1. Zadatak

$1^\circ$ Stavimo $z=x+yi$. Iz prvog uvjeta dobivamo $$\begin{aligned} (x+1)^2+(y+8)^2-(x+2)^2-(y+1)^2&=100\\ 2x+1+16y+64-4x-4-2y-1&=100\\ 14y-2x+60&=100\\ 7y-x&=20\\ y&=\frac17x+\frac{20}7 \end{aligned}$$ $2^\circ$ Iz drugog uvjeta slijedi $$\begin{aligned} (x-5)^2+y^2&=25,\\ (x-5)^2+\left(\frac17x+\frac{20}7\right)^2&=25\\ (x-5)^2\cdot49+(x+20)^2&=49\cdot25\\ 50x^2-450x+400&=0\ /:50\\ x^2-9x+8&=0 \end{aligned}$$ Rješavanjem ove kvadratne jednadžbe dobivamo dva rješenja: $$\begin{aligned} x_1=1&\qquad y_1=3\qquad z_1=1+3i\\ x_2=8&\qquad y_2=4\qquad z_2=8+4i \end{aligned}$$