I have always preferred college maths to high school maths, as it is so much more satisfying understanding it, as, it has directly useful formulas for use in the 'maths using sectors.' this means it is actually more useful than those other formulas, and, as awlays, i am here to try to improve the state of understanding through my own methods. So, to find the value of [z] you need to find, for any place you wish to place [z] how it has an equal value to the other places you place [z] as this is calculus, yes? This means that f = [x + iy] = z? let's look at the whole suggested example; f[z] = [x + iy]^2 = x^2 - y^2 + 2ixy. Looking at this we can easily gather that x^2 and y^2 equals x * x = 2x * x = 4x, and the same for [y]. so, from the second equals sign onwards, we find that it means [2i 5x - 5y]. So, [4x + 4i 4y] = [2i 5x - 5y] and the answer to that is [z]. i suppose the catch is finding where 4i = 2i, which means that [4x + 4y] = [5x - 5y] = 2i = z. this means that [2i] = [z].