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Sorry for taking so long to reply but thank you nonetheless. I had not noticed that at the time I posted my question obviously and I am often easily puzzled when asked to find "all solutions" as I expect there to be imaginary ones in a question such as this. Thank you again and farewell.

I probably shouldn't have posted this under Calculus should I?

Greetings, I'm not likely to post again but I was having trouble with some maths and so I ran a search and found these forums in the hope somebody could answer my question. I understand this: " Given z = root(-9) (note: that's root of minus nine.) then z2 = (-9) .....................(1) (note: that's z squared) Let z = r cis o..................(2) (where o is phi) then z2 = r2 cis 2o Also, -9 = 9 cis p (where p is pi) equation (1) can be rewritten as: r2 cis 2o = 9 cis p r2 (cos 2o + isin 2o) = 9 (cos p+ isin p) hence, r2 = 9 equating the terms inside the brackets gives: 2o = p + 2kp (where k is an integer to account for coterminal angel solutions) Solving: r =root(9) = 3 o= (p/2) + (2kp/2) = (p/2) + kp (where k is an integer) When k = 0, o = (p/2) When k = 1, o = (p/2) + 1(p) = (3p/2) which is coterminal with (-p/2) Substituting these results in (2) gives: z = r cis o z = 3 cis (p/2) or z = 3 cis (-p/2) hence z = 3i or -3i " How then do I find all the solutions for this: " Given z(z+5)=176 find all solutions for z. The real ones are easy: x(x+5) = 176 x2 + 5x = 176 x2 + 5x - 176 = 0 Hence x = {-5 plus or minus root[25-4(1)(-176)]/2(1)} = 11 or -16 Are there any imaginary solutions or was the wording of the question just a trick to get us thinking? If there are any could somebody please show me how to find them? Thank you all.