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Section2.2Polar Form of a Complex Number

A point \((x,y)\) in the plane can be represented in polar form \((r,\theta)\) according to the relationships in Figure 2.2.1.

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Figure2.2.1Polar coordinates of a point in the plane

Using these relationships, we can rewrite \begin{align*} x+yi \amp = r\cos(\theta) + r\sin(\theta) i\\ \amp = r(\cos(\theta) + i \sin(\theta)). \end{align*}


For any real number \(\theta\text{,}\) we define \begin{equation*} e^{i\theta} = \cos(\theta) + i\sin(\theta). \end{equation*}

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

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, see [29]), 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 \(\arg(z\)).

Example2.2.3Exploring the polar form

On the left side of Figure 2.2.4, we plot the points \(z = 2e^{i\pi/4}, w = 3e^{i\pi/2}, v = -2e^{i\pi/6}, u = 3e^{-i\pi}.\)

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Figure2.2.4The polar form of a complex number.

To convert \(z = -3 + 4i\) to polar form, refer to the right side of Figure 2.2.4. We note that \(r = \sqrt{9 + 16} = 5\text{,}\) and \(\tan(\alpha) = 4/3\text{,}\) so \(\theta = \pi - \tan^{-1}(4/3)\approx 2.21\) radians. Thus, \begin{equation*} -3+4i = 5e^{i(\pi-\tan^{-1}(4/3))} \approx 5e^{2.21i}. \end{equation*}

Thus, the product of two complex numbers is obtained by multiplying their magnitudes and adding their arguments, and \begin{equation*} \arg(zw) = \arg(z) + \arg(w), \end{equation*} where the equation is taken modulo \(2\pi\text{.}\) That is, depending on our choices for the arguments, we may have \(\arg(vw) = \arg(v)+ \arg(w) + 2\pi k\) for some integer \(k\text{.}\)

Example2.2.6Expressing \(z\) in 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{align*} re^{i\theta} \amp = - |r|e^{i\theta}\\ \amp = (e^{i\pi})\cdot |r| e^{i\theta} ~~\text{since}~ -1 = e^{i\pi}\\ \amp = |r|e^{i(\theta+\pi)}, ~~\text{by}~{\knowl{./knowl/th_polarmult.html}{\text{Theorem 2.2.5}}}. \end{align*}

Thus, by adding \(\pi\) to the angle if necessary, we may always assume that \(z = re^{i\theta}\) where \(r\) is non-negative.



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


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


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.


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