Section 1.3: Complex Numbers and Quadratic Equations in the Complex Number System

Section 1.3: Complex Numbers and Quadratic Equations in the Complex Number System

Definitions and Core Concepts

The Imaginary Unit

The imaginary unit, denoted by ii, is defined as the number whose square is −1-1: i2=−1i^2 = -1

The Complex Number System

Complex numbers are expressions written in the form a+bia + bi, where aa and bb are real numbers.

  • aa is called the real part of a+bia + bi

  • bb is called the imaginary part of a+bia + bi

  • ii is the imaginary unit (i2=−1i^2 = -1)

Complex Numbers Overview Mindmap

Equality and Basic Arithmetic Operations

Equality of Complex Numbers

Two complex numbers a+bia + bi and c+dic + di are equal if and only if their real parts are equal and their imaginary parts are equal: a + bi = c + di \t\quad \text{if and only if} \quad a = c \text{ and } b = d

Sum of Complex Numbers

(a+bi)+(c+di)=(a+c)+(b+d)i(a + bi) + (c + di) = (a + c) + (b + d)i

Difference of Complex Numbers

(a+bi)−(c+di)=(a−c)+(b−d)i(a + bi) - (c + di) = (a - c) + (b - d)i

Product of Complex Numbers

(a+bi)(c+di)=(ac−bd)+(ad+bc)i(a + bi)(c + di) = (ac - bd) + (ad + bc)i

Worked Examples: Addition, Subtraction, and Multiplication

  • Example 1(a): Add (4+6i)+(−1+2i)(4 + 6i) + (-1 + 2i)   (4+6i)+(−1+2i)=[4+(−1)]+(6+2)i=3+8i(4 + 6i) + (-1 + 2i) = [4 + (-1)] + (6 + 2)i = 3 + 8i

  • Example 1(b): Subtract (5+4i)−(2+8i)(5 + 4i) - (2 + 8i)   (5+4i)−(2+8i)=(5−2)+(4−8)i=3+(−4)i=3−4i(5 + 4i) - (2 + 8i) = (5 - 2) + (4 - 8)i = 3 + (-4)i = 3 - 4i

  • Example 2: Multiply (6+5i)(3+7i)(6 + 5i)(3 + 7i)   (6+5i)(3+7i)=6(3+7i)+5i(3+7i)(6 + 5i)(3 + 7i) = 6(3 + 7i) + 5i(3 + 7i)   =18+42i+15i+35i2= 18 + 42i + 15i + 35i^2   =18+57i+35(−1)= 18 + 57i + 35(-1)   =18−35+57i=−17+57i= 18 - 35 + 57i = -17 + 57i

Step-by-Step Practice Solutions (Problems 11–24)

  • Problem 11: (2−3i)+(6+8i)(2 - 3i) + (6 + 8i)   =2+6−3i+8i=8+5i= 2 + 6 - 3i + 8i = 8 + 5i

  • Problem 12: (4+5i)+(−8+2i)(4 + 5i) + (-8 + 2i)   =4−8+5i+2i=−4+7i= 4 - 8 + 5i + 2i = -4 + 7i

  • Problem 13: (−3+2i)−(4−4i)(-3 + 2i) - (4 - 4i)   =−3−4+2i+4i=−7+6i= -3 - 4 + 2i + 4i = -7 + 6i

  • Problem 14: (3−4i)−(−3−4i)(3 - 4i) - (-3 - 4i)   =3+3−4i+4i=6= 3 + 3 - 4i + 4i = 6

  • Problem 15: (2−5i)−(8+6i)(2 - 5i) - (8 + 6i)   =2−8−5i−6i=−6−11i= 2 - 8 - 5i - 6i = -6 - 11i

  • Problem 16: (−8+4i)−(2−2i)(-8 + 4i) - (2 - 2i)   =−8−2+4i+2i=−10+6i= -8 - 2 + 4i + 2i = -10 + 6i

  • Problem 17: 3(2−6i)3(2 - 6i)   =6−18i= 6 - 18i

  • Problem 18: −4(2+8i)-4(2 + 8i)   =−8−32i= -8 - 32i

  • Problem 19: −3i(7+6i)-3i(7 + 6i)   =−21i−18i2=−21i−18(−1)=18−21i= -21i - 18i^2 = -21i - 18(-1) = 18 - 21i

  • Problem 20: 3i(−3+4i)3i(-3 + 4i)   =−9i+12i2=−9i+12(−1)=−12−9i= -9i + 12i^2 = -9i + 12(-1) = -12 - 9i

  • Problem 21: (3−4i)(2+i)(3 - 4i)(2 + i)   =3(2+i)−4i(2+i)=6+3i−8i−4i2=6−5i+4=10−5i= 3(2 + i) - 4i(2 + i) = 6 + 3i - 8i - 4i^2 = 6 - 5i + 4 = 10 - 5i

  • Problem 22: (5+3i)(2−i)(5 + 3i)(2 - i)   =5(2−i)+3i(2−i)=10−5i+6i−3i2=10+i+3=13+i= 5(2 - i) + 3i(2 - i) = 10 - 5i + 6i - 3i^2 = 10 + i + 3 = 13 + i

  • Problem 23: (−5−i)(−5+i)(-5 - i)(-5 + i)   =(−5)(−5)+(−5)(i)+(−i)(−5)+(−i)(i)=25−5i+5i−i2=25−(−1)=26= (-5)(-5) + (-5)(i) + (-i)(-5) + (-i)(i) = 25 - 5i + 5i - i^2 = 25 - (-1) = 26

  • Problem 24: (−3+i)(3+i)(-3 + i)(3 + i)   =(−3)(3)+(−3)(i)+i(3)+i(i)=−9−3i+3i+i2=−9−1=−10= (-3)(3) + (-3)(i) + i(3) + i(i) = -9 - 3i + 3i + i^2 = -9 - 1 = -10

Complex Conjugates, Reciprocals, and Division

Definition of Complex Conjugate

If z=a+biz = a + bi is a complex number, its conjugate, denoted by zˉ\bar{z}, is defined as: zˉ=a+bi‾=a−bi\bar{z} = \overline{a + bi} = a - bi

Theorem: Product of a Complex Number and Its Conjugate

The product of a complex number and its conjugate is always a non-negative real number: zzˉ=(a+bi)(a−bi)=a2−b2i2=a2+b2z \bar{z} = (a + bi)(a - bi) = a^2 - b^2 i^2 = a^2 + b^2

Properties of Conjugates

  • zˉ‾=z\overline{\bar{z}} = z

  • z+w‾=zˉ+wˉ\overline{z + w} = \bar{z} + \bar{w}

  • z⋅w‾=zˉ⋅wˉ\overline{z \cdot w} = \bar{z} \cdot \bar{w}

  • The conjugate of a real number is the real number itself: aˉ=a\bar{a} = a for any real number aa.

Worked Examples: Conjugates, Reciprocals, and Division

  • Example 3: Find the product of z=2+5iz = 2 + 5i and its conjugate zˉ=2−5i\bar{z} = 2 - 5i.   zzˉ=(2+5i)(2−5i)=22−(5i)2=4−25i2=4+25=29z \bar{z} = (2 + 5i)(2 - 5i) = 2^2 - (5i)^2 = 4 - 25i^2 = 4 + 25 = 29

  • Example 4: Write the reciprocal of 4+3i4 + 3i in standard form 14+3i\frac{1}{4 + 3i}.   14+3i=14+3i⋅4−3i4−3i=4−3i42+32=4−3i16+9=4−3i25=425−325i\frac{1}{4 + 3i} = \frac{1}{4 + 3i} \cdot \frac{4 - 3i}{4 - 3i} = \frac{4 - 3i}{4^2 + 3^2} = \frac{4 - 3i}{16 + 9} = \frac{4 - 3i}{25} = \frac{4}{25} - \frac{3}{25}i

  • Example 5(a): Write 2+3i1−i\frac{2 + 3i}{1 - i} in standard form.   2+3i1−i=(2+3i)(1+i)(1−i)(1+i)=2+2i+3i+3i212−i2=2+5i−31+1=−1+5i2=−12+52i\frac{2 + 3i}{1 - i} = \frac{(2 + 3i)(1 + i)}{(1 - i)(1 + i)} = \frac{2 + 2i + 3i + 3i^2}{1^2 - i^2} = \frac{2 + 5i - 3}{1 + 1} = \frac{-1 + 5i}{2} = -\frac{1}{2} + \frac{5}{2}i

  • Example 5(b): Write 2−12i5−12i\frac{2 - 12i}{5 - 12i} in standard form.   2−12i5−12i=(2−12i)(5+12i)(5−12i)(5+12i)=10+24i−60i−144i252+122=10−36i+14425+144=154−36i169=154169−36169i\frac{2 - 12i}{5 - 12i} = \frac{(2 - 12i)(5 + 12i)}{(5 - 12i)(5 + 12i)} = \frac{10 + 24i - 60i - 144i^2}{5^2 + 12^2} = \frac{10 - 36i + 144}{25 + 144} = \frac{154 - 36i}{169} = \frac{154}{169} - \frac{36}{169}i

  • Example 6: Given z=3−4iz = 3 - 4i and w=6+3iw = 6 + 3i, evaluate:

    1. zw=3−4i6+3i=(3−4i)(6−3i)(6+3i)(6−3i)=18−9i−24i+12i236+9=18−33i−1245=6−33i45=215−1115i\frac{z}{w} = \frac{3 - 4i}{6 + 3i} = \frac{(3 - 4i)(6 - 3i)}{(6 + 3i)(6 - 3i)} = \frac{18 - 9i - 24i + 12i^2}{36 + 9} = \frac{18 - 33i - 12}{45} = \frac{6 - 33i}{45} = \frac{2}{15} - \frac{11}{15}i

    2. z+w‾=(3−4i)+(6+3i)‾=9−i‾=9+i\overline{z + w} = \overline{(3 - 4i) + (6 + 3i)} = \overline{9 - i} = 9 + i

Step-by-Step Practice Solutions (Problems 25–34)

  • Problem 25: 103−4i\frac{10}{3 - 4i}   =10(3+4i)(3−4i)(3+4i)=10(3+4i)9+16=30+40i25=65+85i= \frac{10(3 + 4i)}{(3 - 4i)(3 + 4i)} = \frac{10(3 + 4i)}{9 + 16} = \frac{30 + 40i}{25} = \frac{6}{5} + \frac{8}{5}i

  • Problem 26: 135−12i\frac{13}{5 - 12i}   =13(5+12i)(5−12i)(5+12i)=13(5+12i)25+144=65+156i169=65169+156169i=513+1213i= \frac{13(5 + 12i)}{(5 - 12i)(5 + 12i)} = \frac{13(5 + 12i)}{25 + 144} = \frac{65 + 156i}{169} = \frac{65}{169} + \frac{156}{169}i = \frac{5}{13} + \frac{12}{13}i

  • Problem 27: 2+ii\frac{2 + i}{i}   =(2+i)(−i)i(−i)=−2i−i2−i2=1−2i1=1−2i= \frac{(2 + i)(-i)}{i(-i)} = \frac{-2i - i^2}{-i^2} = \frac{1 - 2i}{1} = 1 - 2i

  • Problem 28: 2−i−2i\frac{2 - i}{-2i}   =(2−i)i−2i(i)=2i−i2−2i2=1+2i2=12+i= \frac{(2 - i)i}{-2i(i)} = \frac{2i - i^2}{-2i^2} = \frac{1 + 2i}{2} = \frac{1}{2} + i

  • Problem 29: 6−i1+i\frac{6 - i}{1 + i}   =(6−i)(1−i)(1+i)(1−i)=6−6i−i+i21+1=5−7i2=52−72i= \frac{(6 - i)(1 - i)}{(1 + i)(1 - i)} = \frac{6 - 6i - i + i^2}{1 + 1} = \frac{5 - 7i}{2} = \frac{5}{2} - \frac{7}{2}i

  • Problem 30: 2+3i1−i\frac{2 + 3i}{1 - i}   =(2+3i)(1+i)(1−i)(1+i)=2+2i+3i+3i21+1=−1+5i2=−12+52i= \frac{(2 + 3i)(1 + i)}{(1 - i)(1 + i)} = \frac{2 + 2i + 3i + 3i^2}{1 + 1} = \frac{-1 + 5i}{2} = -\frac{1}{2} + \frac{5}{2}i

  • Problem 31: (12+32i)2\left(\frac{1}{2} + \frac{\sqrt{3}}{2}i\right)^2   =(12)2+2(12)(32i)+(32i)2=14+32i−34=−12+32i= \left(\frac{1}{2}\right)^2 + 2\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}i\right) + \left(\frac{\sqrt{3}}{2}i\right)^2 = \frac{1}{4} + \frac{\sqrt{3}}{2}i - \frac{3}{4} = -\frac{1}{2} + \frac{\sqrt{3}}{2}i

  • Problem 32: (32−12i)2\left(\frac{\sqrt{3}}{2} - \frac{1}{2}i\right)^2   =(32)2−2(32)(12i)+(12i)2=34−32i−14=12−32i= \left(\frac{\sqrt{3}}{2}\right)^2 - 2\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}i\right) + \left(\frac{1}{2}i\right)^2 = \frac{3}{4} - \frac{\sqrt{3}}{2}i - \frac{1}{4} = \frac{1}{2} - \frac{\sqrt{3}}{2}i

  • Problem 33: (1+i)2(1 + i)^2   =1+2i+i2=1+2i−1=2i= 1 + 2i + i^2 = 1 + 2i - 1 = 2i

  • Problem 34: (1−i)2(1 - i)^2   =1−2i+i2=1−2i−1=−2i= 1 - 2i + i^2 = 1 - 2i - 1 = -2i

Powers of ii

Cyclic Pattern of Powers of ii

  • i1=ii^1 = i

  • i2=−1i^2 = -1

  • i3=i2⋅i=(−1)⋅i=−ii^3 = i^2 \cdot i = (-1) \cdot i = -i

  • i4=(i2)2=(−1)2=1i^4 = (i^2)^2 = (-1)^2 = 1

  • i5=i4⋅i=1⋅i=ii^5 = i^4 \cdot i = 1 \cdot i = i

  • i6=i4⋅i2=1⋅(−1)=−1i^6 = i^4 \cdot i^2 = 1 \cdot (-1) = -1

  • i7=i4⋅i3=1⋅(−i)=−ii^7 = i^4 \cdot i^3 = 1 \cdot (-i) = -i

  • i8=(i4)2=12=1i^8 = (i^4)^2 = 1^2 = 1

To evaluate ini^n for any integer nn, divide nn by 44 to find remainder r∈{0,1,2,3}r \in \{0, 1, 2, 3\}. Then in=iri^n = i^r.

Worked Examples and Practice Problems: Powers of ii

  • Example 7(a): Evaluate i18i^{18}   i18=i16⋅i2=(i4)4⋅i2=(1)4⋅(−1)=−1i^{18} = i^{16} \cdot i^2 = (i^4)^4 \cdot i^2 = (1)^4 \cdot (-1) = -1

  • Example 7(b): Evaluate i103i^{103}   i103=i100⋅i3=(i4)25⋅i3=(1)25⋅(−i)=−ii^{103} = i^{100} \cdot i^3 = (i^4)^{25} \cdot i^3 = (1)^{25} \cdot (-i) = -i

  • Example 8: Expand (3+i)3(3 + i)^3 in standard form using (x+a)3=x3+3ax2+3a2x+a3(x + a)^3 = x^3 + 3ax^2 + 3a^2x + a^3   (3+i)3=33+3(32)(i)+3(3)(i2)+i3(3 + i)^3 = 3^3 + 3(3^2)(i) + 3(3)(i^2) + i^3   =27+27i+9(−1)+(−i)= 27 + 27i + 9(-1) + (-i)   =27−9+(27−1)i=18+26i= 27 - 9 + (27 - 1)i = 18 + 26i

  • Problem 41: Evaluate 6i3−4i56i^3 - 4i^5   6i3−4i5=6(−i)−4(i)=−6i−4i=−10i6i^3 - 4i^5 = 6(-i) - 4(i) = -6i - 4i = -10i

  • Problem 42: Evaluate 4i3−2i2+14i^3 - 2i^2 + 1   4(−i)−2(−1)+1=−4i+2+1=3−4i4(-i) - 2(-1) + 1 = -4i + 2 + 1 = 3 - 4i

  • Problem 47: Evaluate i8+i6−i4−i2i^8 + i^6 - i^4 - i^2   i8=1,i6=−1,i4=1,i2=−1i^8 = 1, \quad i^6 = -1, \quad i^4 = 1, \quad i^2 = -1   1+(−1)−1−(−1)=1−1−1+1=01 + (-1) - 1 - (-1) = 1 - 1 - 1 + 1 = 0

Principal Square Root of a Negative Number

Definition

If NN is a positive real number, the principal square root of −N-N, denoted by −N\sqrt{-N}, is defined as: −N=Ni\sqrt{-N} = \sqrt{N}i

Worked Examples and Practice Problems

  • Example 9(a): −1=1i=i\sqrt{-1} = \sqrt{1}i = i

  • Example 9(b): −25=25i=5i\sqrt{-25} = \sqrt{25}i = 5i

  • Example 9(c): −12=12i=4⋅3i=23i\sqrt{-12} = \sqrt{12}i = \sqrt{4 \cdot 3}i = 2\sqrt{3}i

  • Problem 52: Evaluate −64\sqrt{-64}   −64=64i=8i\sqrt{-64} = \sqrt{64}i = 8i

  • Problem 53: Evaluate −12\sqrt{-12}   −12=12i=4⋅3i=23i\sqrt{-12} = \sqrt{12}i = \sqrt{4 \cdot 3}i = 2\sqrt{3}i

  • Problem 54: Evaluate −18\sqrt{-18}   −18=18i=9⋅2i=32i\sqrt{-18} = \sqrt{18}i = \sqrt{9 \cdot 2}i = 3\sqrt{2}i

Quadratic Equations in the Complex Number System

Square Root Method for Negative Constants

For an equation x2=−Nx^2 = -N (N>0N > 0): x=±−N=±Nix = \pm \sqrt{-N} = \pm \sqrt{N}i

The Quadratic Formula

In the complex number system, the solutions of the quadratic equation ax2+bx+c=0ax^2 + bx + c = 0 (where a,b,ca, b, c are real numbers and a≠0a \neq 0) are given by: x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Worked Examples: Solving Quadratic Equations

  • Example 10(a): Solve x2=16x^2 = 16   x=±16  ⟹  x=±4Solution set: {−4,4}x = \pm \sqrt{16} \implies x = \pm 4 \quad \text{Solution set: } \{-4, 4\}

  • Example 10(b): Solve x2=−25x^2 = -25   x=±−25=±25i  ⟹  x=±5iSolution set: {−5i,5i}x = \pm \sqrt{-25} = \pm \sqrt{25}i \implies x = \pm 5i \quad \text{Solution set: } \{-5i, 5i\}

  • Example 11: Solve x2−2x+5=0x^2 - 2x + 5 = 0   Coefficients: a=1,b=−2,c=5a = 1, b = -2, c = 5   Discriminant: b2−4ac=(−2)2−4(1)(5)=4−20=−16b^2 - 4ac = (-2)^2 - 4(1)(5) = 4 - 20 = -16   x=−(−2)±−162(1)=2±4i2=1±2ix = \frac{-(-2) \pm \sqrt{-16}}{2(1)} = \frac{2 \pm 4i}{2} = 1 \pm 2i   Solution set: {1−2i,1+2i}\{1 - 2i, 1 + 2i\}

  Verification Check:   For x=1+2ix = 1 + 2i:   (1+2i)2−2(1+2i)+5=(1+4i+4i2)−2−4i+5=(1+4i−4)−2−4i+5=−3−2+5=0(1 + 2i)^2 - 2(1 + 2i) + 5 = (1 + 4i + 4i^2) - 2 - 4i + 5 = (1 + 4i - 4) - 2 - 4i + 5 = -3 - 2 + 5 = 0   For x=1−2ix = 1 - 2i:   (1−2i)2−2(1−2i)+5=(1−4i+4i2)−2+4i+5=(1−4i−4)−2+4i+5=−3−2+5=0(1 - 2i)^2 - 2(1 - 2i) + 5 = (1 - 4i + 4i^2) - 2 + 4i + 5 = (1 - 4i - 4) - 2 + 4i + 5 = -3 - 2 + 5 = 0

Step-by-Step Practice Solutions (Problems 61–64)

  • Problem 61: Solve x2−16=0x^2 - 16 = 0   x2=16  ⟹  x=±4x^2 = 16 \implies x = \pm 4

  • Problem 62: Solve x2+25=0x^2 + 25 = 0   x2=−25  ⟹  x=±−25  ⟹  x=±5ix^2 = -25 \implies x = \pm \sqrt{-25} \implies x = \pm 5i

  • Problem 63: Solve x2−6x+13=0x^2 - 6x + 13 = 0   Coefficients: a=1,b=−6,c=13a = 1, b = -6, c = 13   x=−(−6)±(−6)2−4(1)(13)2(1)=6±36−522=6±−162=6±4i2=3±2ix = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(13)}}{2(1)} = \frac{6 \pm \sqrt{36 - 52}}{2} = \frac{6 \pm \sqrt{-16}}{2} = \frac{6 \pm 4i}{2} = 3 \pm 2i

  • Problem 64: Solve x2+4x+8=0x^2 + 4x + 8 = 0   Coefficients: a=1,b=4,c=8a = 1, b = 4, c = 8   x=−4±42−4(1)(8)2(1)=−4±16−322=−4±−162=−4±4i2=−2±2ix = \frac{-4 \pm \sqrt{4^2 - 4(1)(8)}}{2(1)} = \frac{-4 \pm \sqrt{16 - 32}}{2} = \frac{-4 \pm \sqrt{-16}}{2} = \frac{-4 \pm 4i}{2} = -2 \pm 2i

Character of the Solutions of a Quadratic Equation

Character of the Solutions of a Quadratic Equation
Discriminant Analysis (D=b2−4acD = b^2 - 4ac)

For any quadratic equation ax2+bx+c=0ax^2 + bx + c = 0 with real coefficients a,b,ca, b, c (a≠0a \neq 0):

  1. If b2−4ac>0b^2 - 4ac > 0:    The equation has two unequal real solutions (x1≠x2x_1 \neq x_2, both real).    Example: x2−5x+6=0  ⟹  D=25−24=1>0  ⟹  x=2,3x^2 - 5x + 6 = 0 \implies D = 25 - 24 = 1 > 0 \implies x = 2, 3

  2. If b2−4ac=0b^2 - 4ac = 0:    The equation has one repeated real solution (a double root), given by x1=x2=−b2ax_1 = x_2 = -\frac{b}{2a}.    Example: x2−4x+4=0  ⟹  D=16−16=0  ⟹  x=2x^2 - 4x + 4 = 0 \implies D = 16 - 16 = 0 \implies x = 2 (double root)

  3. If b2−4ac<0b^2 - 4ac < 0:    The equation has two non-real complex solutions that are complex conjugates of each other: x1=α+βix_1 = \alpha + \beta i and x2=α−βix_2 = \alpha - \beta i    Example: x2+2x+5=0  ⟹  D=4−20=−16<0  ⟹  x=−1±2ix^2 + 2x + 5 = 0 \implies D = 4 - 20 = -16 < 0 \implies x = -1 \pm 2i

Worked Examples and Practice Problems: Character of Solutions

  • Example 12(a): Determine character of solutions for 2x2−4x+1=02x^2 - 4x + 1 = 0   a=2,b=−4,c=1  ⟹  D=(−4)2−4(2)(1)=16−8=8>0a = 2, b = -4, c = 1 \implies D = (-4)^2 - 4(2)(1) = 16 - 8 = 8 > 0   The equation has two distinct real solutions.

  • Example 12(b): Determine character of solutions for x2−2x+6=0x^2 - 2x + 6 = 0   a=1,b=−2,c=6  ⟹  D=(−2)2−4(1)(6)=4−24=−20<0a = 1, b = -2, c = 6 \implies D = (-2)^2 - 4(1)(6) = 4 - 24 = -20 < 0   The equation has two non-real complex solutions that are complex conjugates of each other.

  • Example 12(c): Determine character of solutions for 4x2+12x+9=04x^2 + 12x + 9 = 0   a=4,b=12,c=9  ⟹  D=122−4(4)(9)=144−144=0a = 4, b = 12, c = 9 \implies D = 12^2 - 4(4)(9) = 144 - 144 = 0   The equation has one repeated real solution (double root).

  • Problem 79: Determine character of solutions for 3x2−3x+4=03x^2 - 3x + 4 = 0   a=3,b=−3,c=4  ⟹  D=(−3)2−4(3)(4)=9−48=−39<0a = 3, b = -3, c = 4 \implies D = (-3)^2 - 4(3)(4) = 9 - 48 = -39 < 0   The equation has two non-real complex solutions.

  • Problem 80: Determine character of solutions for 2x2−4x+1=02x^2 - 4x + 1 = 0   a=2,b=−4,c=1  ⟹  D=(−4)2−4(2)(1)=16−8=8>0a = 2, b = -4, c = 1 \implies D = (-4)^2 - 4(2)(1) = 16 - 8 = 8 > 0   The equation has two distinct real solutions.

Advanced Complex Expressions (Problems 87–90)

Given z=3−4iz = 3 - 4i and w=8+3iw = 8 + 3i:

  • Problem 87: Evaluate z+zˉz + \bar{z}   z+zˉ=(3−4i)+(3+4i)=(3+3)+(−4i+4i)=6+0i=6z + \bar{z} = (3 - 4i) + (3 + 4i) = (3 + 3) + (-4i + 4i) = 6 + 0i = 6

  • Problem 88: Evaluate w−wˉw - \bar{w}   w−wˉ=(8+3i)−(8−3i)=(8−8)+(3i+3i)=0+6i=6iw - \bar{w} = (8 + 3i) - (8 - 3i) = (8 - 8) + (3i + 3i) = 0 + 6i = 6i

  • Problem 89: Evaluate zzˉz \bar{z}   zzˉ=(3−4i)(3+4i)=32−(4i)2=9−16i2=9+16=25z \bar{z} = (3 - 4i)(3 + 4i) = 3^2 - (4i)^2 = 9 - 16i^2 = 9 + 16 = 25

  • Problem 90: Evaluate z−w‾\overline{z - w}   z−w=(3−4i)−(8+3i)=−5−7iz - w = (3 - 4i) - (8 + 3i) = -5 - 7i   z−w‾=−5−7i‾=−5+7i\overline{z - w} = \overline{-5 - 7i} = -5 + 7i