Question 863182
question a seems more complicated than question b.  I'm assuming you meant to indicate exponents in question a, as {{{z^3-5z^2+11z-15}}}.  Your two complex roots with imaginary components are 1-2i and <b>1+2i</b>.  Each of these gives a factor which can be multiplied to give a quadratic factor for p(z).  Polynomial division will give you the other needed factor of p(z).  If you need no help with that and can understand the process, then you would not likely have any confusion about question b...



Question b:
There is a stray mark above the plus sign, but as the quadratic equation which this appears to be, find z using factorization.
{{{z^2+4z+4=(z+2)^2=0}}}.
{{{highlight(z=-2)}}}.
Yes negative 2 IS a complex number.  It is also a real number.  You can express it as {{{highlight(-2-0*i)}}} and {{{highlight(-2+0*i)}}}.