A complex number z satisfies the equation (z)/(z+2) = 2 - i. Express z in the form a+bi, where a and b are real constants.

Here are the solutions for Question 7 and Question 8 (i). Question 7 (i): A complex number z satisfies the equation (z)/(z+2) = 2 - i. Express z in the form a+bi, where a and b are real constants. Step 1: Rearrange the equation to solve for z. z = (2 - i)(z + 2) z = 2z + 4 - iz - 2i Step 2: Group terms containing z on one side and constant terms on the other. z - 2z + iz = 4 - 2i -z + iz = 4 - 2i z(-1 + i) = 4 - 2i Step 3: Isolate z by dividing. z = (4 - 2i)/(-1 + i) Step 4: Multiply the numerator and denominator by the conjugate of the denominator. The conjugate of -1+i is -1-i. z = (4 - 2i)/(-1 + i) × (-1 - i)/(-1 - i) z = ((4)(-1) + (4)(-i) + (-2i)(-1) + (-2i)(-i))/((-1)^2 + (1)^2) z = (-4 - 4i + 2i + 2i^2)/(1 + 1) Step 5: Substitute i^2 = -1 and simplify. z = (-4 - 2i + 2(-1))/(2) z = (-4 - 2i - 2)/(2) z = (-6 - 2i)/(2) z = -3 - i The complex number z in the form a+bi is -3 - i. --- Question 7 (ii): The vector equations of two lines are given by r = 3i + 6j + k + (2i - 3j - k) and r = 3i - j + 4k + (i - 2j + k). Part (a): Find the vector normal to the plane containing both lines. Step 1: Identify the direction vectors of the two lines. The direction vector of the first line is d_1 = 2 \\ -3 \\ -1 . The direction vector of the second line is d_2 = 1 \\ -2 \\ 1 . Step 2: The normal vector to the plane containing both lines is perpendicular to both direction vectors. This can be found by taking the cross product of d_1 and d_2. n = d_1 × d_2 = 2 \\ -3 \\ -1 × 1 \\ -2 \\ 1 n = (-3)(1) - (-1)(-2) \\ (-1)(1) - (2)(1) \\ (2)(-2) - (-3)(1) n = -3 - 2 \\ -1 - 2 \\ -4 + 3 n = -5 \\ -3 \\ -1 The vector normal to the plane containing both lines is -5i - 3j - k. Part (b): Obtain the Cartesian equation of the plane containing both lines. Step 1: Use the normal vector n = -5 \\ -3 \\ -1 from Part (a). Step 2: Choose a point on the plane. We can use the position vector of a point from either line. From the first line, a point on the plane is a = 3 \\ 6 \\ 1 . Step 3: The Cartesian equation of a plane is given by r · n = a · n. Let r = x \\ y \\ z . x \\ y \\ z · -5 \\ -3 \\ -1 = 3 \\ 6 \\ 1 · -5 \\ -3 \\ -1 -5x - 3y - z = (3)(-5) + (6)(-3) + (1)(-1) -5x - 3y - z = -15 - 18 - 1 -5x - 3y - z = -34 Step 4: Multiply the entire equation by -1 to express it with positive leading coefficients (optional, but standard practice). 5x + 3y + z = 34 The Cartesian equation of the plane containing both lines is 5x + 3y + z = 34. --- Question 8 (i): The equation of a line is 3y = x + 5 and the equation of a circle is x^2 + y^2 - 6x - 2y - 15 = 0. Show that the line is a chord to the circle. Step 1: Express x in terms of y from the line equation. x = 3y - 5 Step 2: Substitute this expression for x into the circle equation. (3y - 5)^2 + y^2 - 6(3y - 5) - 2y - 15 = 0 Step 3: Expand and simplify the equation. (9y^2 - 30y + 25) + y^2 - (18y - 30) - 2y - 15 = 0 9y^2 - 30y + 25 + y^2 - 18y + 30 - 2y - 15 = 0 Combine like terms: (9y^2 + y^2) + (-30y - 18y - 2y) + (25 + 30 - 15) = 0