\[w_{2}(z)={\mathrm{e}}^{(b-c+1)\pi\mathrm{i}}\frac{\Gamma\left(2-c\right)\Gamma%
\left(a\right)}{\Gamma\left(a+b-c+1\right)\Gamma\left(1-b\right)}w_{3}(z)+{%
\mathrm{e}}^{(b-c)\pi\mathrm{i}}\frac{\Gamma\left(2-c\right)\Gamma\left(a%
\right)}{\Gamma\left(a-b+1\right)\Gamma\left(b-c+1\right)}w_{5}(z),\]