Revision NotesExam QuestionsFlashcardsPast PapersMock Exams
IBMathsDPApplications & Interpretation (AI)HLRevision Notes1. Number & AlgebraFurther Complex NumbersModulus-Argument (Polar) Form

Modulus-Argument (Polar) Form (DP IB Applications & Interpretation (AI)): Revision Note

Amber

Written by: Amber

Reviewed by: Mark Curtis

Updated on

DP Applications & Interpretation (AI): HLDP Analysis & Approaches (AA): HLDP Analysis & Approaches (AA): HLDP Analysis & Approaches (AA): HLDP Applications & Interpretation (AI): HLDP Applications & Interpretation (AI): HL

Modulus-argument (polar) form

What is modulus-argument (polar) form?

  • The modulus-argument (polar) form of a complex number z is

    • z equals r open parentheses cos space theta plus isin space theta close parentheses

      • sometimes written z equals r space cis space theta

    • where

      • r equals vertical line z vertical line

      • theta equals arg space z

Examiner Tips and Tricks

The modulus-argument (polar) form of a complex number is given in the formula booklet.

  • e.g. z equals 1 plus square root of 3 straight i has a modulus of 2 and an argument of pi over 3

    • so z equals 2 open parentheses cos pi over 3 plus isin pi over 3 close parentheses

  • You can also convert back to Cartesian form

  • e.g. z equals 4 open parentheses cos pi over 4 plus isin pi over 4 close parentheses is 4 open parentheses fraction numerator square root of 2 over denominator 2 end fraction plus fraction numerator square root of 2 over denominator 2 end fraction close parentheses equals 2 square root of 2 plus 2 square root of 2 space straight i

Examiner Tips and Tricks

Negative arguments must be shown clearly and without being further simplified, 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

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

  • The complex conjugate of z equals r open parentheses cos space theta plus isin space theta close parentheses is

    • z to the power of asterisk times equals r open parentheses cos open parentheses negative theta close parentheses plus isin open parentheses negative theta close parentheses close parentheses

      • also written as r space cis open parentheses negative theta close parentheses

  • The modulus is the same

    • but the argument changes sign

  • This works because, in general,

    • sin open parentheses negative theta close parentheses equals sin space theta and cos open parentheses negative theta close parentheses equals cos space theta

    • so r open parentheses cos open parentheses negative theta close parentheses plus isin open parentheses negative theta close parentheses close parentheses equals r open parentheses cos space theta minus isin space theta close parentheses equals x minus straight i y

Examiner Tips and Tricks

The complex conjugate of 2 open parentheses cos space open parentheses pi over 3 close parentheses plus isin space open parentheses pi over 3 close parentheses close parentheses

  • is not 2 open parentheses cos space open parentheses pi over 3 close parentheses minus isin space open parentheses pi over 3 close parentheses close parentheses in modulus-argument (polar) form

    • as you cannot have a negative in front of the sin

  • it is 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

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

  • To multiply two complex numbers in modulus-argument (polar) form

    • multiply their moduli

      • 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

    • and add their arguments

      • 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 z subscript 1 equals r subscript 1 cis space theta subscript 1 and z subscript 2 equals r subscript 2 cis space theta subscript 2

    • then z subscript 1 z subscript 2 equals r subscript 1 r subscript 2 cis open parentheses theta subscript 1 plus theta subscript 2 close parentheses

  • These rules work for

    • multiplying more than two complex numbers, e.g. z subscript 1 z subscript 2 z subscript 3

    • powers of complex numbers, e.g. z squared, z cubed, ...

      • using z squared equals z cross times z etc

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

  • To divide two complex numbers in modulus-argument (polar) form

    • divide their moduli

      • 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

    • and subtract their arguments

      • 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 z subscript 1 equals r subscript 1 cis space theta subscript 1 and z subscript 2 equals r subscript 2 cis space theta subscript 2

    • then z subscript 1 over z subscript 2 equals r subscript 1 over r subscript 2 cis open parentheses theta subscript 1 minus theta subscript 2 close parentheses

What if the new argument is out of range?

  • Sometimes the new argument does not lie in the range negative pi less than theta less or equal than pi

    • so adjust it by either adding or subtracting 2 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 equals fraction numerator 7 pi over denominator 6 end fraction 

    • This is currently not in the range negative pi space less than theta less or equal than pi

    • Subtracting 2 pi from fraction numerator 7 pi over denominator 6 end fraction gives negative fraction numerator 5 pi over denominator 6 end fraction which

      • is in range

      • and represents the same angle

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

You've read 0 of your 5 free revision notes this week

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

    Test yourselfFlashcards
    Previous:Geometry of Complex NumbersNext:Exponential (Euler's) Form
    Author
    Reviewer
    Amber

    Author: Amber

    Expertise: Maths Content Creator

    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.

    Mark Curtis

    Reviewer: Mark Curtis

    Expertise: Maths Content Creator

    Mark graduated twice from the University of Oxford: once in 2009 with a First in Mathematics, then again in 2013 with a PhD (DPhil) in Mathematics. He has had nine successful years as a secondary school teacher, specialising in A-Level Further Maths and running extension classes for Oxbridge Maths applicants. Alongside his teaching, he has written five internal textbooks, introduced new spiralling school curriculums and trained other Maths teachers through outreach programmes.