If $13{\mathrm{e}}^{{\mathrm{itan}}^{-1\left(\frac{5}{12}\right)}} = \mathrm{a} + \mathrm{ib}$ then the ordered pair (a,b) =

$$\begin{gathered} \mathrm{Let} \mathrm{\theta }={\tan }^{-1}\frac{5}{12}\Rightarrow \tan \mathrm{\theta }=\frac{5}{12} \therefore \sin \mathrm{\theta }=\frac{5}{13}, \cos \mathrm{\theta }=\frac{12}{13} \\ \Rightarrow 13 {\mathrm{e}}^{\mathrm{i\theta }}=\mathrm{a}+\mathrm{ib} \\ \Rightarrow 13(\cos \mathrm{\theta }+\mathrm{isin} \mathrm{\theta })=\mathrm{a}+\mathrm{ib} \\ \therefore (\mathrm{a}+\mathrm{ib})=13\left(\frac{12}{13}+\mathrm{i} \frac{5}{13}\right) \\ \Rightarrow a+\mathrm{ib}=12+5\mathrm{i} \\ \Rightarrow (\mathrm{a},\mathrm{b})=(12,5) \end{gathered}$$