No, the sum of all three roots is 0.
deadmotor, you said the 2 roots are 1-2i and 1+2i because they are pair conjugate, but notice the equation z^3 + z + 10 = 0, sum of roots is zero.
if a, b, and c are the roots, then a + b + c = 0,
a + ( 1 + 2i) + (1 - 2i) = 0, surely you know the third root, right? which is a . . . why not evaluate it?
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It can be factored by grouping,
z^3 + z + 10 = 0,
z^3 + 2z^2 - 2z^2 - 4z + 4z + z + 10 = 0,
z^2(z + 2) - 2z(z + 2) + 5(z + 2) = 0,
(z + 2)(z^2 - 2z + 5) = 0.
surely you know how to use the quadratic formula to find the rest of the roots . . .
Another possibility: if 1 + 2i and 1 - 2i are roots then
z - (1 + 2i) and z - (1 - 2i) are factors
So (z-1-2i)(z-1+2i) is a factor
Simply this and you will get a quadratic. then you need one more factor, which should now be obvious (there are not many choices now that you have the 2nd power of z).
From there you should get the third root.