jmc

Let $z=a+bi,$ where $a$ and $b$ are real numbers. Then $$|(a-1)+bi|=|(a+3)+bi|=|a+(b-1)i|.$$Hence, $(a-1{)}^2+b^2=(a+3{)}^2+b^2=a^2+(b-1{)}^2.$

From $(a-1{)}^2+b^2=(a+3{)}^2+b^2,$ $8a=-8,$ so $a=-1.$ Then the equations above become $$4+b^2=1+(b-1{)}^2.$$Solving, we find $b=-1.$ Therefore, $z=\boxed{-1-i}.$