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CIE A-Level Maths Study Notes

2.9.5 Operations in Polar Form

Complex numbers in polar form are a cornerstone of higher mathematics, offering a unique perspective on arithmetic operations. This comprehensive guide delves into the details of multiplication and division in polar form, highlighting the significance of modulus and argument in these processes. Understanding these concepts is crucial for students aiming to master complex number manipulations.

Polar form of a Complex Number

Image courtesy of Cuemath

Multiplication of Complex Numbers in Polar Form

  • Complex Number in Polar Form: Represented as z=reiθz = r e^{i\theta}, where:
    • rr = Modulus (distance from origin)
    • θ\theta = Argument (angle from positive x-axis)
  • Multiplying Two Complex Numbers: For z1=r1eiθ1z_1 = r_1 e^{i\theta_1} and z2=r2eiθ2z_2 = r_2 e^{i\theta_2}:
    • Procedure: Multiply their moduli (r-values) and add their arguments (θ-values).
    • Result: z1z2=r1r2ei(θ1+θ2)z_1 z_2 = r_1 r_2 e^{i(\theta_1 + \theta_2)}
  • Geometric Interpretation:
    • On Argand Plane, multiplication:
      • Stretches by the modulus of one number.
      • Rotates by the angle of the other number.
    • Helps visualize the effect on position and size of complex numbers.

Division of Complex Numbers in Polar Form

  • Complex Number in Polar Form: Represented as z=reiθz = r e^{i\theta}, where:
    • rr = Modulus (distance from origin)
    • θ\theta = Argument (angle from positive x-axis)
  • Dividing Two Complex Numbers: For z1=r1eiθ1z_1 = r_1 e^{i\theta_1} and z2=r2eiθ2z_2 = r_2 e^{i\theta_2}:
    • Procedure: Divide their moduli (r-values) and subtract their arguments (θ-values).
    • Result: z1z2=r1r2ei(θ1θ2)\frac{z_1}{z_2} = \frac{r_1}{r_2} e^{i(\theta_1 - \theta_2)}
  • Geometric Interpretation:
    • On Argand Plane, division:
      • Shrinks by the modulus of the divisor.
      • Rotates by the negative of the divisor's angle.
    • Helps visualize the effect on position and size of complex numbers.

Example Questions

Problem 1: Multiplying Complex Numbers

  • Given: 4eiπ/34 e^{i\pi/3} and 4eiπ/34 e^{i\pi/3}
  • Step 1: Multiply Moduli
    • 4×2=84 \times 2 = 8
  • Step 2: Add Arguments
    • Convert to common denominator: π3=4π12,π4=3π12\frac{\pi}{3} = \frac{4\pi}{12}, \frac{\pi}{4} = \frac{3\pi}{12}
    • Add: 4π12+3π12=7π12\frac{4\pi}{12} + \frac{3\pi}{12} = \frac{7\pi}{12}
  • Result: 8ei(7π/12)8 e^{i(7\pi/12)}

Problem 2: Dividing Complex Numbers

  • Given: 6eiπ/26 e^{i\pi/2} and 3eiπ/63 e^{i\pi/6}
  • Step 1: Divide Moduli
    • 6÷3=26 \div 3 = 2
  • Step 2: Subtract Arguments
    • Convert to common denominator: π2=3π6,π6=1π6\frac{\pi}{2} = \frac{3\pi}{6}, \frac{\pi}{6} = \frac{1\pi}{6}
    • Subtract: 3π61π6=2π6\frac{3\pi}{6} - \frac{1\pi}{6} = \frac{2\pi}{6}, simplify to π3\frac{\pi}{3}
  • Result: 2ei(π/3)2 e^{i(\pi/3)}
Dr Rahil Sachak-Patwa avatar
Written by: Dr Rahil Sachak-Patwa
LinkedIn
Oxford University - PhD Mathematics

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.

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