Croatia_2018

Let $z=x+iy$, where $x,y\in \mathbb{R}$.

Then $$|z-5|=|(x+iy)-5|=|(x-5)+iy|=\sqrt{(x-5{)}^2+y^2}$$ and $$|z-1|=|(x+iy)-1|=|(x-1)+iy|=\sqrt{(x-1{)}^2+y^2}.$$

The given equation is $$\sqrt{(x-5{)}^2+y^2}=\sqrt{(x-1{)}^2+y^2}+4.$$

Let $a=\sqrt{(x-5{)}^2+y^2}$ and $b=\sqrt{(x-1{)}^2+y^2}$, so $a=b+4$.

Then $$a^2=b^2+8b+16.$$ But $a^2=(x-5{)}^2+y^2$ and $b^2=(x-1{)}^2+y^2$.

So $$(x-5{)}^2+y^2=(x-1{)}^2+y^2+8b+16.$$ The $y^2$ terms cancel: $$(x-5{)}^2=(x-1{)}^2+8b+16.$$ Expand: $$(x^2-10x+25)=(x^2-2x+1)+8b+16$$ $$-10x+25=-2x+1+8b+16$$ $$-10x+25+2x-1-16=8b$$ $$-8x+8=8b$$ $$-x+1=b$$ But $b=\sqrt{(x-1{)}^2+y^2}$, so $$-x+1=\sqrt{(x-1{)}^2+y^2}$$

Now, $\sqrt{(x-1{)}^2+y^2}\ge 0$, so $-x+1\ge 0$, i.e., $x\le 1$.

Square both sides: $$(-x+1{)}^2=(x-1{)}^2+y^2$$ But $(-x+1{)}^2=(x-1{)}^2$, so $$(x-1{)}^2=(x-1{)}^2+y^2$$ $$0=y^2$$ $$y=0$$

Therefore, $z=x+i\cdot 0=x$ is a real number.

Thus, $z$ is real.