Complex Numbers Study Notes

4-2.5 Complex Numbers

Vocabulary

  • Definition of i:

    • i is defined as:

    • i=ext√(−1)i = ext{√}(-1)

    • i2=−1i^2 = -1

    • The imaginary number i was invented to solve and explain the concept of negative square root numbers.

    • Example:

      • ext√(−9)=ext√(−1)imesext√(9)=iimes3=3iext{√}(-9) = ext{√}(-1) imes ext{√}(9) = i imes 3 = 3i

Square Roots of Negative Numbers

  • If r is a positive real number, then:

    • ext√(−r)=iext√(r)ext{√}(-r) = i ext{√}(r)

Pure Imaginary Numbers

  • These are numbers of the form bi, where b is a nonzero real number. Examples include:

    • −2i-2i

    • 8iext√158i ext{√}15

    • ii

Simplification Examples

Example 1: Simplify the following expressions:
  1. ext√(ext√(−15))=ext√15ext√(−1)=iext√15ext{√}({ ext{√}}(-15)) = ext{√}15 ext{√}(-1) = i ext{√}15

  2. ext√(−50)=ext√(50)ext√(−1)=5iext√2ext{√}(-50) = ext{√}(50) ext{√}(-1) = 5i ext{√}2

  3. (−6−7i)−(1+3i)=−7−10i(−6 − 7i) − (1 + 3i) = -7 - 10i

  4. (1−i)/(1+i)=rac(1−i)(1−i)(1+i)(1−i)=rac(1−2i−i2)(1+1)=(1−2i+1)=rac(2−2i)2=1−i(1 - i) / (1 + i) = rac{(1 - i)(1 - i)}{(1 + i)(1 - i)} = rac{(1 - 2i - i^2)}{(1 + 1)} = (1 - 2i + 1) = rac{(2 - 2i)}{2} = 1 - i

  5. (−6−5i)(1+3i)=−6−18i−5i−15i2=−6−18i−5i+15=(9−23i)(−6 - 5i)(1 + 3i) = −6 − 18i − 5i − 15i^2 = −6 − 18i − 5i + 15 = (9 - 23i)

  6. iext√(−98)imes(−ext√98)=−1ext√(98)=1ext√(98)=(−1)(−1)ext√(98)i ext{√}(-98) imes (− ext{√}98) = -1 ext{√}(98) = 1 ext{√}(98) = (−1)(−1) ext{√}(98)

Example 2: Find the factored forms
  1. x2+1=(x−i)(x+i)x^2 + 1 = (x - i)(x + i)

  2. −1(9x2+100)−1(9x^2 + 100)

  3. −(3x−10x)(3x+10x)=(−6x)(6x)=−36(−1)=36−(3x - 10x)(3x + 10x) = (−6x)(6x) = −36(−1) = 36

  4. x2+36=(x+6i)(x−6i)x^2 + 36 = (x + 6i)(x - 6i)

Quadratic Equations from Roots

Example 5: Find a quadratic equation for the roots
  1. Roots: a=−6ia = -6i and 6i6i.

    • Sum: −6i+6i=0;-6i + 6i = 0;

    • −b=0-b = 0, implying b=0b = 0;

    • Product: (−6i)(6i)=−36(−1)=36;(−6i)(6i) = −36(−1) = 36; thus the equation is x2+36=0x^2 + 36 = 0.

  2. Roots: −2−2 and 55.

    • Sum: −2+5=3<br>ightarrowb=−3;−2 + 5 = 3 <br>ightarrow b = −3;

    • Product: (−2)(5)=−10;(−2)(5) = −10; hence the equation is x2+3x−10=0x^2 + 3x - 10 = 0.

Finding Sum and Product of Roots

Example 6:
  1. Given the equation: 5x2+2x+15x^2 + 2x + 1

    • Sum of roots: −b/a=−rac25;-b/a = - rac{2}{5};

    • Product of roots: c/a=rac15c/a = rac{1}{5}.

Exercises

  1. Simplify the expressions:

    • ext√(−7)=iext√7ext{√}(-7) = i ext{√}7

    • −81=9i−81 = 9i

    • (12+5i)−(2−i)=10+6i(12 + 5i) - (2 - i) = 10 + 6i

    • Calculate: (8i)(4i)(−9i)=−288i2=288(8i)(4i)(−9i) = -288 i^2 = 288

    • =−ext√(−81)=9i= − ext{√}(-81) = 9i

    • (−6i)2=(−6)2imes(i)2=36(−1)=−36(−6i)^{2} = (-6)^{2} imes (i)^{2} = 36(-1) = -36

  2. Further simplification:

    • Calculate 10+ext√(−9)−(2+ext√(−25))=10+3i−(2+5i)=8−2i10 + ext{√}(-9) - (2 + ext{√}(-25)) = 10 + 3i - (2 + 5i) = 8 - 2i

Absolute Value and Complex Numbers

  • Plotting Complex Numbers:

    • Example: A=3+2iA = 3 + 2i;

    • Absolute Value of a complex number z:
      ∣z∣=ext√(x2+y2)|z| = ext{√}(x^2 + y^2) where x and y are the real and imaginary parts, respectively.

    • For AA: ∣A∣=ext√(32+22)=ext√(9+4)=ext√(13)|A| = ext{√}(3^2 + 2^2) = ext{√}(9 + 4) = ext{√}(13)

Quotients as Complex Numbers

  • Example:
    rac3−2i5i=rac(3−2i)(−i)(5i)(−i)=rac−3i−2(−1)−5=rac2−3i5rac{3 - 2i}{5i} = rac{(3 - 2i)(-i)}{(5i)(-i)} = rac{-3i - 2(-1)}{-5} = rac{2 - 3i}{5}

  • Further simplification may apply to other complex divisions.

Quadratic Equations involving Imaginary Solutions

Quadratic Example 1:x2+2x+3=0x^2 + 2x + 3 = 0
  • Using the quadratic formula:

    • Coefficients:

      • a=1a = 1, b=2b = 2, c=3c = 3

    • Substitute into the formula:
      x=rac−bext±ext√(b2−4ac)2ax = rac{-b ext{±} ext{√}(b^2-4ac)}{2a}

    • Calculation:
      −2ext±ext√((2)2−4(1)(3))=−2ext±ext√(4−12)=−2ext±ext√(−8)=−2ext±2iext√2-2 ext{±} ext{√}((2)^2 - 4(1)(3)) = -2 ext{±} ext{√}(4 - 12) = -2 ext{±} ext{√}(-8) = -2 ext{±} 2i ext{√}2

    • The roots would thus be x=−1ext±iext√2x = -1 ext{± } i ext{√}2

Quadratic Example 2: 2x2−4x+7=02x^2 - 4x + 7 = 0
  • Coefficients:

    • a=2a = 2, b=−4b = -4, c=7c = 7

    • Solve using quadratic formula:
      x=rac4ext±ext√((−4)2−4(2)(7))2(2)x = rac{4 ext{±} ext{√}((-4)^2 - 4(2)(7))}{2(2)}

    • Calculation leads to
      =rac4ext±ext√(16−56)4=rac4ext±ext√(−40)4= rac{4 ext{±} ext{√}(16 - 56)}{4} = rac{4 ext{±} ext{√}(-40)}{4}

    • Results in complex roots: x=1ext±iext√10x = 1 ext{±} i ext{√}10

Inverses and Conjugates of Complex Numbers

  1. Additive Inverse:

    • For AA (let's say points related to previous examples), compute −A-A.

    • For B=3+2iB = 3 + 2i, Additive Inverse is −B=−3−2i-B = -3 - 2i.

  2. Complex Conjugate:

    • Represented as zˉ=a−bi\bar{z} = a - bi if z=a+biz = a + bi. For BB:

    • Conjugate of BB is 3−2i3 - 2i.

  3. Absolute Value:

    • Calculate absolute values of points similarly to before.

Final Example:
  • For the equation (x+3)(x−3i)=34(x + 3)(x - 3i) = 34

    • Rearranging shows complex solutions involved generating (x2−(3i)2=34)(x^2 - (3i)^2 = 34) leading to complex solutions in terms of radicals.

Concluding Notes
  • Understanding complex numbers is essential for advanced mathematics, particularly in fields such as engineering and physics.

  • Practice with sums, products, and the quadratic formula helps reinforce these concepts.