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 root of a complex number?

  • The square roots of a complex number will themselves be complex:

    • i.e. if z squared equals a plus b straight i then z equals c plus d straight i

  • We can then square (c plus d straight i) and equate it to the original complex number (a plus b straight i), as they both describe z squared:

    • a plus b straight i equals open parentheses c plus d straight i close parentheses squared

  • Then expand and simplify:

    • a plus b straight i equals c squared plus 2 c d straight i plus d squared straight i squared

    • a plus b straight i equals c squared plus 2 c d straight i minus d squared

  • As both sides are equal we are able to equate real and imaginary parts:

    • Equating the real components: a equals c squared minus d squared  (1)

    • Equating the imaginary components: b equals 2 c d  (2)

  • These equations can then be solved simultaneously to find the real and imaginary components of the square root

    • In general, we can rearrange (2) to make fraction numerator b over denominator 2 d end fraction equals c and then substitute into (1)

    • This will lead to a quartic equation in terms of d; which can be solved by making a substitution to turn it into a quadratic

  • The values of d can then be used to find the corresponding values of c, so we now have both components of both square roots (c plus d straight i)

  • Note that one root will be the negative of the other root

    • g. c plus d straight i and  negative c minus d straight i

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

  • De Moivre’s Theorem states that a complex number in modulus-argument form can be raised to the power of n by

    • Raising the modulus to the power of n and multiplying the argument by n

  • When in modulus-argument (polar) form de Moivre’s Theorem can then be used to find the roots of a complex number by

    • Taking the nth root of the modulus and dividing the argument by n

    • If z blank equals blank r left parenthesis cos invisible function application theta plus isin invisible function application theta right parenthesis blankthen blank n-th root of z blank equals blank open square brackets r open parentheses cos invisible function application left parenthesis theta plus 2 pi k right parenthesis plus isin invisible function application left parenthesis theta plus 2 pi k right parenthesis close parentheses close square brackets to the power of 1 over n end exponent

      • begin mathsize 16px style k equals 0 comma blank 1 comma blank 2 comma space horizontal ellipsis space comma space n minus 1 end style

      • Recall that adding 2π to the argument of a complex number does not change the complex number

      • Therefore we must consider how different arguments will give the same result

    • This can be rewritten as space n-th root of z blank equals blank r to the power of 1 over n end exponent open parentheses cos invisible function application left parenthesis fraction numerator theta blank plus blank 2 pi k over denominator n end fraction right parenthesis plus isin invisible function application left parenthesis fraction numerator theta blank plus blank 2 pi k over denominator n end fraction right parenthesis close parentheses

  • This can be written in exponential (Euler’s) form as 

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

  • The nth root of complex number will have n roots with the properties:

    • The modulus is n-th root of r for all roots

    • There will be n different arguments spaced at equal intervals on the unit circle

    • This creates some geometrically beautiful results:

      • The five roots of a complex number raised to the power 5 will create a regular pentagon on an Argand diagram

      • The eight roots of a complex number raised to the power 8 will create a regular octagon on an Argand diagram

      • The n roots of a complex number raised to the power n will create a regular n-sided polygon on an Argand diagram

  • Sometimes you may need to use your GDC to find the roots of a complex number

    • Using your GDC’s store function will help when entering complicated modulus and arguments

    • Make sure you choose the correct form to enter your complex number in

    • Your GDC should be able to give you the answer in your preferred form

Examiner Tips and Tricks

  • De Moivre's theorem makes finding roots of complex numbers very easy, but you must be confident converting from Cartesian form into Polar and Euler's form first

    • If you are in a calculator exam your GDC will be able to do this for you but you must clearly show how you got to your answer

    • You must also be prepared to do this by hand in a non-calculator paper

Worked Example

a) Find the square roots of 5 + 12i, giving your answers in the form a + bi.

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 answers in the form r cis θ.

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.

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