Roots of Complex Numbers (DP IB Maths: AA HL)

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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 Tip

  • 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

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.