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Review of complex number arithmetic: addition, subtraction, and multiplication. The imaginary number i is defined where i2 = -1; thus, i = ββ1. All complex numbers can be written in the form a + bi, where a and b are real numbers. Note that if b=0, then the number is real so the set of real numbers is a subset of the complex numbers. Arithmetic operations for the complex numbers parallel those of irrational numbers of the form a+bβπ. Powers of i are cyclic: i = i5 = i9 = β¦ = ββ1 i2 = i6 = i10 = β¦ -1 i3 = i7 = i11 = β¦ = βi ππ β ββ1 i4 = i8 = i12 = β¦ = 1 1. Simplify, writing your answer in the form a + bi: a) i 42 = b) i 113 = c) 4 + 5i β 3i2 +11i3 β 6i4 = d) (3 - 5i) + (8 + 2i) = e) (-1 + i) β 2(-4 + 3i) = f) (4 + 5i)(3 - 5i) = g) (-1 + 2i)3 = 2. Is the set of complex numbers closed with respect to the operation: a. Addition? b. Subtraction? c. Multiplication? d. Division? 3. What is the additive identity for the complex numbers? 4. What is the additive inverse of the number a + bi ? Division, Inverses, & Conjugates When trying to simplify an irrational number of the form 1 π+βπ where aβ 0 and b>0 are integers and bβ π2 , by removing the radical from the denominator (rationalization), we know we must multiply the numerator and denominator by the factor π β βπ so the expression becomes 1 π+βπ = 1 π+βπ β’ πββπ πββπ = πββπ π 2 βπ . This expression now has a non-zero denominator which is an integer and the radical has now moved into the numerator. Similarly, we can do the same with a complex number of the form 1 . π+πi Given a number a + bi, the number a - bi is called the complex conjugate. It is used to rationalize the denominator in the reciprocal of a complex number and in division of complex numbers, thus allowing us to write the number again in the standard form. 5. What is the conjugate of the following numbers? a) 3i ________ b) 4 + 5i ________ c) β 3 β 2i __________ 6. What is the multiplicative identity for the complex numbers? 7. What is the multiplicative inverse of the complex number a + bi ? 8. Write in the form a + bi the following: a) c) 1 2+3π 1 β i 2 + 3π (5 + 3i)2 2 β π = We recall the quadratic formula π₯ = b) 1+i 5β3π d) 1βi (5 + 3i)2 βπ±βπ2 β4ππ 2π ÷ 2 + 3π 2βπ = which gives us the solutions to the quadratic equation ax2+bx+c=0, aβ 0. By the trichotomy principle, the discriminant could be positive, zero, or negative. We recall that the graph of the quadratic equation y = ax2+bx+c is a parabola opening up if a>0 and downward if a<0. If the graph has two x intercepts, this corresponds to the situation where the discriminant is positive. If the graph has one x intercept, this corresponds to the situation where the discriminant is zero. If the graph has NO x intercept, this corresponds to the situation where the discriminant is negative. Thus, imaginary numbers occur βnaturallyβ when using the quadratic formula. 9. Find the solution(s) to the following quadratic equations. Put the answers in the form a+bi: a) x2 +x+1=0 b) x2 = -8 c) 2x2 -x+1=0 10. Find a real polynomial of degree 2 with x=2 and x=3 as roots. 11. Find a real polynomial of degree 3 with x=0, x=-1 and x=6 as roots. Imaginary roots of polynomial equations with real coefficients occur in conjugate pairs; i.e. if a+bi is a root, then so is a-bi. 12. Find a polynomial of degree 2 with real coefficients if x=3-i is a root. 13. Find a polynomial of degree 4 with real coefficients if x=-1 is a double root and x=2+3i is also a root.