A Disguise of Something Else

Multiplying both sides of the equation by z, we get

$iz^3 = z + 2 + \dfrac{3}{z} + \dfrac{4}{z^2} + \cdots$ ,

and subtracting the original equation from this one we get

$iz^2(z-1)=z+1+\dfrac{1}{z}+\dfrac{1}{z^2}+\dfrac{1}{z^3}+\cdots.$

Using the formula for an infinite geometric series, we find

$iz^2(z-1)=\dfrac{z}{1-\dfrac{1}{z}}=\dfrac{z^2}{z-1}.$

Rearranging, we get

$iz^2(z-1)^2=z^2\iff (z-1)^2=\frac{1}{i}=-i\Rightarrow z=1\pm\sqrt{-i}.$ Thus $n=1$ , and the answer is $\lfloor 100n \rfloor = \boxed{100}.$

I solved it using Maclaurin series.

$\dfrac {1}{1-x} = 1 + x + x^2 + x^3 + x^4 + x^5+ ...$

$\Rightarrow \dfrac {d}{dx} \left( \dfrac {1}{1-x} \right) = 1 + 2x + 3x^2 + 4x^3 + 5x^4 + ...$

$\Rightarrow \dfrac {1}{(1-x)^2} = 1 + 2x + 3x^2 + 4x^3 + 5x^4 + ...$

Let $x = \dfrac {1}{z}$ , then we have:

$\Rightarrow \dfrac {1}{(1-\frac{1}{z})^2} = 1 + \dfrac {2}{z} + \dfrac {3}{z^2} + \dfrac {4}{z^3} + \dfrac {5}{z^4} + ...$

$\Rightarrow \dfrac {1}{(1-\frac{1}{z})^2} = iz^2 \quad \Rightarrow \dfrac {z^2}{(z-1)^2} = iz^2 \quad \Rightarrow \dfrac {1}{(z-1)^2} = i$

$\Rightarrow (z-1)^2 = \dfrac {1}{i} = -i \quad \Rightarrow z - 1 = \pm \sqrt{-i}\quad \Rightarrow z = 1 \pm \sqrt{-i}$

$\Rightarrow n = 1\quad \Rightarrow \lfloor 100n \rfloor = \boxed{100}$

Divide by $z$ and obtain

$iz = \dfrac 1z +\dfrac {2}{z^2} + \dfrac{3}{z^3}+\dfrac{4}{z ^4}+\dfrac{5}{z^5}+\cdots$

Since

$\dfrac{z}{z-1} = \sum_{k=0}^{\infty} \dfrac{1}{z^k} = 1 + \dfrac 1z +\dfrac {1}{z^2} + \dfrac{1}{z^3}+\dfrac{1}{z ^4}+\dfrac{1}{z^5}+\cdots$

we have

$iz + \dfrac{z}{z-1} =$

$= \dfrac 1z +\dfrac {2}{z^2} + \dfrac{3}{z^3}+\dfrac{4}{z ^4}+\cdots + 1 + \dfrac 1z +\dfrac {1}{z^2} + \dfrac{1}{z^3}+\dfrac{1}{z ^4}+\cdots =$

$= 1+\dfrac 2z + \dfrac{3}{z^2}+\dfrac{4}{z ^3}+\dfrac{5}{z^4}+\cdots = iz^2$

We have the nice equation

$iz + \dfrac{z}{z-1} = iz^2$

that becames

$z^2 - 2z +(1-i) = 0$

solving it with the second degree polynomial formula we get

$z = 1 \pm \sqrt{-i}$

so that $n = 1$ and the answer is $100$ .