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Number and Algebra - Roots of z^n = a (HL)

Grade 12IB_AA

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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An equation of the form zn=az^n = a, where aa is a complex number, has exactly nn distinct roots in the complex plane.

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To find the roots, express the complex number aa in polar form r(cos⁡(θ+2kπ)+isin⁡(θ+2kπ))r(\cos(\theta + 2k\pi) + i\sin(\theta + 2k\pi)) or exponential form rei(θ+2kπ)re^{i(\theta + 2k\pi)}, where k∈Zk \in \mathbb{Z}.

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By De Moivre's Theorem, the nn-th roots are given by zk=r1/nei(θ+2kπn)z_k = r^{1/n} e^{i\left(\frac{\theta + 2k\pi}{n}\right)} for k=0,1,2,…,n−1k = 0, 1, 2, \dots, n-1.

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Geometrically, the roots of zn=az^n = a lie on a circle of radius ∣a∣n\sqrt[n]{|a|} and form the vertices of a regular nn-gon centered at the origin.

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Roots of unity are the solutions to zn=1z^n = 1. These are given by ωk=ei2kπn\omega_k = e^{i\frac{2k\pi}{n}}. If ω\omega is the root with the smallest positive argument, the roots can be written as 1,ω,ω2,…,ωn−11, \omega, \omega^2, \dots, \omega^{n-1}.

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The sum of the nn-th roots of any complex number aa is zero, provided n>1n > 1.

📐Formulae

zn=rei(θ+2kπ)z^n = r e^{i(\theta + 2k\pi)}

zk=r1/n(cos⁡(θ+2kπn)+isin⁡(θ+2kπn)), for k=0,1,…,n−1z_k = r^{1/n} \left( \cos\left( \frac{\theta + 2k\pi}{n} \right) + i \sin\left( \frac{\theta + 2k\pi}{n} \right) \right), \text{ for } k = 0, 1, \dots, n-1

zk=r1/nei(θ+2kπn)z_k = r^{1/n} e^{i\left( \frac{\theta + 2k\pi}{n} \right)}

∑k=0n−1zk=0(n≥2)\sum_{k=0}^{n-1} z_k = 0 \quad (n \geq 2)

💡Examples

Problem 1:

Solve the equation z3=8iz^3 = 8i, giving your answers in Cartesian form.

Solution:

  1. Express 8i8i in exponential form: 8i=8ei(π2+2kπ)8i = 8 e^{i(\frac{\pi}{2} + 2k\pi)}
  2. Apply the nn-th root formula for n=3n=3: zk=81/3ei(π/2+2kπ3)=2ei(π6+2kπ3)z_k = 8^{1/3} e^{i\left(\frac{\pi/2 + 2k\pi}{3}\right)} = 2 e^{i\left(\frac{\pi}{6} + \frac{2k\pi}{3}\right)}
  3. Find roots for k=0,1,2k=0, 1, 2:
  • For k=0k=0: z0=2eiπ6=2(cos⁡π6+isin⁡π6)=2(32+12i)=3+iz_0 = 2 e^{i\frac{\pi}{6}} = 2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}) = 2(\frac{\sqrt{3}}{2} + \frac{1}{2}i) = \sqrt{3} + i
  • For k=1k=1: z1=2ei5π6=2(cos⁡5π6+isin⁡5π6)=2(−32+12i)=−3+iz_1 = 2 e^{i\frac{5\pi}{6}} = 2(\cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6}) = 2(-\frac{\sqrt{3}}{2} + \frac{1}{2}i) = -\sqrt{3} + i
  • For k=2k=2: z2=2ei9π6=2ei3π2=2(0−i)=−2iz_2 = 2 e^{i\frac{9\pi}{6}} = 2 e^{i\frac{3\pi}{2}} = 2(0 - i) = -2i

Explanation:

We first convert the complex constant to polar/exponential form, ensuring we include the 2kπ2k\pi periodicity. Taking the cube root involves taking the cube root of the modulus and dividing the argument by 3. We then evaluate for successive integers of kk until we have 3 distinct roots.

Problem 2:

Find the roots of unity for z4=1z^4 = 1 and show they sum to zero.

Solution:

  1. Express 11 as ei(0+2kπ)e^{i(0 + 2k\pi)}.
  2. The roots are zk=ei2kπ4=eikπ2z_k = e^{i\frac{2k\pi}{4}} = e^{i\frac{k\pi}{2}} for k=0,1,2,3k=0, 1, 2, 3.
  3. Calculating the values:
  • k=0:z0=e0=1k=0: z_0 = e^0 = 1
  • k=1:z1=eiπ/2=ik=1: z_1 = e^{i\pi/2} = i
  • k=2:z2=eiπ=−1k=2: z_2 = e^{i\pi} = -1
  • k=3:z3=ei3π/2=−ik=3: z_3 = e^{i3\pi/2} = -i
  1. Sum of roots: 1+i+(−1)+(−i)=01 + i + (-1) + (-i) = 0.

Explanation:

Roots of unity are found by solving zn=1z^n = 1. They are spaced evenly around the unit circle. The sum of these roots is always zero because of the symmetry of the regular nn-gon they form.