Seems tough....But its Real Easy !!

$We\quad know\quad that,\quad Odd+Even=Odd\\ \\ Then\quad definitely,\quad a\quad must\quad be\quad odd\quad \& \quad b\quad must\quad be\quad even\quad (OR)\quad a\quad must\quad be\quad even\quad \& \quad b\quad must\quad be\quad odd\\ \\ Let\quad us\quad take,\quad a\quad is\quad odd\quad and\quad b\quad is\quad even\quad \\ Then,\\ \quad \quad \quad \quad \quad \quad { (-1) }^{ odd\quad times }+{ (-1) }^{ even\quad times }\\ \quad \quad \quad \quad \quad (-1)+1\quad =0$

An alternating way to solve this is to view $b = 2015-a$ , then we will get $(-1)^a+\frac{(-1)^{2015}}{(-1)^a} = (-1)^a-(\frac{1}{-1})^{2015} = (-1)^a-(-1)^a = 0$