n\underset{4}\equiv0$ then it becomes $$\bigg|1-1-ni(1+i)\bigg|=36\sqrt{2}$$ $$n\bigg|1-i\bigg|=36\sqrt{2}$$ $$n\sqrt{2}=36\sqrt{2}$$ $$n=36.$$

If $n\underset{4}\equiv1$ then it becomes $$\bigg|i-1+n(1+i)\bigg|=36\sqrt{2}$$ $$\bigg|n-1+i(n+1)\bigg|=36\sqrt{2}$$ $$\sqrt{(n-1)^2+(n+1)^2}=36\sqrt{2}$$ $$2(n^2+1)=2\cdot 36^2$$ $$n^2+1=36^2$$ $$n\in\emptyset.$$

If $n\underset{4}\equiv2$ then it becomes $$\bigg|-1-1+ni(1+i)\bigg|=36\sqrt{2}$$ $$\bigg|-n-2+ni\bigg|=36\sqrt{2}$$ $$\sqrt{(n+2)^2+n^2}=36\sqrt{2}$$ $$2n^2+4n+4=36^2\cdot2$$ $$n^2+2n+2=36^2$$ $$n\in\emptyset.$$

If $n\underset{4}\equiv3$ then it becomes $$\bigg|-i-1-n(1+i)\bigg|=36\sqrt{2}$$ $$\bigg|-n-1-i(1+n)\bigg|=36\sqrt{2}$$ $$(n+1)\sqrt{2}=36\sqrt2$$ $$n=35.$$

The answer is: $n=35$ or $36$.

score 0 · Answer 3 · answered Mar 19 '24 at 08:15

0

Using $i+2i^2+3i^3+4i^4=i-2-3i+4=2-2i=2(1-i)$

For every $4$ pair, we get value $2(1-i)$

So for $36\sqrt{2}$, We have $9$ pairs

Means $n=36$

answered Mar 19 '24 at 08:15

jacky

5,194

score -1 · Answer 4 · answered Mar 16 '24 at 08:18

$$S(1-i) = i +...+i^{n} - ni^{n+1}$$

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

we know that $|1-i| = \sqrt{2}$, so:

$$|S| = 18 \sqrt 2 \iff |S(1-i)| = 36 \iff$$

$$\iff | \frac{i(i^n-1)}{i-1} - ni^{n+1}| = 36 \iff 36\sqrt2 = |i^{n+1}-i-ni^{n+1}(i-1)|$$

I think now you must analyze each case: $n \equiv 0, \pm 1,2 \mod 4$ and see which one works

Find $n$ where $\displaystyle \bigg|i+2i^2+3i^3+\cdots \cdots +ni^n\bigg|=18\sqrt{2}$.

4 Answers4