[tex]\dfrac{2+6i}{2-4i}\\\\\\=\dfrac{2(1+3i)}{2(1-2i)}\\\\\\=\dfrac{1+3i}{1-2i}\\\\\\=\dfrac{(1+3i)(1+2i)}{(1-2i)(1+2i)}\\\\\\=\dfrac{1+2i+3i+6i^2}{1-(2i)^2}\\\\\\=\dfrac{1+5i-6}{1+4}~~~~;[i^2=-1]\\\\\\=\dfrac{5i-5}{5}\\\\\\=\dfrac{5(i-1)}5\\\\=i-1\\\\=-1+i\\\\\text{It is in a form of a +bi so,}~ a = -1~ \text{and}~ b = 1[/tex]
