(FP1) Given k = q + 3i and z = w^2 - 8w* - 18q^2 i, and if w is purely imaginary, show that there is only one possible non-zero value of z

k = q + 3i and z = k² - 8k* - 18q² i. Plug k into w to give: z = (q + 3i)²- 8*(q - 3i) - 18q² i. k means the compliment of k which just means that the sign before the imaginary value is switched to negative. Multiplying this out and splitting into value with i (i.e. imaginary) and values without i (i.e. real) gives: z = [q²- 8q -9] + i[-18q²+6q+24]. The question says that the real part is zero, meaning q²- 8q -9 = 0. Splitting this into brackets gives the values for q, (q-9)(q+1) meaning q = 9 or -1. This will help us find the value of z that is non-zero. Do this by plugging these possible q values into what we have left of z which is z = 0 + i[-18q²+6q+24]. This gives w = -1380i for q = 9 and w = 0 for q = -1 meaning q = 9 is the only possible value that can make w non-zero and the value w is is w= -1380i.