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Let $z=a+bi$ and $w=c+di,$ where $a,$ $b,$ $c,$ and $d$ are complex numbers. Then from $|z|=1,$ $a^2+b^2=1,$ and from $|w|=1,$ $c^2+d^2=1.$ Also, from $z\overline{w}+\overline{z}w=1,$ $$(a+bi)(c-di)+(a-bi)(c+di)=1,$$so $2ac+2bd=1.$

Then $$\begin{array}{rl}(a+c{)}^2+(b+d{)}^2& =a^2+2ac+c^2+b^2+2bd+d^2\\ & =(a^2+b^2)+(c^2+d^2)+(2ac+2bd)\\ & =3.\end{array}$$The real part of $z+w$ is $a+c,$ which can be at most $\sqrt{3}.$ Equality occurs when $z=\frac{\sqrt{3}}{2}+\frac{1}{2}i$ and $w=\frac{\sqrt{3}}{2}-\frac{1}{2}i,$ so the largest possible value of $a+c$ is $\boxed{\sqrt{3}}.$