Exercise 1.2: Complex Numbers - Simplification and Additive Inverses

Theoretical Foundations of Complex Numbers

  • Definition of Complex Numbers:

    • A complex number is any expression of the form z=a+biz = a + bi, where aa and bb are real numbers (a,b∈Ra, b \in \mathbb{R}), and ii represents the imaginary unit.
    • The component aa is defined as the real part of zz, written as Re(z)=a\text{Re}(z) = a
    • The component bb is defined as the imaginary part of zz, written as Im(z)=b\text{Im}(z) = b

  • Properties of the Imaginary Unit (ii):

    • The imaginary unit is defined by the property i=−1i = \sqrt{-1}.
    • Squaring the imaginary unit yields i2=−1i^2 = -1

  • Fundamental Algebraic Operations:

    • Addition: (a+bi)+(c+di)=(a+c)+(b+d)i(a + bi) + (c + di) = (a + c) + (b + d)i
    • Subtraction: (a+bi)−(c+di)=(a−c)+(b−d)i(a + bi) - (c + di) = (a - c) + (b - d)i
    • Multiplication: (a+bi)(c+di)=(ac−bd)+(ad+bc)i(a + bi)(c + di) = (ac - bd) + (ad + bc)i
    • Division: a+bic+di=(a+bi)(c−di)(c+di)(c−di)=(ac+bd)+(bc−ad)ic2+d2\frac{a + bi}{c + di} = \frac{(a + bi)(c - di)}{(c + di)(c - di)} = \frac{(ac + bd) + (bc - ad)i}{c^2 + d^2}
    • Additive Inverse: The additive inverse of z=a+biz = a + bi is −z=−a−bi-z = -a - bi, which satisfies z+(−z)=0+0i=0z + (-z) = 0 + 0i = 0

Exercise 1.2: Simplification of Complex Numbers

Exercise 1.2 problem set on complex numbers

Section 1: Simplification into Standard Form a+bia + bi

Problem 1 (i): (2+5i)+(3−7i)(2 + 5i) + (3 - 7i)

  • Problem Statement:

    • Simplify (2+5i)+(3−7i)(2 + 5i) + (3 - 7i) and write in the form a+bia + bi

  • Step-by-Step Solution:

    • Group real components and imaginary components together:     (2+3)+(5i−7i)(2 + 3) + (5i - 7i)
    • Perform real component addition:     2+3=52 + 3 = 5
    • Perform imaginary component combination:     5i−7i=(5−7)i=−2i5i - 7i = (5 - 7)i = -2i

  • Final Answer:

    • 5−2i5 - 2i
Problem 1 (ii): (16−3i)+(9+2i)(16 - 3i) + (9 + 2i)

  • Problem Statement:

    • Simplify (16−3i)+(9+2i)(16 - 3i) + (9 + 2i) and write in the form a+bia + bi

  • Step-by-Step Solution:

    • Group real and imaginary components:     (16+9)+(−3i+2i)(16 + 9) + (-3i + 2i)
    • Evaluate the real part:     16+9=2516 + 9 = 25
    • Evaluate the imaginary part:     −3i+2i=(−3+2)i=−1i=−i-3i + 2i = (-3 + 2)i = -1i = -i

  • Final Answer:

    • 25−i25 - i
Problem 1 (iii): (9−2i)−(7−3i)(9 - 2i) - (7 - 3i)

  • Problem Statement:

    • Simplify (9−2i)−(7−3i)(9 - 2i) - (7 - 3i) and write in the form a+bia + bi

  • Step-by-Step Solution:

    • Distribute the minus sign through the second complex number:     9−2i−7+3i9 - 2i - 7 + 3i
    • Rearrange into real and imaginary terms:     (9−7)+(−2i+3i)(9 - 7) + (-2i + 3i)
    • Compute real and imaginary results:     9−7=29 - 7 = 2−2i+3i=1i=i-2i + 3i = 1i = i

  • Final Answer:

    • 2+i2 + i
Problem 1 (iv): (11+9i)−(9−7i)(11 + 9i) - (9 - 7i)

  • Problem Statement:

    • Simplify (11+9i)−(9−7i)(11 + 9i) - (9 - 7i) and write in the form a+bia + bi

  • Step-by-Step Solution:

    • Distribute the negative sign:     11+9i−9+7i11 + 9i - 9 + 7i
    • Group real and imaginary components:     (11−9)+(9i+7i)(11 - 9) + (9i + 7i)
    • Simplify each part:     11−9=211 - 9 = 29i+7i=16i9i + 7i = 16i

  • Final Answer:

    • 2+16i2 + 16i
Problem 1 (v): (3+4i)(2−3i)(3 + 4i)(2 - 3i)

  • Problem Statement:

    • Simplify (3+4i)(2−3i)(3 + 4i)(2 - 3i) and write in the form a+bia + bi

  • Step-by-Step Solution:

    • Apply the distributive expansion method:     (3×2)+(3×−3i)+(4i×2)+(4i×−3i)(3 \times 2) + (3 \times -3i) + (4i \times 2) + (4i \times -3i)
    • Multiply terms:     6−9i+8i−12i26 - 9i + 8i - 12i^2
    • Substitute i2=−1i^2 = -16−i−12(−1)6 - i - 12(-1)=6−i+12= 6 - i + 12
    • Combine real numbers:     6+12=186 + 12 = 18

  • Final Answer:

    • 18−i18 - i
Problem 1 (vi): (5−2i)(3−4i)(5 - 2i)(3 - 4i)

  • Problem Statement:

    • Simplify (5−2i)(3−4i)(5 - 2i)(3 - 4i) and write in the form a+bia + bi

  • Step-by-Step Solution:

    • Expand using the distributive property:     (5×3)+(5×−4i)+(−2i×3)+(−2i×−4i)(5 \times 3) + (5 \times -4i) + (-2i \times 3) + (-2i \times -4i)
    • Compute individual products:     15−20i−6i+8i215 - 20i - 6i + 8i^2
    • Substitute i2=−1i^2 = -115−26i+8(−1)15 - 26i + 8(-1)=15−26i−8= 15 - 26i - 8
    • Combine real terms:     15−8=715 - 8 = 7

  • Final Answer:

    • 7−26i7 - 26i
Problem 1 (vii): (3−5i)÷(2−4i)(3 - 5i) \div (2 - 4i)

  • Problem Statement:

    • Simplify (3−5i)÷(2−4i)(3 - 5i) \div (2 - 4i) and write in the form a+bia + bi

  • Step-by-Step Solution:

    • Express division as a fraction:     3−5i2−4i\frac{3 - 5i}{2 - 4i}
    • Multiply numerator and denominator by the complex conjugate of the denominator, (2+4i)(2 + 4i):     (3−5i)(2+4i)(2−4i)(2+4i)\frac{(3 - 5i)(2 + 4i)}{(2 - 4i)(2 + 4i)}
    • Expand numerator:     (3−5i)(2+4i)=6+12i−10i−20i2=6+2i−20(−1)=6+2i+20=26+2i(3 - 5i)(2 + 4i) = 6 + 12i - 10i - 20i^2 = 6 + 2i - 20(-1) = 6 + 2i + 20 = 26 + 2i
    • Expand denominator using standard difference of squares (a−bi)(a+bi)=a2+b2(a - bi)(a + bi) = a^2 + b^2(2−4i)(2+4i)=22−(4i)2=4−16i2=4−16(−1)=4+16=20(2 - 4i)(2 + 4i) = 2^2 - (4i)^2 = 4 - 16i^2 = 4 - 16(-1) = 4 + 16 = 20
    • Divide real and imaginary terms by denominator:     26+2i20=2620+220i\frac{26 + 2i}{20} = \frac{26}{20} + \frac{2}{20}i
    • Simplify fractions to lowest terms:     2620=1310\frac{26}{20} = \frac{13}{10}220=110\frac{2}{20} = \frac{1}{10}

  • Final Answer:

    • 1310+110i\frac{13}{10} + \frac{1}{10}i
Problem 1 (viii): (5+2i)÷(6−3i)(5 + 2i) \div (6 - 3i)

  • Problem Statement:

    • Simplify (5+2i)÷(6−3i)(5 + 2i) \div (6 - 3i) and write in the form a+bia + bi

  • Step-by-Step Solution:

    • Express division in fractional form:     5+2i6−3i\frac{5 + 2i}{6 - 3i}
    • Multiply numerator and denominator by the complex conjugate (6+3i)(6 + 3i)(5+2i)(6+3i)(6−3i)(6+3i)\frac{(5 + 2i)(6 + 3i)}{(6 - 3i)(6 + 3i)}
    • Expand numerator:     (5+2i)(6+3i)=30+15i+12i+6i2=30+27i+6(−1)=30+27i−6=24+27i(5 + 2i)(6 + 3i) = 30 + 15i + 12i + 6i^2 = 30 + 27i + 6(-1) = 30 + 27i - 6 = 24 + 27i
    • Expand denominator:     (6−3i)(6+3i)=62−(3i)2=36−9i2=36−9(−1)=36+9=45(6 - 3i)(6 + 3i) = 6^2 - (3i)^2 = 36 - 9i^2 = 36 - 9(-1) = 36 + 9 = 45
    • Separate real and imaginary components over 45:     24+27i45=2445+2745i\frac{24 + 27i}{45} = \frac{24}{45} + \frac{27}{45}i
    • Reduce both fractions by their greatest common divisor:     24÷345÷3=815\frac{24 \div 3}{45 \div 3} = \frac{8}{15}27÷945÷9=35\frac{27 \div 9}{45 \div 9} = \frac{3}{5}

  • Final Answer:

    • 815+35i\frac{8}{15} + \frac{3}{5}i

Exercise 1.2: Additive Inverses of Complex Numbers

Section 2: Determination of Additive Inverses

Theoretical Principle
  • For any complex number z=a+biz = a + bi, its additive inverse −z-z is defined as:   −z=−(a+bi)=−a−bi-z = -(a + bi) = -a - bi
  • The fundamental property of the additive inverse states:   z+(−z)=0+0i=0z + (-z) = 0 + 0i = 0
Problem 2 (i): Additive Inverse of 3+2i3 + 2i
  • Given Number:z=3+2iz = 3 + 2i
  • Calculation:−z=−(3+2i)=−3−2i-z = -(3 + 2i) = -3 - 2i
  • Verification:(3+2i)+(−3−2i)=(3−3)+(2−2)i=0+0i=0(3 + 2i) + (-3 - 2i) = (3 - 3) + (2 - 2)i = 0 + 0i = 0
  • Final Answer:
    • −3−2i-3 - 2i
Problem 2 (ii): Additive Inverse of 4−3i4 - 3i
  • Given Number:z=4−3iz = 4 - 3i
  • Calculation:−z=−(4−3i)=−4+3i-z = -(4 - 3i) = -4 + 3i
  • Verification:(4−3i)+(−4+3i)=(4−4)+(−3+3)i=0+0i=0(4 - 3i) + (-4 + 3i) = (4 - 4) + (-3 + 3)i = 0 + 0i = 0
  • Final Answer:
    • −4+3i-4 + 3i
Problem 2 (iii): Additive Inverse of 5−7i5 - 7i
  • Given Number:z=5−7iz = 5 - 7i
  • Calculation:−z=−(5−7i)=−5+7i-z = -(5 - 7i) = -5 + 7i
  • Verification:(5−7i)+(−5+7i)=(5−5)+(−7+7)i=0+0i=0(5 - 7i) + (-5 + 7i) = (5 - 5) + (-7 + 7)i = 0 + 0i = 0
  • Final Answer:
    • −5+7i-5 + 7i
Problem 2 (iv): Additive Inverse of −23+54i-\frac{2}{3} + \frac{5}{4}i
  • Given Number:z=−23+54iz = -\frac{2}{3} + \frac{5}{4}i
  • Calculation:−z=−(−23+54i)=23−54i-z = -\left(-\frac{2}{3} + \frac{5}{4}i\right) = \frac{2}{3} - \frac{5}{4}i
  • Verification:(−23+54i)+(23−54i)=(−23+23)+(54−54)i=0+0i=0\left(-\frac{2}{3} + \frac{5}{4}i\right) + \left(\frac{2}{3} - \frac{5}{4}i\right) = \left(-\frac{2}{3} + \frac{2}{3}\right) + \left(\frac{5}{4} - \frac{5}{4}\right)i = 0 + 0i = 0
  • Final Answer:
    • 23−54i\frac{2}{3} - \frac{5}{4}i