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CIE A-Level Maths Study Notes

2.9.2 Arithmetic Operations with Complex Numbers

Complex numbers, integral to advanced mathematics, are numbers consisting of a real part and an imaginary part. They are pivotal in various scientific and engineering fields, offering deep insights into mathematical concepts.

Addition and Subtraction of Complex Numbers

Complex Number Form:

  • z=a+biz = a + bi , where 'aa' is the real part, 'bibi' is the imaginary part.

Addition:

  • For z1=a1+b1iz_1 = a_1 + b_1i and z2=a2+b2iz_2 = a_2 + b_2i, add real and imaginary parts separately:
  • z1+z2=(a1+a2)+(b1+b2)iz_1 + z_2 = (a_1 + a_2) + (b_1 + b_2)i.

Subtraction:

  • z1z2=(a1a2)+(b1b2)iz_1 - z_2 = (a_1 - a_2) + (b_1 - b_2)i .
  • Used in various mathematical and engineering contexts.

Multiplication of Complex Numbers

Distributive Property:

  • Multiply (a1+b1i)(a2+b2i)(a_1 + b_1i)(a_2 + b_2i) to get a1a2+a1b2i+b1a2i+b1b2i2a_1a_2 + a_1b_2i + b_1a_2i + b_1b_2i^2.
  • Simplify using i2=1:a1a2b1b2+(a1b2+b1a2)ii^2 = -1 : a_1a_2 - b_1b_2 + (a_1b_2 + b_1a_2)i.
  • Key for understanding rotation and scaling in the complex plane.

Division of Complex Numbers

Using the Conjugate:

  • The conjugate of z=a+biz = a + bi is z=abiz^* = a - bi.
  • Divide z1z2\frac{z_1}{z_2} by multiplying numerator and denominator with z2z_2^* : z1z2\frac{z_1}{z_2} = z1z2z2z2\frac{z_1 \cdot z_2^*}{z_2 \cdot z_2^*}.
  • Eliminates the imaginary part in the denominator.
  • Essential in complex equations and electrical engineering.

Example Questions

Addition Example

Problem Statement:

Compute (3+4i)+(52i) (3 + 4i) + (5 - 2i) .

Solution:

1. Separate Real and Imaginary Parts:

  • Real: 3+5 3 + 5 .
  • Imaginary: 4i2i 4i - 2i .

2. Combine and Simplify:

  • Real: 3+5=8 3 + 5 = 8 .
  • Imaginary: 4i2i=2i 4i - 2i = 2i .

3. Result:

  • Sum is 8+2i 8 + 2i .

Multiplication Example

Problem Statement:

Determine the product of (2+3i) (2 + 3i) and (45i) (4 - 5i) .

Solution:

1. Apply Distributive Property:

  • Perform (2+3i)(45i) (2 + 3i)(4 - 5i) .

2. Multiply Each Pair of Terms:

  • Calculate 24 2 \cdot 4 , 25i2 \cdot -5i , 3i4 3i \cdot 4 , and 3i5i3i \cdot -5i .

3. Combine Results:

  • Sum up 810i+12i15i2 8 - 10i + 12i - 15i^2 .

4. Use i2=1 i^2 = -1 :

  • Convert 15i2-15i^2 to +15+15.

5. Combine Like Terms:

  • Add 810i+12i+158 - 10i + 12i + 15 together.

6. Final Product:

  • Result is 23+2i23 + 2i.

Division Example

Problem Statement:

Divide (1+2i)(1 + 2i) by (34i)(3 - 4i).

Solution:

1. Multiply by Conjugate:

  • Use 3+4i3+4i\frac{3 + 4i}{3 + 4i} to clear the imaginary part in the denominator.
  • Thus, 1+2i34i3+4i3+4i\frac{1 + 2i}{3 - 4i} \cdot \frac{3 + 4i}{3 + 4i}.

2. Simplify Numerator:

  • Calculate (1+2i)(3+4i)(1 + 2i)(3 + 4i).
  • Expand to get 3+4i+6i+83 + 4i + 6i + 8 (using i2=1i^2 = -1).

3. Simplify Numerator Further:

  • Combine like terms: 11+10i11 + 10i.

4. Simplify Denominator:

  • Calculate (34i)(3+4i)(3 - 4i)(3 + 4i).
  • Expand to 9+169 + 16 (imaginary terms cancel out).

5. Complete Division:

  • Divide 11+10i25\frac{11 + 10i}{25}.

6. Final Answer:

  • Separate into real and imaginary parts: 1125+1025i\frac{11}{25} + \frac{10}{25}i
Dr Rahil Sachak-Patwa avatar
Written by: Dr Rahil Sachak-Patwa
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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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