Roots of Complex Numbers (DP IB Analysis & Approaches (AA)): Revision Note

Amber

Written by: Amber

Reviewed by: Mark Curtis

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Roots of complex numbers

How do I find the square roots of a complex number?

  • A complex number z has two square roots, w and negative w, that are also complex

    • To find them, let w equals a plus b straight i where a comma space b element of straight real numbers

    • and form simultaneous equations from the relationship w squared equals z

      • by expanding w squared

      • and equating the real and imaginary parts with z

  • E.g. find the square roots of 3 plus 4 straight i

    • Let a plus b straight i be one of the square roots of 3 plus 4 straight i

    • This means open parentheses a plus b straight i close parentheses squared equals 3 plus 4 straight i

    • Expand the left-hand side

      • open parentheses a plus b straight i close parentheses open parentheses a plus b straight i close parentheses equals a squared plus a b straight i plus b a straight i plus b squared straight i squared equals a squared plus 2 a b straight i minus b squared

      • so a squared plus 2 a b straight i minus b squared equals 3 plus 4 straight i

    • Equate the real parts on both sides

      • a squared minus b squared equals 3

    • Equate the imaginary parts on both sides

      • 2 a b equals 4

    • Solve these two equations simultaneously

      • e.g. make b the subject of 2 a b equals 4 to get b equals 2 over a

      • Substitute b equals 2 over a into a squared minus b squared equals 3 to get a squared minus open parentheses 2 over a close parentheses squared equals 3

      • Rearrange to a to the power of 4 minus 3 a squared minus 4 equals 0 which gives a squared equals 4 or a squared equals negative 1 (but a element of straight real numbers)

      • so a equals plus-or-minus 2 which gives b equals plus-or-minus 1

    • The two square roots are 2 plus straight i and space minus 2 minus straight i

How do I use de Moivre’s theorem to find the nth roots of a complex number?

  • De Moivre's theorem to find the nth roots of a complex number, z equals r open parentheses cos theta plus isin theta close parentheses, is

    • n-th root of z equals z to the power of 1 over n end exponent equals r to the power of 1 over n end exponent open parentheses cos open parentheses fraction numerator theta plus 2 k pi over denominator n end fraction close parentheses plus straight i space sin open parentheses fraction numerator theta plus 2 k pi over denominator n end fraction close parentheses close parentheses

      • Letting k equals 0 comma space 1 comma space 2 comma space... comma space n minus 1 gives each nth root

  • In exponential (Euler's) form, this is

    • n-th root of z equals z to the power of 1 over n end exponent equals r to the power of 1 over n end exponent straight e to the power of fraction numerator theta plus 2 k pi over denominator n end fraction end exponent

      • where k equals 0 comma space 1 comma space 2 comma space... comma space n minus 1

Examiner Tips and Tricks

This formula is not given in the formula booklet so it must be learnt.

  • e.g. find the fourth roots of 16 open parentheses cos pi over 2 plus straight i space sin pi over 2 close parentheses

    • fourth root of z equals z to the power of 1 fourth end exponent equals 16 to the power of 1 fourth end exponent open parentheses cos open parentheses fraction numerator pi over 2 plus 2 k pi over denominator 4 end fraction close parentheses plus straight i space sin open parentheses fraction numerator pi over 2 plus 2 k pi over denominator 4 end fraction close parentheses close parentheses

      • k equals 0 gives 2 open parentheses cos pi over 8 plus straight i space sin pi over 8 close parentheses

      • k equals 1 gives 2 open parentheses cos fraction numerator 5 pi over denominator 8 end fraction plus straight i space sin fraction numerator 5 pi over denominator 8 end fraction close parentheses

      • k equals 0 gives 2 open parentheses cos fraction numerator 9 pi over denominator 8 end fraction plus straight i space sin fraction numerator 9 pi over denominator 8 end fraction close parentheses

      • k equals 0 gives 2 open parentheses cos fraction numerator 13 pi over denominator 8 end fraction plus straight i space sin fraction numerator 13 pi over denominator 8 end fraction close parentheses

Examiner Tips and Tricks

Your GDC can find roots of complex numbers.

  • If you plot the n complex roots of of a complex number on an Argand diagram

    • they form a regular n-sided polygon

  • e.g. in the example above

    • 2 open parentheses cos pi over 8 plus straight i space sin pi over 8 close parentheses, 2 open parentheses cos fraction numerator 5 pi over denominator 8 end fraction plus straight i space sin fraction numerator 5 pi over denominator 8 end fraction close parentheses, 2 open parentheses cos fraction numerator 9 pi over denominator 8 end fraction plus straight i space sin fraction numerator 9 pi over denominator 8 end fraction close parentheses and 2 open parentheses cos fraction numerator 13 pi over denominator 8 end fraction plus straight i space sin fraction numerator 13 pi over denominator 8 end fraction close parentheses

      • form a square

      • whose vertices lies on a circle of radius r equals 2

Worked Example

(a) Find both square roots of 5 plus 12 straight i, giving your answers in the form a plus b straight i.

k-4_ksf4_1-9-3-ib-aa-hl-de-moivres-theorem-we-solution-2

(b) Solve the equation z cubed equals negative 4 plus 4 square root of 3 straight i, giving your solutions in the form r space cis space theta.

1-9-3-ib-aa-hl-roots-of-cn-we-solution

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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.