Class 11 Mathematics Notes By Adil Aslam Unit # 1 : Solution Contact Me Whatsapp: 0346-6248138 Email:
[email protected] Notes By Adil Aslam 28 Mathematics Class 11 Unit # 1 Physics Class 10 Notes By Adil Aslam = 4α + 10 + 5αi − 8i [Rearrange] = (4α + 10) + i(5α − 8) Re (z1z2 ) = (4α + 10) ... ... ... ... ... ... ... (ii) By comparing equation (i) and (ii),we get 4α + 10 = 20 4α = 20 − 10 4α = 10 α = 10 4 α = 5 2 Complex Polynomial: A complex polynomial P(z) is a polynomial function of the complex variable z with complex coefficients. It is expressed in the general form as: P(z) = anz n + an−1z n−1 + ... ... + a1z + z0 Where an, an−1, ... ... , a1, a0 are complex numbers (an ≠ 0) and n ≥ 0 is an integer representing the degree of the polynomial. Examples: o P(z) = (1 − i)z + 3i is linear complex polynomial. o P(z) = (5 − 4i)z 2 + (2 + i)z + (3 − 4i) is quadratic complex polynomial. o P(z) = (2 − i)z 3 + 2z 2 i + (5 + 3i) is cubic complex polynomial. Fundamental Property of Complex Polynomial: A fundamental property of complex polynomials is that they can always be factored into a product of linear factors. Example: P(z) = z 2 − 1 is quadratic complex polynomial and it can be factored as: P(z) = (z + 1)(z − 1) is quadratic complex polynomial. Fundamental Theorem of Algebra: According to the Fundamental theorem of algebra, a polynomial of degree n ≥ 1 has exactly n roots in complex numbers system C.
Notes By Adil Aslam 29 Mathematics Class 11 Unit # 1 Physics Class 10 Notes By Adil Aslam Example: The polynomial P(z) = z 2 + 1 has degree 2, And its 2 complex roots: z = i and z = −i. Corollary to the Fundamental Theorem of Algebra: Any polynomial P(z) of degree n can be completely factored into a product of a constant a and n linear factors over the complex numbers C: P(z) = a(z − z1)(z − z2) ... (z − zn) Where z1, z2, z3, ... , zn are the complex roots of the polynomial. o Once the roots of a polynomial are known, we can factor it into linear terms corresponding to those roots. Example: The polynomial P(z) = z 2 + 4 has complex roots z = 2i and z = −2i, So it can be factorized as (z − 2i)(z + 2i). Zeros of a Polynomial Function: If P(z) is a polynomial function, the values of z that satisfy P(z) = 0 are called the zeros of the function P(z) and also the roots of the polynomial equation P(z) = 0. Example: Let P(z) = z 2 + 4. To find the zeros, solve P(z) = 0: z 2 + 4 = 0 ⟹ z = ±2i Zeros: z = 2i and z = −2i. Rational Root Theorem: If a polynomial P(x) = anx n + an−1x n−1 + ⋯ + a1x + a0 has integer coefficients, then every rational root p q (in simplest terms) satisfies: o p is a factor of the constant term a0. o q is a factor of the leading coefficient an. Completing the Square: Completing the square is a method of solving quadratic equations by rewriting them into a perfect square trinomial, allowing the equation to be solved by taking the square root of both sides. o This method is especially useful when factoring is difficult and works even for irrational or complex roots.
Notes By Adil Aslam 30 Mathematics Class 11 Unit # 1 Physics Class 10 Notes By Adil Aslam Exercise 1. 3 1. Factorize the following: (i) a 2 + 4b 2 (ii) 9a 2 + 16b 2 (iii) 3x 2 + 3y 2 (iv) 144x 2 + 225y 2 (v) z 2 − 2zi − 1 (vi) z 2 + 6z + 13 (vii) z 2 + 4z + 5 (viii) 2z 2 − 22z + 65 Solution: (i) a 2 + 4b 2 = a 2 − (−1)4b 2 = a 2 − i 22 2b 2 [i 2 = −1 and 4 = 2 2 ] = (a) 2 − (2bi) 2 = (a + 2bi)(a − 2bi) [Using a 2 − b 2 = (a + b)(a − b)] (ii) 9a 2 + 16b 2 = 9a 2 − (−1)16b 2 = 9a 2 − i 24 2b 2 [16 = 4 2 ] = (3a) 2 − (4bi) 2 = (3a + 4bi)(3a − 4bi) (iii) 3x2 + 3y 2 = 3(x 2 + y 2 ) = 3(x 2 − (−1)y 2 ) = 3(x 2 − i 2y 2 ) = 3[(x) 2 − (yi) 2 ] = 3(x + yi)(x − yi) (iv) 144x2 + 225y 2 = 9(16x 2 + 25y 2 ) = 9[16x 2 − (−1)25y 2 ] = 9(16x 2 − i 225y 2 ) = 9(4 2x 2 − i 25 2y 2 ) [144 = 122 and 225 = 152 ] = 9(4x) 2 − (5yi) 2 = 9(4x + 5yi)(4x − 5yi) (v) z 2 − 2zi − 1
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