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:
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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
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How then do I find all the solutions for this:
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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.
