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

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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.