If z = a + ib lies in third quadrant then a and b both are less than zero

a² – b² < 0 and ab > 0 because a² + b² is always greater than zero

(a – b)(a + b) < 0

Here a and b both are less than zero that means (a + b) is always less than zero

So, a – b > 0 ⇒ a > b

Then, final answer is b < a < 0