Skip to main content
Mathematics LibreTexts

2.2: Polar Form of a Complex Number

  • Page ID
    23302
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    A point \((x,y)\) in the plane can be represented in polar form \((r,\theta)\) according to the relationships in Figure \(\PageIndex{1}\).

    im-polar.svg

    Figure \(\PageIndex{1}\): Polar coordinates of a point in the plane. (Copyright; author via source)

    Using these relationships, we can rewrite

    \[\begin{align*} x+yi &= r\cos(\theta) + r\sin(\theta) i\\ &= r(\cos(\theta) + i \sin(\theta))\text{.} \end{align*}\]

    This leads us to make the following definition. For any real number \(\theta\text{,}\) we define

    \[ e^{i\theta} = \cos(\theta) + i\sin(\theta)\text{.} \]

    For instance, \(e^{i\pi/2} = \cos(\dfrac{\pi}{2}) + i\sin(\dfrac{\pi}{2}) = 0 + i\cdot 1 = i\text{.}\)

    Similarly, \(e^{i0} = \cos(0) + i\sin(0) = 1\text{,}\) and it's a quick check to see that \(e^{i\pi} = -1\text{,}\) which leads to a simple equation involving the most famous numbers in mathematics (except \(8\)), truly an all-star equation:

    \[ e^{i \pi} + 1 = 0\text{.} \]

    If \(z = x+yi\) and \((x,y)\) has polar form \((r,\theta)\) then \(z = re^{i\theta}\) is called the polar form of \(z\text{.}\) The non-negative scalar \(|r|\) is the modulus of \(z\text{,}\) and the angle \(\theta\) is called the argument of \(z\), denoted \(\arg(z\)).

    Example \(\PageIndex{1}\): Exploring the Polar Form

    On the left side of the following diagram, we plot the points

    \(z = 2e^{i\pi/4}, w = 3e^{i\pi/2}, v = -2e^{i\pi/6}, u = 3e^{-i\pi/3}.\)

    im-polarform.svg

    To convert \(z = -3 + 4i\) to polar form, refer to the right side of the diagram. We note that \(r = \sqrt{9 + 16} = 5\text{,}\) and \(\tan(\alpha) =\dfrac{4}{3} \text{,}\) so \(\theta = \pi - \tan^{-1}(\dfrac{4}{3})\approx 2.21\) radians. Thus,

    \[ -3+4i = 5e^{i(\pi-\tan^{-1}(4/3))} \approx 5e^{2.21i}\text{.} \]

    Theorem \(\PageIndex{1}\)

    The product of two complex numbers in polar form is given by

    \[ \displaystyle re^{i\theta}\cdot se^{i\beta} = (rs)e^{i(\theta+\beta)}\text{.} \]

    Proof

    We use the definition of the complex exponential and some trigonometric identities. \[\begin{array} d re^{i\theta}\cdot se^{i\beta} &= r(\cos\theta + i\sin\theta)\cdot s(\cos\beta+i\sin\beta)\\ &= (rs)(\cos\theta + i\sin\theta)\cdot (\cos\beta+i\sin\beta)\\ &= rs[\cos\theta\cos\beta - \sin\theta\sin\beta + (\cos\theta\sin\beta+\sin\theta\cos\beta)i]\\ &= rs[\cos(\theta+\beta) + \sin(\theta+\beta)i]\\ &=rs[e^{i(\theta+\beta)}]\text{.} \end{array}\]

    Thus, the product of two complex numbers is obtained by multiplying their magnitudes and adding their arguments, and

    \[ \arg(zw) = \arg(z) + \arg(w)\text{,} \]

    where the equation is taken modulo \(2\pi\text{.}\) That is, depending on our choices for the arguments, we have \(\arg(vw) = \arg(v)+ \arg(w) + 2\pi k\) for some integer \(k\text{.}\)

    Example \(\PageIndex{2}\): Polar Form with \(r \geq 0\)

    When representing a complex number \(z\) in polar form as \(z = re^{i\theta}\text{,}\) we may assume that \(r\) is non-negative. If \(r \lt 0\text{,}\) then \[\begin{array}d re^{i\theta} &= - |r|e^{i\theta}\\ &= (e^{i\pi})\cdot |r| e^{i\theta} \;\; \text{since} \; -1 = e^{i\pi}\\ &= |r|e^{i(\theta+\pi)}, \;\; \text{by Theorem 2.2.1} \end{array}\] Thus, by adding \(\pi\) to the angle if necessary, we may always assume that \(z = re^{i\theta}\) where \(r\) is non-negative.

    Exercises

    Exercise \(\PageIndex{1}\)

    Convert the following points to polar form and plot them: \(3 + i\text{,}\) \(-1 - 2i\text{,}\) \(3 - 4i\text{,}\) \(7,002,001\text{,}\) and \(-4i\text{.}\)

    Exercise \(\PageIndex{2}\)

    Express the following points in Cartesian form and plot them: \(z = 2e^{i\pi/3}\text{,}\) \(w = -2e^{i\pi/4}\text{,}\) \(u = 4e^{i5\pi/3},\) and \(z\cdot u\text{.}\)

    Exercise \(\PageIndex{3}\)

    Modify the all-star equation to involve \(8\). In particular, write an expression involving \(e, i, \pi, 1,\) and \(8\), that equals \(0\). You may use no other numbers, and certainly not \(3\).

    Exercise \(\PageIndex{4}\)

    If \(z = re^{i\theta}\text{,}\) prove that \(\overline{z} = re^{-i\theta}\text{.}\)


    This page titled 2.2: Polar Form of a Complex Number is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Michael P. Hitchman via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

    • Was this article helpful?