Please, do not give me answers to these questions, I just want to have some hints/tips to point me in the correct direction, or if you're so inclined, you can use another similar/relevant to help me understand this concept. Thanks! 1. The problem statement, all variables and given/known data a.) Given that z/(z + 2) = 2-i, z ∈ ℂ, find z in the form a+bi b.) Given that (a + bi)2 = 3 + 4i obtain a pair of simultaneous equations involving a and b. Hence find the two square roots of 3 + 4i. 2. Relevant equations None, should be basic logic/algebra. 3. The attempt at a solution a.) z = (z + 2)(2 - i) = 2z - zi + 4 - 2i = 2z + 4 - zi - 2i Is it then right to say that: z = a + bi ∴ a = 2z + 4, b = -z - 2 ? However, that's not very helpful. I could plug z into the original equation and try to do some long algebra: 2 - i = z/(z + 2) = (2z + 4 - zi - 2i)/(2z + 4 - zi - 2i + 2) 2 - i = (2z + 4 - zi - 2i) / (2z + 6 - zi - 2i) 2 - i = [2z + 4 - (z + 2)i] / [2z + 6 - (z - 2)i] I could at this point multiply by the conjugate to get reals out of the denominator but that's a lot of work and I feel like I've already missed the point of this, as the question isn't worth too many points. So, the most meaningful information I've presented relevant to this question (I think) is: z = 2z + 4 - zi - 2i 2 - i = [2z + 4 - (z + 2)i] / [2z + 6 - (z - 2)i] b.) (a + bi)2 = (a + bi)(a + bi) = a2 + 2abi + b2 a2 + 2abi + b2 = 3 + 4i a2 + b2 + 2abi = 3 + 4i Hence, is it right to say that: a2 + b2 = 3, 2ab = 4, so ab = 2 ? This would mean that a = 2/b and likewise, b = 2/a I can try substituting this into the other equation: (2/b)2 + b2 = 3 (4/b2) + b2 = 3 4 + b4 = 3b2 4 = 3b2 - b4 Unfortunately, from here I'm completely lost, even though it's basic algebra, I can't quite put my finger on the next move. In fact, I'm not even sure if I've been doing anything remotely correct to this point. Thank you all for any help you can offer me!