DP IB Maths: AI HL

Revision Notes

1.6.2 Forms of Complex Numbers

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Modulus-Argument (Polar) Form

How do I write a complex number in modulus-argument (polar) form?

  • The Cartesian form of a complex number, z equals x plus straight i y, is written in terms of its real part, x, and its imaginary part, y
  • If we let r equals vertical line z vertical line and theta equals arg space z, then it is possible to write a complex number in terms of its modulus, r, and its argument, theta, called the modulus-argument (polar) form, given by...
    • z equals r open parentheses cos space theta plus isin space theta close parentheses
    • This is often written as z = r cis θ
    • This is given in the formula book under Modulus-argument (polar) form and exponential (Euler) form
  • It is usual to give arguments in the range negative pi space less than space theta space less or equal than space pi  or  0 space less or equal than space theta space less than space 2 pi
    • Negative arguments should be shown clearly
    • e.g. z equals 2 open parentheses cos space open parentheses negative pi over 3 close parentheses plus isin space open parentheses negative pi over 3 close parentheses close parentheses space equals space 2 space cis space open parentheses negative straight pi over 3 close parentheses
      • without simplifying cos invisible function application left parenthesis negative pi over 3 right parenthesis  to either cos invisible function application open parentheses pi over 3 close parentheses or 1 half
  • The complex conjugate of r cis θ is r cis (-θ )
  • If a complex number is given in the form z equals r open parentheses cos space theta minus isin space theta close parentheses, then it is not in modulus-argument (polar) form due to the minus sign
    • It can be converted by considering transformations of trigonometric functions
      • negative sin invisible function application theta space equals space sin invisible function application left parenthesis negative theta right parenthesis and cos invisible function application theta space equals space cos invisible function application left parenthesis negative theta right parenthesis
    • So  z equals r open parentheses cos invisible function application theta minus isin invisible function application theta close parentheses space equals space z equals r open parentheses cos invisible function application open parentheses negative theta close parentheses plus isin invisible function application open parentheses negative theta close parentheses close parentheses space equals space r space cis space open parentheses negative theta close parentheses  
  • To convert from modulus-argument (polar) form back to Cartesian form, evaluate the real and imaginary parts
    • E.g. z equals 2 open parentheses cos invisible function application open parentheses negative pi over 3 close parentheses plus isin invisible function application open parentheses negative pi over 3 close parentheses close parentheses becomes z equals 2 open parentheses 1 half plus straight i open parentheses negative fraction numerator square root of 3 over denominator 2 end fraction close parentheses close parentheses equals 1 minus square root of 3 blank straight i

How do I multiply complex numbers in modulus-argument (polar) form?

  • The main benefit of writing complex numbers in modulus-argument (polar) form is that they multiply and divide very easily 
  • To multiply two complex numbers in modulus-argument (polar) form we multiply their moduli and add their arguments
    • open vertical bar z subscript 1 z subscript 2 close vertical bar equals open vertical bar z subscript 1 close vertical bar open vertical bar z subscript 2 close vertical bar
    • arg space left parenthesis z subscript 1 z subscript 2 right parenthesis equals arg space z subscript 1 plus arg space z subscript 2
  • So if z1 = r1 cis (θ1) and z2 = r2 cis (θ2)
    • z1 z2 = r1r2 cis (θ1 + θ2)
  • Sometimes the new argument, theta subscript 1 plus theta subscript 2, does not lie in the range negative pi space less than space theta space less or equal than space pi (or  0 space less or equal than space theta space less than space 2 pi  if this is being used)
    • An out-of-range argument can be adjusted by either adding or subtracting 2 straight pi
    • E.g. If theta subscript 1 equals fraction numerator 2 pi over denominator 3 end fraction and theta subscript 2 equals pi over 2  then  theta subscript 1 plus theta subscript 2 space equals space fraction numerator 7 straight pi over denominator 6 end fraction 
    • This is currently not in the range negative pi space less than space theta space less or equal than space pi
    • Subtracting 2 straight pi from fraction numerator 7 straight pi over denominator 6 end fraction to give negative fraction numerator 5 straight pi over denominator 6 end fraction, a new argument is formed
      •  This lies in the correct range and represents the same angle on an Argand diagram
  • The rules of multiplying the moduli and adding the arguments can also be applied when…
    • …multiplying three complex numbers together, z subscript 1 z subscript 2 z subscript 3, or more
    • …finding powers of a complex number (e.g. z squared can be written as z z)
  • The rules for multiplication can be proved algebraically by multiplying z1 = r1 cis (θ1) by z2 = r2 cis (θ2), expanding the brackets and using compound angle formulae

How do I divide complex numbers in modulus-argument (polar) form?

  • To divide two complex numbers in modulus-argument (polar) form, we divide their moduli and subtract their arguments
    • open vertical bar z subscript 1 over z subscript 2 close vertical bar blank equals fraction numerator open vertical bar z subscript 1 close vertical bar over denominator vertical line z subscript 2 vertical line end fraction
    • arg space open parentheses z subscript 1 over z subscript 2 close parentheses equals arg space z subscript 1 minus arg space z subscript 2
  • So if z1 = r1 cis (θ1) and z2 = r2 cis (θ2) then 
    • z subscript 1 over z subscript 2 equals r subscript 1 over r subscript 2 cis space open parentheses theta subscript 1 minus theta subscript 2 close parentheses blank
  • Sometimes the new argument, theta subscript 1 minus theta subscript 2, can lie out of the range negative pi space less than space theta space less or equal than space pi (or the range 0 space less than space theta space less or equal than space 2 pi if this is being used)
    • You can add or subtract 2 straight pi to bring out-of-range arguments back in range
  • The rules for division can be proved algebraically by dividing z1 = r1 cis (θ1) by z2 = r2 cis (θ2) using complex division and the compound angle formulae

Exam Tip

  • Remember that r cis θ only refers to  r open parentheses cos space theta plus isin space theta close parentheses
    • If you see a complex number written in the form z equals r open parentheses cos space theta minus isin space theta close parentheses then you will need to convert it to the correct form first
    • Make sure you are confident with basic trig identities to help you do this

Worked example

Let z subscript 1 equals 4 square root of 2 blank cis blank fraction numerator 3 pi over denominator 4 end fraction  and z subscript 2 equals square root of 8 open parentheses cos invisible function application open parentheses pi over 2 close parentheses minus isin invisible function application open parentheses pi over 2 close parentheses close parentheses

a)
Find z subscript 1 z subscript 2, giving your answer in the form r open parentheses cos invisible function application theta plus isin invisible function application theta close parentheses where 0 less or equal than theta less than 2 pi

1-9-2-ib-aa-hl-forms-of-cn-we-solution-1-a

b)
Find z subscript 1 over z subscript 2, giving your answer in the form r open parentheses cos invisible function application theta plus isin invisible function application theta close parentheses where negative straight pi less or equal than theta less than pi

1-9-2-ib-aa-hl-forms-of-cn-we-solution-1-b

Exponential (Euler's) Form

How do we write a complex number in Euler's (exponential) form?

  • A complex number can be written in Euler's form as z equals r straight e to the power of straight i theta end exponent 
    • This relates to the modulus-argument (polar) form as z equals r straight e to the power of straight i theta end exponent equals r blank cis blank theta
    • This shows a clear link between exponential functions and trigonometric functions
    • This is given in the formula booklet under 'Modulus-argument (polar) form and exponential (Euler) form'
  • The argument is normally given in the range 0 ≤ θ < 2π
    • However in exponential form other arguments can be used and the same convention of adding or subtracting 2π can be applied

How do we multiply and divide complex numbers in Euler's form?

  • Euler's form allows for quick and easy multiplication and division of complex numbers
  • If z subscript 1 equals r subscript 1 straight e to the power of straight i theta subscript 1 end exponent spaceand z subscript 2 equals r subscript 2 straight e to the power of straight i theta subscript 2 end exponent then 
    • z subscript 1 cross times z subscript 2 equals r subscript 1 r subscript 2 straight e to the power of straight i open parentheses theta subscript 1 plus theta subscript 2 close parentheses end exponent
      • Multiply the moduli and add the arguments
    • z subscript 1 over z subscript 2 equals r subscript 1 over r subscript 2 straight e to the power of straight i open parentheses theta subscript 1 minus theta subscript 2 close parentheses end exponent
      • Divide the moduli and subtract the arguments
  • Using these rules makes multiplying and dividing more than two complex numbers much easier than in Cartesian form
  • When a complex number is written in Euler's form it is easy to raise that complex number to a power
    • If z equals r straight e to the power of straight i theta end exponent,  z squared equals r squared straight e to the power of 2 straight i theta end exponent  and  z to the power of n equals r to the power of n straight e to the power of ni theta end exponent

What are some common numbers in exponential form?

  • As cos space left parenthesis 2 pi right parenthesis equals 1 and sin space left parenthesis 2 pi right parenthesis equals 0 you can write:
    • 1 equals straight e to the power of 2 pi straight i end exponent
  • Using the same idea you can write:
    • 1 equals straight e to the power of 0 equals straight e to the power of 2 pi straight i end exponent equals straight e to the power of 4 pi straight i end exponent equals straight e to the power of 6 pi straight i end exponent equals straight e to the power of 2 k pi straight i end exponent
    • where k is any integer
  • As cos invisible function application open parentheses pi close parentheses equals negative 1 and sin invisible function application left parenthesis pi right parenthesis equals 0 you can write:
    • straight e to the power of pi straight i end exponent equals negative 1
    • Or more commonly written as straight e to the power of iπ plus 1 equals 0
      • This is known as Euler's identity and is considered by some mathematicians as the most beautiful equation
  • As cos invisible function application open parentheses pi over 2 close parentheses equals 0 and sin invisible function application open parentheses pi over 2 close parentheses equals 1 you can write:
    • straight i equals straight e to the power of pi over 2 straight i end exponent

Exam Tip

  • Euler's form allows for easy manipulation of complex numbers, in an exam it is often worth the time converting a complex number into Euler's form if further calculations need to be carried out
    • Familiarise yourself with which calculations are easier in which form, for example multiplication and division are easiest in Euler's form but adding and subtracting are easiest in Cartesian form

Worked example

Consider the complex number z equals 2 straight e to the power of pi over 3 straight i end exponent. Calculate z squared giving your answer in the form r straight e to the power of straight i theta end exponent.

1-9-2-ib-aa-hl-forms-of-cn-we-solution-2-eulers

Conversion of Forms

Converting from Cartesian form to modulus-argument (polar) form or exponential (Euler's) form

  • To convert from Cartesian form to modulus-argument (polar) form or exponential (Euler) form use 
    • r equals open vertical bar z close vertical bar equals square root of x squared plus y squared end root
  • and  
    • theta equals arg invisible function application z


Converting from modulus-argument (polar) form or exponential (Euler's) form to Cartesian form

  • To convert from modulus-argument (polar) form to Cartesian form
    • You may need to use your knowledge of trig exact values
    • a = r cosθ and b = r sinθ
    • Write z = r (cosθ + isinθ ) as z = r cosθ + (r sinθ )i
    • Find the values of the trigonometric ratios r sinθ and r cosθ
    • Rewrite as z = a + bi where
  • To convert from exponential (Euler’s) form to Cartesian form first rewrite z = r e  in the form z = r cosθ + (r sinθ)i and then follow the steps above


Converting between complex number forms using your GDC

  • Your GDC may also be able to convert complex numbers between the various forms
    • TI calculators, for example, have 'Convert to Polar' and 'Convert to Rectangular' (i.e. Cartesian) as options in the 'Complex Number Tools' menu
    • Make sure you are familiar with your GDC and what it can (and cannot) do with complex numbers

Exam Tip

  • When converting from Cartesian form into Polar or Euler's form, always leave your modulus and argument as an exact value
    • Rounding values too early may result in inaccuracies later on

Worked example

Two complex numbers are given by z subscript 1 equals 2 plus 2 straight i and z subscript 2 equals 3 straight e to the power of fraction numerator 2 pi over denominator 3 end fraction straight i end exponent.

a)
Write z subscript 1 in the form r straight e to the power of straight i theta end exponent.

1-9-2-ib-aa-hl-forms-of-cn-we-solution-3-a

b)
Write z subscript 2 in the form r open parentheses cos invisible function application theta plus isin invisible function application theta close parentheses and then convert it to Cartesian form.

1-9-2-ib-aa-hl-forms-of-cn-we-solution-3-b

 

 

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Amber

Author: Amber

Amber gained a first class degree in Mathematics & Meteorology from the University of Reading before training to become a teacher. She is passionate about teaching, having spent 8 years teaching GCSE and A Level Mathematics both in the UK and internationally. Amber loves creating bright and informative resources to help students reach their potential.