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Prove some properties of nth roots of complex numbers

by RoRi

Let z = re^{i \theta} be a nonzero complex number where \theta = \arg (z). Then, let R = r^{\frac{1}{n}} and \alpha = \frac{\theta}{n} and define another complex number z_1 = Re^{i \alpha}. Now, let \varepsilon = e^{\frac{2 \pi i}{n}} for n \in \mathbb{Z}_{>0}.

  1. Prove that z_1^n = z. We say z_1 is an nth root of z.
  2. Prove that z has exactly n distinct nth roots given by

        \[ z_1, \varepsilon z_1, \varepsilon^2 z_1, \ldots, \varepsilon^{n-1} z_1, \]

    and that they are equally spaced on a circle of radius R.

  3. Find the three cube roots of i.
  4. Find the four fourth roots of i.
  5. Find the four fourth roots of -i.
  1. Proof. Using the definitions of R and \alpha we compute,

        \[ z_1^n = \left( Re^{i \alpha} \right)^n = (R)^n (e^{i \alpha})^n = (r^{\frac{1}{n}})^n (e^{\frac{i \theta}{n}} )^n = re^{i \theta} = z. \qquad \blacksquare \]

  2. Proof. From part (a) we know z_1 = Re^{i \alpha} is an nth root of z. Then, if 1 \leq m < n we have

        \[ \left( \varepsilon^m z_1 \right)^n = e^{\frac{2 \pi i m}{n} \cdot n} \cdot z_1^n = e^{2 \pi i m} \cdot z = z \]

    (since e^{2 \pi i m} = 1 for all m \in \mathbb{Z}_{>0}). Hence, \varepsilon^m z_1 is an nth root of z.

    Then, if k \in \mathbb{Z}_{>0} and k \geq n we have k = j \cdot n + m for 1 \leq m < n. This implies

        \[ \varepsilon^k = e^{2 \pi i(nj+m)} = 1 \cdot e^{2 \pi i m} = e^{2 \pi i m}, \]

    for 1 \leq m < n; hence, there are n-1 distinct values of \varepsilon^k. Therefore, z has exactly n distinct nth roots. By the Fundamental Theorem of Algebra we know it cannot have more than n roots; hence, we have shown that they are all of the form \varepsilon^m z_1. \qquad \blacksquare

  3. For the complex number i we have

        \[ i = e^{ \frac{\pi i}{2}} \quad \implies \quad r = 1, \ \ \theta = \frac{\pi}{2}. \]

    Therefore, from parts (a) and (b) we have

        \[ R = 1, \qquad \alpha = \frac{\pi}{6} \qquad \varepsilon = e^{\frac{2 \pi i}{3}}. \]

    Hence, the three cube roots of i are

        \begin{align*} z_1 &= e^{\frac{\pi i}{6}} \\[9pt] &= \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \\[9pt] &= \frac{1}{2} \sqrt{3} + \frac{1}{2} i. \\[9pt] z_2 &= e^{\frac{\pi i}{6}} \cdot e^{\frac{2 \pi i}{3}} = e^{\frac{5 \pi i}{6}} \\[9pt] &= \cos \frac{5 \pi}{6} + i \sin \frac{5 \pi}{6} \\[9pt] &= -\frac{1}{2} \sqrt{3} + \frac{1}{2} i. \\[9pt] z_3 &= e^{ \frac{\pi i}{6}} \cdot e^{\frac{4 \pi i}{3}} = e^{\frac{3 \pi i}{2}} \\[9pt] &= \cos \frac{3 \pi}{2} + i \sin \frac{3 \pi}{2} \\[9pt] &= -i. \end{align*}

  4. This time we have

        \[ R = 1, \quad \alpha = \frac{\pi}{8}, \quad \implies \quad \varepsilon = e^{\frac{\pi i}{2}}. \]

    Therefore the four fourth roots of i are

        \begin{align*} z_1 &= e^{\frac{\pi i}{8}} \\[9pt] &= \cos \frac{\pi}{8} + i \sin \frac{\pi}{8} \\[9pt] &= \frac{1}{2} \left( \sqrt{2 + \sqrt{2}} \right) + \frac{i}{2} \left( \sqrt{2 - \sqrt{2}} \right). \\[9pt] z_2 &= e^{\frac{5 \pi}{8}} \\[9pt] &= \cos \frac{5 \pi}{8} + i \sin \frac{5 \pi}{8} \\[9pt] &= -\frac{1}{2} \sqrt{2 - \sqrt{2}} + \frac{i}{2} \sqrt{2 + \sqrt{2}}. \\[9pt] z_3 &= e^{\frac{9 \pi i}{8}} \\[9pt] &= \cos \frac{9 \pi}{8} + i \sin \frac{9 \pi}{8} \\[9pt] &= -\frac{1}{2} \sqrt{2+ \sqrt{2}} - \frac{i}{2} \sqrt{2 - \sqrt{2}}. \\[9pt] z_4 &= e^{\frac{13 \pi i}{8}} \\[9pt] &= \cos \frac{13 \pi}{8} + i \sin \frac{13 \pi}{8} \\[9pt] &= \frac{1}{2}\sqrt{2 - \sqrt{2}} - \frac{i}{2} \sqrt{2 + \sqrt{2}}. \end{align*}

  5. For -i we have

        \[ -i = e^{\frac{3 \pi i}{2}} \quad \implies \quad r = 1, \ \ \theta = \frac{3 \pi}{2}. \]

    Therefore,

        \[ R = 1, \ \ \alpha = \frac{3 \pi}{8} \quad \implies \quad \varepsilon = e^{\pi i}{2}. \]

    Thus,

        \begin{align*} z_1 &= e^{\frac{3 \pi i}{8}} \\[9pt] &= \cos \frac{3 \pi}{8} + i \sin \frac{3 \pi}{8} \\[9pt] &= \frac{1}{2} \left( \sqrt{2 - \sqrt{2}} \right) + \frac{i}{2} \left( \sqrt{2 + \sqrt{2}} \right). \\[9pt] z_2 &= e^{\frac{7 \pi}{8}} \\[9pt] &= \cos \frac{7 \pi}{8} + i \sin \frac{7 \pi}{8} \\[9pt] &= -\frac{1}{2} \sqrt{2 + \sqrt{2}} + \frac{i}{2} \sqrt{2 - \sqrt{2}}. \\[9pt] z_3 &= e^{\frac{11 \pi i}{8}} \\[9pt] &= \cos \frac{11 \pi}{8} + i \sin \frac{11 \pi}{8} \\[9pt] &= \frac{1}{2} \sqrt{2 - \sqrt{2}} - \frac{i}{2} \sqrt{2 - \sqrt{2}}. \\[9pt] z_4 &= e^{\frac{15 \pi i}{8}} \\[9pt] &= \cos \frac{15 \pi}{8} + i \sin \frac{15 \pi}{8} \\[9pt] &= \frac{1}{2}\sqrt{2 + \sqrt{2}} - \frac{i}{2} \sqrt{2 - \sqrt{2}}. \end{align*}

3 comments

  1. Mohammad Azad says:

    To show that the n roots corresponding to powers of epsilon are actually distinct, you can assume that two of them are equal that is (q^k1)z1=(q^k2)z1 where q is epsilon and 0≤k1,k2≤n-1, then since z1≠0 we have q^k1=q^k2 which implies that
    exp(i2pi(k1-k2)/n)=1 ==> (k1-k2)/n=m ,an integer, but |k1-k2|0 there are integers t and s such that m=tn+s with 0≤s<n thus a=2tpi+2spi/n+d/n so
    z2=(r^1/n)(e^ia)=(r^1/n)(e^i2tpi)(e^i2spi/n)(e^id/n)=
    z1q^s which is one of the n roots we found since 0≤s<n.

    • Mohammad Azad says:

      Something went wrong, hopefully it works this time:
      To show that the n roots corresponding to powers of epsilon are actually distinct, you can assume that two of them are equal that is (q^k1)z1=(q^k2)z1 where q is epsilon and 0≤k1,k2≤n-1, then since z1≠0 we have q^k1=q^k2 which implies that
      exp(i2pi(k1-k2)/n)=1 ==> (k1-k2)/n=m ,an integer, but abs(k1-k2)0 there are integers t and s such that m=tn+s with 0≤s<n thus a=2tpi+2spi/n+d/n so
      z2=(r^1/n)(e^ia)=(r^1/n)(e^i2tpi)(e^i2spi/n)(e^id/n)=
      z1q^s which is one of the n roots we found since 0≤s<n.

      • Mohammad Azad says:

        It got messed up again, basically you take two of them with exponents k1 and k2 and show that k1=k2 using the fact that their difference cannot exceed n-1. To show there aren’t more roots you assume there is one.. you prove it has the same modulus.. you prove the difference of exponents must be 2mpi for some integer m and then use the division algorithm (exercise 10 page 36) to carry out the rest of the proof.

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