Question

Given that $i{z}^2$ =$1+\frac{2}{z}+\frac{3}{z^2}+\frac{4}{z^3}+\frac{5}{z^4}+....$ and $z=n\pm \sqrt{-i}$, find $\lfloor 100n\rfloor$

Answer 1

$i{z}^2=1+\frac{2}{z}+\frac{3}{z^2}+\frac{4}{z^3}+....\infty$ ....(1)

also

$iz=\frac{1}{2}+\frac{2}{z^2}+\frac{3}{z^3}+\frac{4}{z^4}....\infty$ ....(2)

(1) - (2)

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

$iz(z-1)=\frac{1}{1-\frac{1}{z}}$

$iz(z-1)=\frac{z}{z-1}$

$z[iz-i-\frac{1}{z-1}]=0$

$\frac{z}{z-1}[i{z}^2-iz-iz+i-1]=0$

$\frac{z}{z-1}[i{z}^2-2iz+i-1]=0$

$\underset{\left(notpossible\right)}{z=0}$ or $z=\frac{2i\pm \sqrt{-4-4i(i-1)}}{2i}$

$=1\pm \frac{2}{2i}\sqrt{-1+1}$

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

so $n=\perp$ (comparing with $n\pm \sqrt{-i}$)

$\therefore \left[100n\right]=100$