jmc

Let $z=x+yi,$ where $x$ and $y$ are real numbers. Since $|z|=1,$ $x^2+y^2=1.$ Then $$\begin{array}{rl}|1+z|+|1-z+z^2|& =|1+x+yi|+|1-x-yi+x^2+2xyi-y^2|\\ & =|(1+x)+yi|+|(1-x+x^2-1+x^2)+(-y+2xy)i|\\ & =|(1+x)+yi|+|(-x+2x^2)+(-y+2xy)i|\\ & =\sqrt{(1+x{)}^2+y^2}+\sqrt{(-x+2x^2{)}^2+(-y+2xy{)}^2}\\ & =\sqrt{(1+x{)}^2+y^2}+\sqrt{(-x+2x^2{)}^2+y^2(1-2x{)}^2}\\ & =\sqrt{(1+x{)}^2+1-x^2}+\sqrt{(-x+2x^2{)}^2+(1-x^2)(1-2x{)}^2}\\ & =\sqrt{2+2x}+\sqrt{1-4x+4x^2}\\ & =\sqrt{2+2x}+|1-2x|.\end{array}$$Let $u=\sqrt{2+2x}.$ Then $u^2=2+2x,$ so $$\sqrt{2+2x}+|1-2x|=u+|3-u^2|.$$Since $-1\le x\le 1,$ $0\le u\le 2.$

If $0\le u\le \sqrt{3},$ then $$u+|3-u^2|=u+3-u^2=\frac{13}{4}-{\left(u-\frac{1}{2}\right)}^2\le \frac{13}{4}.$$Equality occurs when $u=\frac{1}{2},$ or $x=-\frac{7}{8}.$

If $\sqrt{3}\le u\le 2,$ then $$u+u^2-3={\left(u+\frac{1}{2}\right)}^2-\frac{13}{4}\le {\left(2+\frac{1}{2}\right)}^2-\frac{13}{4}=3<\frac{13}{4}.$$Therefore, the maximum value is $\boxed{\frac{13}{4}}.$