Question 1 – Factorize the following

In this question, we have eight different parts. Each part has an expression that we need to factorize. Factorization means breaking it into smaller parts that multiply together to give the original expression.

For parts (i) to (iv), we have expressions like a² + 4b². These are sums of squares. Normally, a sum of squares does not factorize with real numbers, but with complex numbers it does. We use the formula a² + b² = (a + ib)(a – ib). So we will write each expression in this form and factorize.

For parts (v) to (viii), we have quadratic expressions like z² + 6z + 13. These are in the form of a standard quadratic. To factorize them, we will first find the roots of the quadratic using the quadratic formula. Once we have the roots, we can write the expression as (z – root1)(z – root2). This will be the factorized form.

Question 2 – Factorize polynomials into linear factors

In this question, we have to factorize polynomials completely into linear factors. Linear factors mean factors of the form (z – a) where a can be a real or complex number.

For some parts, the polynomial is simple like z² + 8. We will write it as z² – (–8) and then use the formula a² – b² = (a – b)(a + b). But since –8 is negative, the square root will involve i.

For higher degree polynomials like z⁴ – 16, we will first write it as (z²)² – (4)², which is a difference of squares. Then we factor it further. Each factor may break down again until all factors are linear.

For cubic polynomials like z³ – 2z² + 16z – 32, we will first find one root by trying possible values. Once we find one root, we divide the polynomial by (z – that root). Then we get a quadratic, which we can factorize further using the quadratic formula.

Question 3 – Find the roots and express as product of linear factors

In this question, we are given an equation like z² – 72z – 144 = 0. First, we solve this equation to find the roots. We use the quadratic formula. The roots may be real or complex.

After we find the roots, we write the polynomial as a product of linear factors. If the roots are r1 and r2, then the polynomial becomes (z – r1)(z – r2). This is the answer.

For higher degree polynomials, we will find all the roots first. Then we write the polynomial as (z – r1)(z – r2)(z – r3) and so on.

Question 4 – Solve complex quadratic equations by completing square method

In this question, we have six quadratic equations. All of them are of the form az² + bz + c = 0. We have to solve them using the completing square method, not the quadratic formula.

The completing square method works like this: First, we make the coefficient of z² equal to 1 by dividing the whole equation by a. Then we move the constant term to the other side. Then we take half of the coefficient of z, square it, and add it to both sides. This makes the left side a perfect square. Then we take the square root of both sides. This gives us two equations. We solve them to find the two roots.

The roots will be complex numbers because the discriminant will be negative in most cases.

Question 5 – Solve the following equations

In this question, we have six different equations. They are not all quadratic. Some are cubic, some have higher powers. We have to solve each equation for z.

For equations like 2z⁴ – 32 = 0, we first take 2 common, then we get z⁴ – 16 = 0. Then we write it as (z²)² – (4)² = 0. This is a difference of squares. We factor it step by step until we get linear factors. Then each factor is set to zero and we solve for z.

For equations like z³ + z² + z + 1 = 0, we will first try to find one root by checking simple values like 1, –1, i, –i. Once we find one root, we divide the polynomial and then solve the remaining quadratic.

For cubic equations like 5z³ – 5z = 0, we first take 5z common. Then we get 5z(z² – 1) = 0. Then we factor z² – 1 further. Then set each factor to zero and find all roots.

Question 6 – Find a polynomial of degree 3 with given zeros

In this question, we are told that the polynomial has degree 3. Its zeros are given: 3, –2i, and 2i. Zeros are the values of z for which the polynomial becomes zero.

If a polynomial has a zero at z = a, then (z – a) is a factor. So here the factors are (z – 3), (z + 2i), and (z – 2i). We multiply these three factors together to get the polynomial. But we also have a condition: P(1) = 20. This means when we put z = 1 in the polynomial, the value should be 20. So first we multiply the factors to get a polynomial with some constant. Then we use P(1) = 20 to find the value of that constant.

Question 7 – Find a polynomial of degree 4 with given zeros

This question is similar to question 6. Here the polynomial has degree 4. The zeros given are 2i, –2i, 1, and –1.

The factors will be (z – 2i), (z + 2i), (z – 1), and (z + 1). We multiply all these factors together. First, multiply the pairs that are conjugates. (z – 2i)(z + 2i) = z² + 4. Similarly, (z – 1)(z + 1) = z² – 1. Then multiply these two results. This gives us the polynomial with some constant. Then we use the condition P(2) = 240 to find the value of the constant.

Question 8 – Find a polynomial of degree 4 with given zeros

This question is also the same type. The polynomial has degree 4. The zeros given are 4, –4, 1 + i, and 1 – i.

The factors are (z – 4), (z + 4), (z – (1 + i)), and (z – (1 – i)). First, multiply (z – 4)(z + 4) = z² – 16. Then multiply the conjugate pair: (z – (1 + i))(z – (1 – i)). This will give us a quadratic with real coefficients. Then multiply the two quadratics together to get the polynomial. Then use the condition P(2) = 72 to find the constant.