${\text{If x=a+ib is a complex number such that x}}^2=3+4i and x^3=2+11i where i=\sqrt{-1}, then (a+b)=$

$$\begin{gathered} x=\frac{x^3}{x^2}=\frac{2+11i}{3+4i} \\ =\frac{2+11i}{3+4i}\times \frac{3-4i}{3-4i} \\ =\frac{6-8i+33i-44i^2}{9-16i^2} \\ =\frac{50+25i}{25} \\ \Rightarrow 2+i=a+ib \\ \therefore a=2, b=1 \\ \Rightarrow a+b=2+1=3 \end{gathered}$$