1. Zadatak

Prvo rješenje.

Neka je $z_1=a+bi$ i $z_2=c+di$ ($a,b,c,d\in\mathbb R$).

Dani uvjet $|z_1+2z_2|=|2z_1+z_2|$ ekvivalentan je s: $$\begin{aligned} |(a+bi)+2(c+di)|&=|2(a+bi)+(c+di)|,\\ |(a+2c)+(b+2d)i|&=|(2a+c)+(2b+d)i|,\\ (a+2c)^2+(b+2d)^2&=(2a+c)^2+(2b+d)^2,\\ a^2+4ac+4c^2+b^2+4bd+4d^2&=4a^2+4ac+c^2+4b^2-4bd+d^2,\\ 3c^2+3d^2&=3a^2+3b^2,\\ a^2+b^2&=c^2+d^2. \end{aligned}$$

Neka je $\alpha\in\mathbb R$. Množenjem gornje jednakosti s $1-\alpha^2$ dobivamo $$\begin{aligned} (1-\alpha^2)a^2+(1-\alpha^2)b^2&=(1-\alpha^2)c^2+(1-\alpha^2)d^2,\\ a^2+\alpha^2c^2+b^2+\alpha^2d^2&=\alpha^2a^2+c^2+\alpha^2b^2+d^2. \end{aligned}$$

Sada na obje strane jednakosti dodamo $2\alpha ac+2\alpha bd$: $$\begin{gathered} a^2+2\alpha ac+\alpha^2c^2+b^2+2\alpha bd+\alpha^2d^2\\ =\alpha^2a^2+2\alpha ac+c^2+\alpha^2b^2+2\alpha bd+d^2,\\ (a+\alpha c)^2+(b+\alpha d)^2=(\alpha a+c)^2+(\alpha b+d)^2. \end{gathered}$$

Konačno, $$\begin{aligned} |(a+\alpha c)+(b-\alpha d)i|&=|(\alpha a+c)+(\alpha b+d)i|,\\ |(a+bi)+\alpha(c+di)|&=|\alpha(a+bi)+(c+di)|, \end{aligned}$$ odnosno $|z_1+\alpha z_2|=|\alpha z_1+z_2|$, što je i trebalo dokazati.

Drugo rješenje.

Za bilo koji realni broj $\alpha$ vrijedi: $$\begin{aligned} &|z_1+\alpha z_2|^2-|\alpha z_1+z_2|^2\\ &=(z_1+\alpha z_2)\overline{(z_1+\alpha z_2)}-(\alpha z_1+z_2)\overline{(\alpha z_1+z_2)}\\ &=(z_1+\alpha z_2)(\overline{z_1}+\alpha\overline{z_2})-(\alpha z_1+z_2)(\alpha\overline{z_1}+\overline{z_2})\\ &=|z_1|^2+\alpha^2|z_2|^2+\alpha(z_1\overline{z_2}+\overline{z_1}z_2)\\ &\qquad-\alpha^2|z_1|^2-|z_2|^2-\alpha(z_1\overline{z_2}+\overline{z_1}z_2)\\ &=(1-\alpha^2)(|z_1|^2-|z_2|^2). \end{aligned}$$

Kako je $|z_1+2z_2|=|2z_1+z_2|$, odnosno $|z_1+2z_2|^2-|2z_1+z_2|^2=0$, i kako je (prema gornjoj formuli) $$|z_1+2z_2|^2-|2z_1+z_2|^2=(1-2^2)(|z_1|^2-|z_2|^2),$$ vrijedi $|z_1|^2-|z_2|^2=0$.

Zato je (opet prema gornjoj formuli) $|z_1+\alpha z_2|^2-|\alpha z_1+z_2|^2=0$, za sve $\alpha\in\mathbb R$ odnosno $|z_1+\alpha z_2|=|\alpha z_1+z_2|$.