No you can not. Complex roots appear as conjugates. if a root is complex so is its conjugate. so either the roots are real or are both coplex.
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Yes, but in this case, the coefficients of the polynomial can not all be real.
The "i" part is the imaginary square root of negative one.
The fourth square root is the 16th root of a number. On a computer, to find the 16th root of a number, say 5.6, enter 5.6^(1/16). If the number you start with is positive, you will have 2 real roots (one positive and one negative) and 14 complex roots. If it is negative, you will have 16 complex roots.
Provided some of the coefficients and the constant were imaginary (complex) as well, yes. For example, (x + 2)(x - 3+i) has both a real and an imaginary root, and has coefficients that are also both real and imaginary, i.e. 1, -1+i, and -6+2i.
A positive real number, such as 17, has two square roots. One is the one your calculator gives you, if you use the square root function. The other is the same number, with a minus sign in front. None of these has an imaginary part. There are no additional complex roots that have a non-zero imaginary part.
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There is not enough information. You can't calculate one root on the basis of another root. HOWEVER, if we assume that all the polynomial's coefficients are real, then if the polynomial has a complex root, then the complex conjugate of that root (in this case, 4 - 17i) must also be a root.
One is always one of the roots. There are n-l other complex roots, evenly spaced around the unit circle in the complex plane.
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