SB Cyclic Groups of Complex Numbers

Section 2.3 Cyclic Groups of Complex Numbers

View Source for section

<section>
    <title>Cyclic Groups of Complex Numbers</title>
    <p>
      The <term>complex numbers</term> are defined as
      <me>
        {\mathbb C} = \{ a + bi : a, b \in {\mathbb R} \}
      </me>,
      where <m>i^2 = -1</m>.
      If <m>z = a + bi</m>, then <m>a</m> is the <term>real part</term>
      of <m>z</m> and <m>b</m> is the
      <term>imaginary part</term> of <m>z</m>.
    </p>
    <p>
      To add two complex numbers <m>z=a+bi</m> and <m>w= c+di</m>,
      we just add the corresponding real and imaginary parts:
      <me>
        z + w=(a + bi ) + (c + di)  =  (a + c) + (b + d)i
      </me>.
      Remembering that <m>i^2 = -1</m>,
      we multiply complex numbers just like polynomials.
      The product of <m>z</m> and <m>w</m> is
      <me>
        (a + bi )(c + di)  =   ac + bdi^2 + adi + bci =  (ac -bd) +(ad + bc)i
      </me>.
    </p>
    <p>
      Every nonzero complex number
      <m>z = a +bi</m> has a multiplicative inverse;
      that is, there exists a <m>z^{-1} \in {\mathbb C}^\ast</m> such that <m>z z^{-1} = z^{-1} z = 1</m>.
      If <m>z = a + bi</m>, then
      <me>
        z^{-1} = \frac{a-bi}{ a^2 + b^2  }
      </me>.
      The <term>complex conjugate</term>
          <idx><h>Conjugate, complex</h></idx>
      of a complex number <m>z = a + bi</m> is defined to be <m>\overline{z} = a- bi</m>.
      The <term>absolute value</term> or <term>modulus</term>
      of  <m>z = a + bi</m>  is <m>|z| = \sqrt{a^2 + b^2}</m>.
    </p>
    <example xml:id="example-cyclic-complex-add">
      <title>Complex Number Operations</title>
      <p>
        Let <m>z = 2 + 3i</m> and <m>w = 1-2i</m>.
        Then
        <me>
          z + w = (2 + 3i) + (1 - 2i) = 3 + i
        </me>
        and
        <me>
          z  w = (2 + 3i)(1 - 2i ) = 8 - i
        </me>.
        Also,
        <md>
          <mrow>z^{-1} &amp; = \frac{2}{13} - \frac{3}{13}i</mrow>
          <mrow>|z| &amp; = \sqrt{13}</mrow>
          <mrow>\overline{z} &amp; = 2-3i</mrow>
        </md>.
      </p>
    </example>
    <figure xml:id="figure-rectcoord">
      <caption>Rectangular coordinates of a complex number</caption>

      <image xml:id="cyclic-complex-rectangular">
<latex-image>
\begin{tikzpicture}[scale=0.5]
\draw [-&gt;]  (0,-5) -- (0,5);
\draw  [-&gt;] (-8,0) -- (8,0);
\node [right] at (0,5) {$y$};
\node [below] at (8,0) {$x$};
\node [below] at (0.5,0) {$0$};
\filldraw[fill=black, draw=black] (2,3) circle (0.05cm);
\node [right] at (2,3) {$z_1 = 2 + 3i$};
\filldraw[fill=black, draw=black] (-3,2) circle (0.05cm);
\node [left] at (-3, 2) {$z_3 = -3 + 2i$};
\filldraw[fill=black, draw=black] (1,-2) circle (0.05cm);
\node [right] at (1, -2) {$z_2 = 1 -  2i$};
\end{tikzpicture}
</latex-image>
      </image>
    </figure>
    <p>
      There are several ways of graphically representing complex numbers.
      We can represent a complex number
      <m>z = a +bi</m> as an ordered pair on the <m>xy</m> plane where <m>a</m> is the <m>x</m>
      (or real)
      coordinate and <m>b</m> is the <m>y</m>
      (or imaginary)
      coordinate.
      This is called the <term>rectangular</term>
      or <term>Cartesian</term> representation.
      The rectangular representations of
      <m>z_1 = 2 + 3i</m>, <m>z_2 = 1 - 2i</m>,
      and <m>z_3 = - 3 + 2i</m> are depicted in <xref ref="figure-rectcoord" />.
    </p>
    <figure xml:id="figure-polarcoord">
      <caption>Polar coordinates of a complex number</caption>

      <image xml:id="cyclic-complex-polar">
<latex-image>
\begin{tikzpicture}[scale=0.5]
\draw [-&gt;]  (0,-5) -- (0,5);
\draw  [-&gt;] (-8,0) -- (8,0);
\node [right] at (0,5) {$y$};
\node [below] at (8,0) {$x$};
\node [below] at (0.5,0) {$0$};
\draw (0,0) -- (35:6);
\draw (2,0) arc (0:35:2);
\filldraw[fill=black, draw=black] (35:6) circle (0.05cm);
\node [right] at (35:6) {$a + bi$};
\node [above] at (35:3) {$r$};
\node [right] at (17:2) {$\theta$};
\end{tikzpicture}
</latex-image>
      </image>
    </figure>
    <p>
      Nonzero complex numbers can also be represented using
      <term>polar coordinates</term>.
      To specify  any nonzero point on the plane,
      it suffices to give an angle <m>\theta</m> from the positive <m>x</m> axis in the counterclockwise direction and a distance <m>r</m> from the origin,
      as in <xref ref="figure-polarcoord" />.
      We can see that
      <me>
        z = a + bi = r( \cos \theta + i \sin \theta)
      </me>.
      Hence,
      <me>
        r = |z| = \sqrt{a^2 + b^2}
      </me>
      and
      <md>
        <mrow>a &amp; = r \cos \theta</mrow>
        <mrow>b &amp; = r \sin \theta</mrow>
      </md>.
      We sometimes abbreviate <m>r( \cos \theta + i \sin \theta)</m> as <m>r \cis \theta</m>.
      To assure that the representation of <m>z</m> is  well-defined,
      we also require that <m>0^{\circ} \leq \theta \lt 360^{\circ}</m>.
      If the measurement is in radians,
      then <m>0 \leq \theta \lt2 \pi</m>.
      <notation>
        <usage><m>\cis \theta</m></usage>
        <description><m>\cos \theta + i \sin \theta</m></description>
      </notation>
    </p>
    <example xml:id="example-cyclic-polar">
      <title>Complex Numbers in Polar Form</title>
      <p>
        Suppose that <m>z = 2 \cis  60^{\circ}</m>.
        Then
        <me>
          a = 2 \cos 60^{\circ} = 1
        </me>
        and
        <me>
          b = 2 \sin 60^{\circ} = \sqrt{3}
        </me>.
        Hence, the rectangular representation is <m>z = 1+\sqrt{3}\, i</m>.
      </p>
      <p>
        Conversely, if we are given a rectangular representation of a complex number,
        it is often useful to know the number's polar representation.
        If <m>z = 3 \sqrt{2} - 3 \sqrt{2}\, i</m>, then
        <me>
          r = \sqrt{a^2 + b^2} = \sqrt{36 } = 6
        </me>
        and
        <me>
          \theta = \arctan \left( \frac{b}{a} \right) = \arctan( - 1) = 315^{\circ}
        </me>,
        so <m>3 \sqrt{2} - 3 \sqrt{2}\, i=6 \cis  315^{\circ}</m>.
      </p>
    </example>
    <p>
      The polar representation of a complex number makes it easy to find products and powers of complex numbers.
      The proof of the following proposition is straightforward and is left as an exercise.
    </p>
    <proposition xml:id="proposition-polar-mult">
      <statement>
        <p>
          Let <m>z = r \cis \theta</m> and
          <m>w = s \cis \phi</m> be two nonzero complex numbers.
          Then
          <me>
            zw = r s \cis( \theta + \phi)
          </me>.
        </p>
      </statement>
    </proposition>
    <example xml:id="example-cyclic-polar-mult">
      <title>Multiplication of Complex Numbers in Polar Form</title>
      <p>
        If <m>z = 3 \cis( \pi / 3 )</m> and <m>w = 2 \cis(\pi / 6 )</m>,
        then <m>zw = 6 \cis( \pi / 2 ) = 6i</m>.
      </p>
    </example>
    <theorem xml:id="theorem-demoivre">
      <title>DeMoivre</title>
      <idx><h>DeMoivre's Theorem</h></idx>
      <statement>
        <p>
          Let <m>z = r \cis  \theta</m> be a nonzero complex number.
          Then
          <me>
            [r \cis \theta  ]^n = r^n \cis( n \theta)
          </me>
          for <m>n = 1, 2, \ldots</m>.
        </p>
      </statement>
      <proof>
        <p>
          We will use induction on <m>n</m>
          (see <xref ref="section-math-induction" />).
          For <m>n = 1</m> the theorem is trivial.
          Assume that the theorem is true for all <m>k</m> such that <m>1  \leq k \leq n</m>.
          Then
          <md>
            <mrow>z^{n+1} &amp; = z^n z</mrow>
            <mrow>&amp; = r^n( \cos  n \theta + i \sin n \theta ) r( \cos \theta + i\sin \theta )</mrow>
            <mrow>&amp; = r^{n+1} [( \cos n \theta \cos \theta - \sin n \theta \sin \theta ) + i ( \sin n \theta \cos \theta + \cos n \theta \sin \theta)]</mrow>
            <mrow>&amp; = r^{n+1} [ \cos( n \theta + \theta) + i \sin( n \theta + \theta) ]</mrow>
            <mrow>&amp; = r^{n+1} [ \cos( n +1) \theta + i \sin( n+1) \theta  ]</mrow>
          </md>.
        </p>
      </proof>
    </theorem>
    <example>
      <title>Powers of Complex Numbers</title>
      <p>
        Suppose that <m>z= 1+i</m> and we wish to compute <m>z^{10}</m>.
        Rather than computing <m>(1 + i)^{10}</m> directly,
        it is much easier to switch to polar coordinates and calculate <m>z^{10}</m> using DeMoivre's Theorem:
        <md>
          <mrow>z^{10} &amp; = (1+i)^{10}</mrow>
          <mrow>&amp; = \left( \sqrt{2} \cis \left( \frac{\pi }{4} \right) \right)^{10}</mrow>
          <mrow>&amp; = ( \sqrt{2}\, )^{10} \cis \left( \frac{5\pi }{2} \right)</mrow>
          <mrow>&amp; = 32  \cis \left( \frac{\pi }{2} \right)</mrow>
          <mrow>&amp; = 32i</mrow>
        </md>.
      </p>
    </example>
    <p>
      The multiplicative group of the complex numbers,
      <m>{\mathbb C}^*</m>, possesses some interesting subgroups.
      Whereas <m>{\mathbb Q}^*</m> and
      <m>{\mathbb R}^*</m> have no interesting subgroups of finite order,
      <m>{\mathbb C}^*</m> has many.
      We first consider the <term>circle group</term>,
          <idx><h>Group</h><h>circle</h></idx>
      <me>
        {\mathbb T} = \{ z \in {\mathbb C} : |z| = 1 \}
      </me>.
      The following proposition is a direct result of <xref ref="proposition-polar-mult" />.
      <notation>
        <usage><m>\mathbb T</m></usage>
        <description>the circle group</description>
      </notation>
    </p>
    <proposition>
      <statement>
        <p>
          The circle group is a subgroup of <m>{\mathbb C}^*</m>.
        </p>
      </statement>
    </proposition>
    <p>
      Although the circle group has infinite order,
      it has many interesting finite subgroups.
      Suppose that <m>H = \{ 1, -1, i, -i \}</m>.
      Then <m>H</m> is a subgroup of the circle group.
      Also, <m>1</m>, <m>-1</m>, <m>i</m>,
      and <m>-i</m> are exactly those complex numbers that satisfy the equation <m>z^4 = 1</m>.
      The complex numbers satisfying the equation <m>z^n=1</m> are called the
      <term><m>n</m>th roots of unity</term>.
          <idx><h><m>n</m>th root of unity</h></idx>
    </p>
    <theorem>
      <statement>
        <p>
          If <m>z^n = 1</m>, then the <m>n</m>th roots of unity are
          <me>
            z = \cis\left( \frac{2 k \pi}{n } \right)
          </me>,
          where <m>k = 0, 1, \ldots, n-1</m>.
          Furthermore,
          the <m>n</m>th roots of unity form a cyclic subgroup of <m>{\mathbb T}</m> of order <m>n</m>
        </p>
      </statement>
      <proof>
        <p>
          By DeMoivre's Theorem,
          <me>
            z^n = \cis \left( n \frac{2 k \pi}{n } \right) = \cis( 2 k \pi ) = 1
          </me>.
          The <m>z</m>'s are distinct since the numbers
          <m>2 k \pi /n</m> are all distinct and are greater than or equal to 0 but less than <m>2 \pi</m>.
          We will leave the proof that the <m>n</m>th roots of unity form a cyclic subgroup of <m>{\mathbb T}</m> as an exercise.
        </p>
      </proof>
    </theorem>
    <p>
      A generator for the group of the <m>n</m>th roots of unity is called a
      <term>primitive <m>n</m>th root of unity</term>.
          <idx><h>Primitive <m>n</m>th root of unity</h></idx>
    </p>
    <example xml:id="example-roots_unity">
      <title>Roots of Unity</title>
      <p>
        The 8th roots of unity can be represented as eight equally spaced points on the unit circle
        (<xref ref="figure-rtsunity" />).
        The primitive 8th roots of unity are
        <md>
          <mrow>\omega &amp; = \frac{\sqrt{2}}{2}  + \frac{\sqrt{2}}{2} i</mrow>
          <mrow>\omega^3 &amp; = -\frac{\sqrt{2}}{2}  + \frac{\sqrt{2}}{2} i</mrow>
          <mrow>\omega^5 &amp; = -\frac{\sqrt{2}}{2}  - \frac{\sqrt{2}}{2} i</mrow>
          <mrow>\omega^7 &amp; = \frac{\sqrt{2}}{2}  - \frac{\sqrt{2}}{2}i</mrow>
        </md>.
      </p>
    </example>
    <figure xml:id="figure-rtsunity">
      <caption>8th roots of unity</caption>

      <image xml:id="cyclic-roots-unity">
<latex-image>% Cyclic 8-th roots of unity
% NB: comments are in TeX, not XML
%
\begin{tikzpicture}[scale=1.65]

\draw [-&gt;]  (0,-1.5) -- (0,1.5);
\draw  [-&gt;] (-1.75,0) -- (1.75,0);
\node [right] at (0,1.5) {$y$};
\node [below] at (1.75,0) {$x$};
\node [below] at (0.1,0) {$0$};

\draw (0,0) circle (1);

\filldraw[fill=black, draw=black] (0:1) circle (0.03);
\filldraw[fill=black, draw=black] (45:1) circle (0.03);
\filldraw[fill=black, draw=black] (90:1) circle (0.03);
\filldraw[fill=black, draw=black] (135:1) circle (0.03);
\filldraw[fill=black, draw=black] (180:1) circle (0.03);
\filldraw[fill=black, draw=black] (225:1) circle (0.03);
\filldraw[fill=black, draw=black] (270:1) circle (0.03);
\filldraw[fill=black, draw=black] (315:1) circle (0.03);


\node [right] at (1,-0.15) {1};
\node [right] at (45:1) {$\omega$};
\node [left] at (0,1.15) {$i$};
\node [left] at (135:1) {$\omega^3$};
\node [left] at (-1,-0.15) {$-1$};
\node [left] at (225:1) {$\omega^5$};
\node [left] at (0,-1.15) {$-i$};
\node [right] at (315:1) {$\omega^7$};

\end{tikzpicture}
</latex-image>
      </image>
    </figure>
    <p>
      We interrupt this exposition to repeat the previous diagram,
      wrapped as different figure with a different caption.
      The TikZ code to produce these diagrams lives in an external file,
      <c>tikz/cyclic-roots-unity.tex</c>,
      which is pure text,
      freed from any need to format for XML processing.
      So, in particular,
      there is no need to escape ampersands and angle brackets,
      nor is there employment of the <c>CDATA</c> mechanism.
      But the real value is that there is just one version to edit,
      and any changes will be reflected in both copies.
    </p>
    <figure xml:id="figure-rtsunity-copy">
      <caption>Repeat: 8th roots of unity</caption>
      <image xml:id="cyclic-roots-unity-copy">
<latex-image>% Cyclic 8-th roots of unity
% NB: comments are in TeX, not XML
%
\begin{tikzpicture}[scale=1.65]

\draw [-&gt;]  (0,-1.5) -- (0,1.5);
\draw  [-&gt;] (-1.75,0) -- (1.75,0);
\node [right] at (0,1.5) {$y$};
\node [below] at (1.75,0) {$x$};
\node [below] at (0.1,0) {$0$};

\draw (0,0) circle (1);

\filldraw[fill=black, draw=black] (0:1) circle (0.03);
\filldraw[fill=black, draw=black] (45:1) circle (0.03);
\filldraw[fill=black, draw=black] (90:1) circle (0.03);
\filldraw[fill=black, draw=black] (135:1) circle (0.03);
\filldraw[fill=black, draw=black] (180:1) circle (0.03);
\filldraw[fill=black, draw=black] (225:1) circle (0.03);
\filldraw[fill=black, draw=black] (270:1) circle (0.03);
\filldraw[fill=black, draw=black] (315:1) circle (0.03);


\node [right] at (1,-0.15) {1};
\node [right] at (45:1) {$\omega$};
\node [left] at (0,1.15) {$i$};
\node [left] at (135:1) {$\omega^3$};
\node [left] at (-1,-0.15) {$-1$};
\node [left] at (225:1) {$\omega^5$};
\node [left] at (0,-1.15) {$-i$};
\node [right] at (315:1) {$\omega^7$};

\end{tikzpicture}
</latex-image>
      </image>
    </figure>
  </section>

The complex numbers are defined as

\begin{equation*} {\mathbb C} = \{ a + bi : a, b \in {\mathbb R} \}\text{,} \end{equation*}

where \(i^2 = -1\text{.}\) If \(z = a + bi\text{,}\) then \(a\) is the real part of \(z\) and \(b\) is the imaginary part of \(z\text{.}\)

To add two complex numbers \(z=a+bi\) and \(w= c+di\text{,}\) we just add the corresponding real and imaginary parts:

\begin{equation*} z + w=(a + bi ) + (c + di) = (a + c) + (b + d)i\text{.} \end{equation*}

Remembering that \(i^2 = -1\text{,}\) we multiply complex numbers just like polynomials. The product of \(z\) and \(w\) is

\begin{equation*} (a + bi )(c + di) = ac + bdi^2 + adi + bci = (ac -bd) +(ad + bc)i\text{.} \end{equation*}

Every nonzero complex number \(z = a +bi\) has a multiplicative inverse; that is, there exists a \(z^{-1} \in {\mathbb C}^\ast\) such that \(z z^{-1} = z^{-1} z = 1\text{.}\) If \(z = a + bi\text{,}\) then

\begin{equation*} z^{-1} = \frac{a-bi}{ a^2 + b^2 }\text{.} \end{equation*}

The complex conjugate of a complex number \(z = a + bi\) is defined to be \(\overline{z} = a- bi\text{.}\) The absolute value or modulus of \(z = a + bi\) is \(|z| = \sqrt{a^2 + b^2}\text{.}\)

Example 2.3.1. Complex Number Operations.

View Source for example

<example xml:id="example-cyclic-complex-add">
  <title>Complex Number Operations</title>
  <p>
    Let <m>z = 2 + 3i</m> and <m>w = 1-2i</m>.
    Then
    <me>
      z + w = (2 + 3i) + (1 - 2i) = 3 + i
    </me>
    and
    <me>
      z  w = (2 + 3i)(1 - 2i ) = 8 - i
    </me>.
    Also,
    <md>
      <mrow>z^{-1} &amp; = \frac{2}{13} - \frac{3}{13}i</mrow>
      <mrow>|z| &amp; = \sqrt{13}</mrow>
      <mrow>\overline{z} &amp; = 2-3i</mrow>
    </md>.
  </p>
</example>

Let \(z = 2 + 3i\) and \(w = 1-2i\text{.}\) Then

\begin{equation*} z + w = (2 + 3i) + (1 - 2i) = 3 + i \end{equation*}

and

\begin{equation*} z w = (2 + 3i)(1 - 2i ) = 8 - i\text{.} \end{equation*}

Also,

\begin{align*} z^{-1} & = \frac{2}{13} - \frac{3}{13}i\\ |z| & = \sqrt{13}\\ \overline{z} & = 2-3i\text{.} \end{align*}

Figure 2.3.2. Rectangular coordinates of a complex number

There are several ways of graphically representing complex numbers. We can represent a complex number \(z = a +bi\) as an ordered pair on the \(xy\) plane where \(a\) is the \(x\) (or real) coordinate and \(b\) is the \(y\) (or imaginary) coordinate. This is called the rectangular or Cartesian representation. The rectangular representations of \(z_1 = 2 + 3i\text{,}\) \(z_2 = 1 - 2i\text{,}\) and \(z_3 = - 3 + 2i\) are depicted in Figure 2.3.2.

Figure 2.3.3. Polar coordinates of a complex number

Nonzero complex numbers can also be represented using polar coordinates. To specify any nonzero point on the plane, it suffices to give an angle \(\theta\) from the positive \(x\) axis in the counterclockwise direction and a distance \(r\) from the origin, as in Figure 2.3.3. We can see that

\begin{equation*} z = a + bi = r( \cos \theta + i \sin \theta)\text{.} \end{equation*}

Hence,

\begin{equation*} r = |z| = \sqrt{a^2 + b^2} \end{equation*}

and

\begin{align*} a & = r \cos \theta\\ b & = r \sin \theta\text{.} \end{align*}

We sometimes abbreviate \(r( \cos \theta + i \sin \theta)\) as \(r \cis \theta\text{.}\) To assure that the representation of \(z\) is well-defined, we also require that \(0^{\circ} \leq \theta \lt 360^{\circ}\text{.}\) If the measurement is in radians, then \(0 \leq \theta \lt2 \pi\text{.}\)

Example 2.3.4. Complex Numbers in Polar Form.

View Source for example

<example xml:id="example-cyclic-polar">
  <title>Complex Numbers in Polar Form</title>
  <p>
    Suppose that <m>z = 2 \cis  60^{\circ}</m>.
    Then
    <me>
      a = 2 \cos 60^{\circ} = 1
    </me>
    and
    <me>
      b = 2 \sin 60^{\circ} = \sqrt{3}
    </me>.
    Hence, the rectangular representation is <m>z = 1+\sqrt{3}\, i</m>.
  </p>
  <p>
    Conversely, if we are given a rectangular representation of a complex number,
    it is often useful to know the number's polar representation.
    If <m>z = 3 \sqrt{2} - 3 \sqrt{2}\, i</m>, then
    <me>
      r = \sqrt{a^2 + b^2} = \sqrt{36 } = 6
    </me>
    and
    <me>
      \theta = \arctan \left( \frac{b}{a} \right) = \arctan( - 1) = 315^{\circ}
    </me>,
    so <m>3 \sqrt{2} - 3 \sqrt{2}\, i=6 \cis  315^{\circ}</m>.
  </p>
</example>

Suppose that \(z = 2 \cis 60^{\circ}\text{.}\) Then

\begin{equation*} a = 2 \cos 60^{\circ} = 1 \end{equation*}

and

\begin{equation*} b = 2 \sin 60^{\circ} = \sqrt{3}\text{.} \end{equation*}

Hence, the rectangular representation is \(z = 1+\sqrt{3}\, i\text{.}\)

Conversely, if we are given a rectangular representation of a complex number, it is often useful to know the number’s polar representation. If \(z = 3 \sqrt{2} - 3 \sqrt{2}\, i\text{,}\) then

\begin{equation*} r = \sqrt{a^2 + b^2} = \sqrt{36 } = 6 \end{equation*}

and

\begin{equation*} \theta = \arctan \left( \frac{b}{a} \right) = \arctan( - 1) = 315^{\circ}\text{,} \end{equation*}

so \(3 \sqrt{2} - 3 \sqrt{2}\, i=6 \cis 315^{\circ}\text{.}\)

The polar representation of a complex number makes it easy to find products and powers of complex numbers. The proof of the following proposition is straightforward and is left as an exercise.

Proposition 2.3.5.

View Source for proposition

<proposition xml:id="proposition-polar-mult">
  <statement>
    <p>
      Let <m>z = r \cis \theta</m> and
      <m>w = s \cis \phi</m> be two nonzero complex numbers.
      Then
      <me>
        zw = r s \cis( \theta + \phi)
      </me>.
    </p>
  </statement>
</proposition>

Let \(z = r \cis \theta\) and \(w = s \cis \phi\) be two nonzero complex numbers. Then

\begin{equation*} zw = r s \cis( \theta + \phi)\text{.} \end{equation*}

Example 2.3.6. Multiplication of Complex Numbers in Polar Form.

View Source for example

<example xml:id="example-cyclic-polar-mult">
  <title>Multiplication of Complex Numbers in Polar Form</title>
  <p>
    If <m>z = 3 \cis( \pi / 3 )</m> and <m>w = 2 \cis(\pi / 6 )</m>,
    then <m>zw = 6 \cis( \pi / 2 ) = 6i</m>.
  </p>
</example>

If \(z = 3 \cis( \pi / 3 )\) and \(w = 2 \cis(\pi / 6 )\text{,}\) then \(zw = 6 \cis( \pi / 2 ) = 6i\text{.}\)

Theorem 2.3.7. DeMoivre.

View Source for theorem

<theorem xml:id="theorem-demoivre">
  <title>DeMoivre</title>
  <idx><h>DeMoivre's Theorem</h></idx>
  <statement>
    <p>
      Let <m>z = r \cis  \theta</m> be a nonzero complex number.
      Then
      <me>
        [r \cis \theta  ]^n = r^n \cis( n \theta)
      </me>
      for <m>n = 1, 2, \ldots</m>.
    </p>
  </statement>
  <proof>
    <p>
      We will use induction on <m>n</m>
      (see <xref ref="section-math-induction" />).
      For <m>n = 1</m> the theorem is trivial.
      Assume that the theorem is true for all <m>k</m> such that <m>1  \leq k \leq n</m>.
      Then
      <md>
        <mrow>z^{n+1} &amp; = z^n z</mrow>
        <mrow>&amp; = r^n( \cos  n \theta + i \sin n \theta ) r( \cos \theta + i\sin \theta )</mrow>
        <mrow>&amp; = r^{n+1} [( \cos n \theta \cos \theta - \sin n \theta \sin \theta ) + i ( \sin n \theta \cos \theta + \cos n \theta \sin \theta)]</mrow>
        <mrow>&amp; = r^{n+1} [ \cos( n \theta + \theta) + i \sin( n \theta + \theta) ]</mrow>
        <mrow>&amp; = r^{n+1} [ \cos( n +1) \theta + i \sin( n+1) \theta  ]</mrow>
      </md>.
    </p>
  </proof>
</theorem>

Let \(z = r \cis \theta\) be a nonzero complex number. Then

\begin{equation*} [r \cis \theta ]^n = r^n \cis( n \theta) \end{equation*}

for \(n = 1, 2, \ldots\text{.}\)

Proof.

View Source for proof

<proof>
  <p>
    We will use induction on <m>n</m>
    (see <xref ref="section-math-induction" />).
    For <m>n = 1</m> the theorem is trivial.
    Assume that the theorem is true for all <m>k</m> such that <m>1  \leq k \leq n</m>.
    Then
    <md>
      <mrow>z^{n+1} &amp; = z^n z</mrow>
      <mrow>&amp; = r^n( \cos  n \theta + i \sin n \theta ) r( \cos \theta + i\sin \theta )</mrow>
      <mrow>&amp; = r^{n+1} [( \cos n \theta \cos \theta - \sin n \theta \sin \theta ) + i ( \sin n \theta \cos \theta + \cos n \theta \sin \theta)]</mrow>
      <mrow>&amp; = r^{n+1} [ \cos( n \theta + \theta) + i \sin( n \theta + \theta) ]</mrow>
      <mrow>&amp; = r^{n+1} [ \cos( n +1) \theta + i \sin( n+1) \theta  ]</mrow>
    </md>.
  </p>
</proof>

We will use induction on \(n\) (see Section I.2.1). For \(n = 1\) the theorem is trivial. Assume that the theorem is true for all \(k\) such that \(1 \leq k \leq n\text{.}\) Then

\begin{align*} z^{n+1} & = z^n z\\ & = r^n( \cos n \theta + i \sin n \theta ) r( \cos \theta + i\sin \theta )\\ & = r^{n+1} [( \cos n \theta \cos \theta - \sin n \theta \sin \theta ) + i ( \sin n \theta \cos \theta + \cos n \theta \sin \theta)]\\ & = r^{n+1} [ \cos( n \theta + \theta) + i \sin( n \theta + \theta) ]\\ & = r^{n+1} [ \cos( n +1) \theta + i \sin( n+1) \theta ]\text{.} \end{align*}

Example 2.3.8. Powers of Complex Numbers.

View Source for example

<example>
  <title>Powers of Complex Numbers</title>
  <p>
    Suppose that <m>z= 1+i</m> and we wish to compute <m>z^{10}</m>.
    Rather than computing <m>(1 + i)^{10}</m> directly,
    it is much easier to switch to polar coordinates and calculate <m>z^{10}</m> using DeMoivre's Theorem:
    <md>
      <mrow>z^{10} &amp; = (1+i)^{10}</mrow>
      <mrow>&amp; = \left( \sqrt{2} \cis \left( \frac{\pi }{4} \right) \right)^{10}</mrow>
      <mrow>&amp; = ( \sqrt{2}\, )^{10} \cis \left( \frac{5\pi }{2} \right)</mrow>
      <mrow>&amp; = 32  \cis \left( \frac{\pi }{2} \right)</mrow>
      <mrow>&amp; = 32i</mrow>
    </md>.
  </p>
</example>

Suppose that \(z= 1+i\) and we wish to compute \(z^{10}\text{.}\) Rather than computing \((1 + i)^{10}\) directly, it is much easier to switch to polar coordinates and calculate \(z^{10}\) using DeMoivre’s Theorem:

\begin{align*} z^{10} & = (1+i)^{10}\\ & = \left( \sqrt{2} \cis \left( \frac{\pi }{4} \right) \right)^{10}\\ & = ( \sqrt{2}\, )^{10} \cis \left( \frac{5\pi }{2} \right)\\ & = 32 \cis \left( \frac{\pi }{2} \right)\\ & = 32i\text{.} \end{align*}

The multiplicative group of the complex numbers, \({\mathbb C}^*\text{,}\) possesses some interesting subgroups. Whereas \({\mathbb Q}^*\) and \({\mathbb R}^*\) have no interesting subgroups of finite order, \({\mathbb C}^*\) has many. We first consider the circle group,

\begin{equation*} {\mathbb T} = \{ z \in {\mathbb C} : |z| = 1 \}\text{.} \end{equation*}

The following proposition is a direct result of Proposition 2.3.5.

Proposition 2.3.9.

View Source for proposition

<proposition>
  <statement>
    <p>
      The circle group is a subgroup of <m>{\mathbb C}^*</m>.
    </p>
  </statement>
</proposition>

The circle group is a subgroup of \({\mathbb C}^*\text{.}\)

Although the circle group has infinite order, it has many interesting finite subgroups. Suppose that \(H = \{ 1, -1, i, -i \}\text{.}\) Then \(H\) is a subgroup of the circle group. Also, \(1\text{,}\) \(-1\text{,}\) \(i\text{,}\) and \(-i\) are exactly those complex numbers that satisfy the equation \(z^4 = 1\text{.}\) The complex numbers satisfying the equation \(z^n=1\) are called the \(n\)th roots of unity.

Theorem 2.3.10.

View Source for theorem

<theorem>
  <statement>
    <p>
      If <m>z^n = 1</m>, then the <m>n</m>th roots of unity are
      <me>
        z = \cis\left( \frac{2 k \pi}{n } \right)
      </me>,
      where <m>k = 0, 1, \ldots, n-1</m>.
      Furthermore,
      the <m>n</m>th roots of unity form a cyclic subgroup of <m>{\mathbb T}</m> of order <m>n</m>
    </p>
  </statement>
  <proof>
    <p>
      By DeMoivre's Theorem,
      <me>
        z^n = \cis \left( n \frac{2 k \pi}{n } \right) = \cis( 2 k \pi ) = 1
      </me>.
      The <m>z</m>'s are distinct since the numbers
      <m>2 k \pi /n</m> are all distinct and are greater than or equal to 0 but less than <m>2 \pi</m>.
      We will leave the proof that the <m>n</m>th roots of unity form a cyclic subgroup of <m>{\mathbb T}</m> as an exercise.
    </p>
  </proof>
</theorem>

If \(z^n = 1\text{,}\) then the \(n\)th roots of unity are

\begin{equation*} z = \cis\left( \frac{2 k \pi}{n } \right)\text{,} \end{equation*}

where \(k = 0, 1, \ldots, n-1\text{.}\) Furthermore, the \(n\)th roots of unity form a cyclic subgroup of \({\mathbb T}\) of order \(n\)

Proof.

View Source for proof

<proof>
  <p>
    By DeMoivre's Theorem,
    <me>
      z^n = \cis \left( n \frac{2 k \pi}{n } \right) = \cis( 2 k \pi ) = 1
    </me>.
    The <m>z</m>'s are distinct since the numbers
    <m>2 k \pi /n</m> are all distinct and are greater than or equal to 0 but less than <m>2 \pi</m>.
    We will leave the proof that the <m>n</m>th roots of unity form a cyclic subgroup of <m>{\mathbb T}</m> as an exercise.
  </p>
</proof>

By DeMoivre’s Theorem,

\begin{equation*} z^n = \cis \left( n \frac{2 k \pi}{n } \right) = \cis( 2 k \pi ) = 1\text{.} \end{equation*}

The \(z\)’s are distinct since the numbers \(2 k \pi /n\) are all distinct and are greater than or equal to 0 but less than \(2 \pi\text{.}\) We will leave the proof that the \(n\)th roots of unity form a cyclic subgroup of \({\mathbb T}\) as an exercise.

A generator for the group of the \(n\)th roots of unity is called a primitive \(n\)th root of unity.

Example 2.3.11. Roots of Unity.

View Source for example

<example xml:id="example-roots_unity">
  <title>Roots of Unity</title>
  <p>
    The 8th roots of unity can be represented as eight equally spaced points on the unit circle
    (<xref ref="figure-rtsunity" />).
    The primitive 8th roots of unity are
    <md>
      <mrow>\omega &amp; = \frac{\sqrt{2}}{2}  + \frac{\sqrt{2}}{2} i</mrow>
      <mrow>\omega^3 &amp; = -\frac{\sqrt{2}}{2}  + \frac{\sqrt{2}}{2} i</mrow>
      <mrow>\omega^5 &amp; = -\frac{\sqrt{2}}{2}  - \frac{\sqrt{2}}{2} i</mrow>
      <mrow>\omega^7 &amp; = \frac{\sqrt{2}}{2}  - \frac{\sqrt{2}}{2}i</mrow>
    </md>.
  </p>
</example>

The 8th roots of unity can be represented as eight equally spaced points on the unit circle (Figure 2.3.12). The primitive 8th roots of unity are

\begin{align*} \omega & = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i\\ \omega^3 & = -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i\\ \omega^5 & = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i\\ \omega^7 & = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i\text{.} \end{align*}

We interrupt this exposition to repeat the previous diagram, wrapped as different figure with a different caption. The TikZ code to produce these diagrams lives in an external file, tikz/cyclic-roots-unity.tex, which is pure text, freed from any need to format for XML processing. So, in particular, there is no need to escape ampersands and angle brackets, nor is there employment of the CDATA mechanism. But the real value is that there is just one version to edit, and any changes will be reflected in both copies.

Figure 2.3.13. Repeat: 8th roots of unity

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