Solution. At \(x_0 = 2\) synthetic division gives \(P(2) = -5\) (row \(1, 0, 0, -5\)) and \(P'(2) = 4\) (row \(1, 2, 4\)), so \(x_1 = 2+\frac 54 = 3.25\). Continuing, \(x_2 = 2.811037\), \(x_3 = 2.697990\), \(x_4 = 2.690677\), \(x_5 = 2.690647\). Deflating by \(\hat x = 2.6906474\) leaves \(Q(x) = x^2+0.6906474x+1.8582888\), whose roots are \begin {align*} x &= \frac {-0.6906474\pm \sqrt {0.4769938-7.4331552}}{2}\\ &= -0.3453237\pm 1.3187268i . \end {align*}

Solution. \(u = e^x\cos y\), \(v = e^x\sin y\). Then \(u_x = e^x\cos y = v_y\) and \(u_y = -e^x\sin y = -v_x\). So \(f'(z) = u_x+iv_x = e^x\cos y+ie^x\sin y = e^z\), as expected.

Solution. \(u = x^2+y^2\), \(v = 0\): \(u_x = 2x\), \(v_y = 0\), \(u_y = 2y\), \(v_x = 0\). The equations \(2x = 0\) and \(2y = 0\) hold only at the origin, so \(f\) is not differentiable anywhere else, and \(f'(z_n)\) in Newton’s formula does not exist. (The equation \(|z|^2 = 1\) is satisfied by the whole unit circle, a curve rather than isolated points: one real equation in two real unknowns.)

Solution. \(z^3 = 8\) gives \(z = 2e^{2\pi ik/3}\): \(2\) and \(-1\pm \sqrt 3i = -1\pm 1.7321i\). With \(z_0 = -1+2i\), \(z_0^2 = -3-4i\) and \(z_0^3 = 11-2i\). So \[z_1 = z_0-\frac {z_0^3-8}{3z_0^2} = -1+2i-\frac {3-2i}{-9-12i} = -1+2i-\frac {(3-2i)(-9+12i)}{225} = -1+2i-\frac {-3+54i}{225},\] that is, \(z_1 = -0.98667+1.76i\), already close to the root \(-1+1.7321i\).

Solution. \(f(0) = -1\), \(f(1) = -1\), \(f(2) = 5\). Then \(h_1 = h_2 = 1\), \(\delta _1 = 0\), \(\delta _2 = 6\), \(a = \frac {6-0}{2} = 3\), \(b = 6+1\cdot 3 = 9\), \(c = 5\). So \(b^2-4ac = 81-60 = 21\), and with the sign of \(b\), \[x_3 = 2-\frac {10}{9+\sqrt {21}} = 2-\frac {10}{13.5826} = 1.26376 .\] The root is \(1.32472\), so one step from three rough guesses has already reduced the error to \(0.06\).

Solution. By Corollary 3.1.3, \(P\) has \(n\) roots counted with multiplicity, \(n\) odd. By Theorem 3.1.4 the non-real roots come in conjugate pairs, with equal multiplicities (apply the theorem to the factors \((x-z)^m\)), so they account for an even number of the \(n\) roots. Since \(n\) is odd, at least one root is real. (This also follows from the Intermediate Value Theorem, since \(P(x)\) takes both signs for large \(|x|\).)