(b) We have that p (z) = z 3 + z 2 + z and z0 = Thus (a) We have that p (z) = z 2 and z0 = i, thus p (z0 ) = p (i) = Thus p (z) h (z) = z I.1.9 Find the polynomial h (z) in the preceding exercise for the following choices of p (z) and z0 (a) p (z) = z 2 and z0 = i (b) p (z) = z 3 + z 2 + z and z0 = 1 (c) p (z) = 1 + z + z 2 + + z m and z0 = 1 Solution From the preceding exercise we have p (z) = (z Since deg q (z) n 2, deg r (z) Plug in z0, get p (z0 ) = c.

By using the induction hypothesis, we can assume that r (z) = q (z) (z where deg q (z) Proof by induction on degree n of p (z), set p (z) = an z n + an 1 z n where an 6= 0. = (z z0 ) bn 1 z n 1 + bn 2 z n 2 + bn 3 z n 3 + : : : + b2 z 2 + b1 z + b0 +p (z0 ) = bn 1 z n + bn 2 z n 1 + bn 3 z n 2 + : : : + b2 z 3 + b1 z 2 + b0 z + p (z0 ) bn 1 z0 z n 1 bn 2 z0 z n 2 bn 3 z0 z n 3 : : : b2 z0 z 2 b1 z0 z b0 z0 = bn 1 z n +(bn 2 bn 1 z0 ) z n 1 +(bn 3 bn 2 z0 ) z n 2 +(bn 4 bn 3 z0 ) z n 3 + + (b2 b3 z0 ) z 3 + (b1 b2 z0 ) z 2 + (b0 b1 z0 ) z + p (z0 ) b0 z0 : Equate and solve for the bj ´s in terms of aj ´s. We equate coe¢ cients in the polynomial identity p (z) = (z p (z0 ), and get an z n + an 1 z n In particular, if p (z0 ) = 0, then p (z) = (z z0 ) h (z). Show that there is a polynomial h (z) of degree n 1 such that p (z) = (z z0 ) h (z)+p (z0 ). I.1.8 Let p (z) be a polynomial of degree n 1 and let z0 2 C. When z0 = 0 z1 = 1, this is the line Re z = 21. Which is the line bisecting the two points. If z0 = 0, and z1 = 1, we have 2ġ 0, and when = 2, When = 21 we have a circle of radius 23 centered at 3 4 2 we have a circle of radius 3 centered at 3 0. Now complete the squares for both the x and y terms. Thus our equation becomes x0 )2 + (yĮxpanding the squares, and grouping terms, we have 1 Let z = x + iy, z0 = x0 + iy0 and z1 = x1 + iy1. The solutions set of the equation above remains the same if we square both sides, z0 j2 =

What happens when = 1? Solution Recall that a circle in R2 centered at (a b) with radius r is given by the equation a)2 + (y Sketch it for = 12 and = 2, with z0 = 0 and z1 = 1. Show that the set of z satisfying jz z0 j = jz z1 j is a circle. (f) Horizontal strip, 0 0 (i) Half plane y 0, Re (i (x + iy) + 2) > 0 ,

Complex analysis gamelin solutions