Sullivan, 8th ed.: 1.3, 5.6 MAC1140/1147 1.3 Complex Numbers (condensed)The complex number system enables us to take even roots of negative numbers by means of the imaginary uniti, which is equal to the square root of –1; that is i2 = -1 and i = 1-. By factoring –1 out of a negativeexpression, it becomes positive and an even root can be taken: -b=i b. Standard form for complexexpression is a + bi, where a is the real part and bi is the imaginary part. All properties of exponents hold whenthe base is i, thus i1 = i, i2 = -1, i3 = i2(i) = -1i = -i, i4 = i2(i2) = -1(-1) = 1. In general, for in, divide n by 4: if theremainder is 0, in= 1; if the remainder is 1, in = i, if the remainder is 2, in= -1; if the remainder is 3, in= -i. Theproduct of a complex number (a + bi) and its conjugate (a – bi) is a nonnegative real number (a2 + b2).Write in a + bi form:1.25-2. a.50-b.- 503.3 16 2- +4 (2 + 6i)(2 – 6i) 5. [x – (3 – 5i)] [x – (3 + 5i)]5.6 COMPLEX ZEROS; FUNDAMENTAL THEOREM OF ALGEBRAFundamental Theorem of Algebra: Every complex polynomial function of degree n ≥ 1 or higher has at least one complex zero. Theorem: Every complex polynomial of degree n ≥ or higher can be factored into n linear factors (not necessarily distinct). Conjugate Pairs Theorem: Given a polynomial with real coefficients, if r = a + bi is a zero, then r a bi= - is also a zero of the polynomial. Information is given about a polynomial f(x) whose coefficients are real numbers. Find the remaining zeros: 1. Degree 4; zeros: 3, 5, 4 + i 2. Degree 7; zeros: 2, 3i, 5 – i, 3 + i Form a polynomial f(x) with real coefficients having the given degree and zeros:3. Degree 4; zeros: 4 – 3i, -3 multiplicity 2 4. Degree 5; zeros: 3 multiplicity 3, 2 – iUse the given zero to find the remaining zeros:5. f(x) = x4 – 7x3 + 14x2 – 38x – 60; zero: 1 + 3iFind the complex zeros of the polynomial function. Write f in factored form:6. f(x) = 2x4 + x3 – 35x2 – 113x + 65 Max. Zeros: Positive Zeros: Negative Zeros: Potential Zeros: Real Zeros: Try These (5.6)Use the given zero to find the remaining zeros:1. 2x5 – 3x4 - 5x3 - 15x2 – 207x + 108; zero: 3iFind the complex zeros of each polynomial function; write f in factored form:2. f(x) = x4 + 3x3 – 19x2 + 27x -