(a) Substitute \(u = a + ib\) and \(w = c + id\). The conjugate of \(u + w\) is \((a + c) + i(b + d)\)^* = (a + c) - i(b + d)\). The conjugate of \(u\) is \(a - ib\) and the conjugate of \(w\) is \(c - id\). Therefore, \(u^* + w^* = (a - ib) + (c - id) = (a + c) - i(b + d)\). Thus, \((u + w)^* = u^* + w^*\).

(b) Let \(z = x + iy\). Then \((z + 2 + i)^* = (x + 2 + i(y + 1))^* = x + 2 - i(y + 1)\). Substitute into the equation: \(x + 2 - i(y + 1) + (2 + i)(x + iy) = 0\). Simplify: \(x + 2 - i(y + 1) + (2x - y) + i(2y + x) = 0\). Equate real and imaginary parts to zero: \(3x - y + 2 = 0\) and \(x + y - 1 = 0\). Solve these equations: From \(x + y - 1 = 0\), we get \(y = 1 - x\). Substitute into \(3x - y + 2 = 0\): \(3x - (1 - x) + 2 = 0\) gives \(4x + 1 = 0\), so \(x = -\frac{1}{4}\). Then \(y = 1 - (-\frac{1}{4}) = \frac{5}{4}\). Therefore, \(z = -\frac{1}{4} + \frac{5}{4}i\).

(a) Substitute u = 1 + 2i into the polynomial p(x) = 2x^3 + ax^2 + 4x + b. Calculate u^2 = -3 + 4i and u^3 = -11 - 2i. Substitute these into the polynomial and equate real and imaginary parts to zero:

\(-18 - 3a + b = 0\)

\(4 + 4a = 0\)

\(Solving these equations gives a = -1 and b = 15.\)

(b) Since the coefficients are real, the complex roots occur in conjugate pairs. Therefore, the second root is 1 - 2i.

(c) The polynomial can be factored as (x^2 - 2x + 5)(2x + 3).

(d)(i) The region is a circle centered at 1 + 2i with radius √5, intersecting the line y = x in the first quadrant. Shade the region satisfying |z - u| ≤ √5 and arg z ≤ 1/4 π.

(d)(ii) The least value of Im z is 2 - √5.

(a) Substitute \(z = -1 + \sqrt{2}i\) into the equation:

\((-1 + \sqrt{2}i)^4 + 3(-1 + \sqrt{2}i)^2 + 2(-1 + \sqrt{2}i) + 12 = 0\).

Calculate \((-1 + \sqrt{2}i)^2 = 1 - 2\sqrt{2}i - 2 = -1 - 2\sqrt{2}i\).

Calculate \((-1 + \sqrt{2}i)^4 = ((-1 + \sqrt{2}i)^2)^2 = (-1 - 2\sqrt{2}i)^2 = 1 + 4\cdot2 - 4\sqrt{2}i = 9 - 4\sqrt{2}i\).

Substitute back: \(9 - 4\sqrt{2}i + 3(-1 - 2\sqrt{2}i) + 2(-1 + \sqrt{2}i) + 12 = 0\).

Simplify: \(9 - 4\sqrt{2}i - 3 - 6\sqrt{2}i - 2 + 2\sqrt{2}i + 12 = 0\).

Combine terms: \(16 - 8\sqrt{2}i = 0\), which is true.

(b) The conjugate root is \(-1 - \sqrt{2}i\).

Find a quadratic factor: \((z + 1 - \sqrt{2}i)(z + 1 + \sqrt{2}i) = z^2 + 2z + 3\).

Divide \(z^4 + 3z^2 + 2z + 12\) by \(z^2 + 2z + 3\) to find the other quadratic factor:

\(z^4 + 3z^2 + 2z + 12 = (z^2 + 2z + 3)(z^2 - 2z + 4)\).

Solve \(z^2 - 2z + 4 = 0\) using the quadratic formula:

\(z = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} = \frac{2 \pm \sqrt{-12}}{2} = 1 \pm \sqrt{3}i\).

The other roots are \(1 + \sqrt{3}i\) and \(1 - \sqrt{3}i\).

Let the square root of u be a + ib. Then, \\((a + ib)^2 = u = 10 - 4\sqrt{6}i \\).

Expanding the left side, we have \\(a^2 - b^2 + 2abi = 10 - 4\sqrt{6}i \\).

Equating real and imaginary parts, we get:

1. \\(a^2 - b^2 = 10 \\)

2. \\(2ab = -4\sqrt{6} \\)

From equation 2, \\(ab = -2\sqrt{6} \\).

Substitute \\(b = \frac{-2\sqrt{6}}{a} \\) into equation 1:

\\(a^2 - \left(\frac{-2\sqrt{6}}{a}\right)^2 = 10 \\)

\\(a^2 - \frac{24}{a^2} = 10 \\)

Multiply through by \\(a^2 \\) to clear the fraction:

\\(a^4 - 10a^2 - 24 = 0 \\)

Let \\(x = a^2 \\), then \\(x^2 - 10x - 24 = 0 \\).

Solving this quadratic equation using the quadratic formula:

\\(x = \frac{10 \pm \sqrt{100 + 96}}{2} = \frac{10 \pm \sqrt{196}}{2} = \frac{10 \pm 14}{2} \\)

Thus, \\(x = 12 \\) or \\(x = -2 \\) (discard \\(x = -2 \\) as it is not possible for \\(a^2 \\)).

So, \\(a^2 = 12 \\), giving \\(a = \pm \sqrt{12} = \pm 2\sqrt{3} \\).

Substitute back to find \\(b \\):

For \\(a = 2\sqrt{3} \\), \\(b = \frac{-2\sqrt{6}}{2\sqrt{3}} = -\sqrt{2} \\).

For \\(a = -2\sqrt{3} \\), \\(b = \frac{-2\sqrt{6}}{-2\sqrt{3}} = \sqrt{2} \\).

Thus, the square roots are \\(\pm (2\sqrt{3} - \sqrt{2}i) \\).

(a) Multiply both sides by \(a + 2i\) to clear the fraction:

\(2 + 3ai = \lambda(2 - i)(a + 2i)\).

Expand the right-hand side:

\(2 + 3ai = \lambda(2a + 4i - ai - 2)\).

Separate real and imaginary parts:

Real: \(2 = \lambda(2a - 2)\)

Imaginary: \(3a = \lambda(4 - a)\).

From the real part: \(2 = \lambda(2a - 2) \Rightarrow \lambda = \frac{2}{2a - 2}\).

From the imaginary part: \(3a = \lambda(4 - a) \Rightarrow \lambda = \frac{3a}{4 - a}\).

Equating the two expressions for \(\lambda\):

\(\frac{2}{2a - 2} = \frac{3a}{4 - a}\).

Cross-multiply and simplify:

\(2(4 - a) = 3a(2a - 2)\).

\(8 - 2a = 6a^2 - 6a\).

Rearrange to form a quadratic equation:

\(6a^2 - 4a - 8 = 0\).

Divide by 2:

\(3a^2 + 4a - 4 = 0\).

(b) Solve the quadratic equation \(3a^2 + 4a - 4 = 0\) using the quadratic formula:

\(a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3, b = 4, c = -4\).

\(a = \frac{-4 \pm \sqrt{16 + 48}}{6}\).

\(a = \frac{-4 \pm \sqrt{64}}{6}\).

\(a = \frac{-4 \pm 8}{6}\).

\(a = \frac{4}{6} = \frac{2}{3}\) or \(a = \frac{-12}{6} = -2\).

For \(a = -2\), substitute back to find \(\lambda\):

\(3(-2) = \lambda(4 + 2) \Rightarrow -6 = 6\lambda \Rightarrow \lambda = -1\).

For \(a = \frac{2}{3}\), substitute back to find \(\lambda\):

\(3\left(\frac{2}{3}\right) = \lambda\left(4 - \frac{2}{3}\right) \Rightarrow 2 = \lambda\left(\frac{10}{3}\right) \Rightarrow \lambda = \frac{3}{5}\).