[tex]^\Big{16+5i)}\dfrac{15+6i}{16-5i} = \dfrac{(16+5i)(15+6i)}{(16+5i)(16-5i)} = \dfrac{16\cdot 15+96i+75i+30i^2}{16^2-(5i)^2} = \\ \\ = \dfrac{240+171i+30\cdot(-1)}{256-(25\cdot (-1))} = \dfrac{240-30+171i}{256+25} = \dfrac{210+171i}{281} = \\ \\ =\dfrac{210}{281}+\dfrac{171}{281}i \\ \\\dfrac{15+6i}{16-5i} = a+bi \Rightarrow \dfrac{210}{281}+\dfrac{171}{281}i = a+bi \Rightarrow
\left\{ \begin{array}{c} a=\dfrac{210}{281} \\\\ b= \dfrac{171}{281} \end{array} \right |[/tex]
