1. The problem statement, all variables and given/known data Find all pairs of values a and b that satisfy (a + bi)2 = 48 + 14i 2. The attempt at a solution I managed to solve it, but it took a while and I was wondering if there is an easier/quicker way. What I did was expanded (a + bi)2 into (a2 - b2) + 2abi From there, I can now say: a2 = b2 = 48 2ab = 14 I then got a value for b from the 2ab = 14 equation (b = 7/a) and substituted this into the a2 - b2 = 48 equation, and I got this polynomial: a4 - 48a2 - 49 = 0 To solve the polynomial, I used the factor theorem and found that (a + 7) and (a - 7) are two factors, then I used synthetic division to divide the polynomial by these two factors (a + 7) and (a - 7), to get the quadratic a2 + 1 = 0 Solving this quadratic resulted in the last two factors being (a - i) and (a + i) So now I have the factorised polynomial (a - 7)(a + 7)(a - i)(a + i) and can get the values for a: a = 7, -7, i, -i Now I substitute those values into the equation used earlier (a2 - b2 = 48) to get the values for b, resulting in these pairs: a = 7, b = ±1 a = -7, b = ±1 a = i, b = ±7i a = -i, b = ±7i And from there I substitute all pairs into the original equation to check their validity and remove the invalid ones. It's quite a procedure for a relatively simple problem, is there any quicker way?