6.2: Polar Form
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Outcomes
- Convert a complex number from standard form to polar form, and from polar form to standard form.
In the previous section, we identified a complex number \(z=a+bi\) with a point \(\left( a, b\right)\) in the coordinate plane. There is another form in which we can express the same number, called the polar form . The polar form is the focus of this section. It will turn out to be very useful if not crucial for certain calculations as we shall soon see.
Suppose \(z=a+bi\) is a complex number, and let \(r=\sqrt{a^{2}+b^{2}} = |z|\). Recall that \(r\) is the modulus of \(z\). Note first that \[\left( \frac{a}{r} \right) ^{2}+\left( \frac{b}{r}\right) ^{2}= \frac{a^2+b^2}{r^2}=1\nonumber\] and so \(\left( \frac{a}{r},\frac{b}{r}\right)\) is a point on the unit circle. Therefore, there exists an angle \(\theta\) (in radians) such that \[\cos \theta =\frac{a}{r},\ \sin \theta =\frac{b}{r}\nonumber\] In other words \(\theta\) is an angle such that \(a = r\cos \theta\) and \(b=r \sin \theta\), that is \(\theta = \cos^{-1}(a/r)\) and \(\theta = \sin^{-1}(b/r)\). We call this angle \(\theta\) the argument of \(z\).
We often speak of the principal argument of \(z\). This is the unique angle \(\theta \in (-\pi, \pi]\) such that \[\cos \theta =\frac{a}{r},\ \sin \theta =\frac{b}{r}\nonumber\]
The polar form of the complex number \(z=a+bi = r \left( \cos \theta +i\sin \theta \right)\) is for convenience written as: \[z = r e^{i \theta}\nonumber\] where \(\theta\) is the argument of \(z\).
When given \(z = re^{i\theta}\), the identity \(e^{i\theta} = \cos\theta + i \sin\theta\) will convert \(z\) back to standard form. Here we think of \(e^{i \theta}\) as a short cut for \(\cos \theta +i\sin \theta\). This is all we will need in this course, but in reality \(e^{i \theta}\) can be considered as the complex equivalent of the exponential function where this turns out to be a true equality.
Thus we can convert any complex number in the standard (Cartesian) form \(z = a+bi\) into its polar form. Consider the following example.
Example \(\PageIndex{1}\): Standard to Polar Form
Let \(z = 2 + 2i\) be a complex number. Write \(z\) in the polar form \[z = re^{i \theta}\nonumber\]
Solution
First, find \(r\). By the above discussion, \(r=\sqrt{ a^{2}+b^{2}} = |z|\). Therefore, \[r = \sqrt{2^{2} + 2^{2}} = \sqrt{8} =2\sqrt{2}\nonumber \]
Now, to find \(\theta\), we plot the point \(\left( 2, 2 \right)\) and find the angle from the positive \(x\) axis to the line between this point and the origin. In this case, \(\theta = 45^{\circ} = \frac{\pi}{4}\). That is we found the unique angle \(\theta\) such that \(\theta = \cos^{-1}(1/\sqrt{2})\) and \(\theta = \sin^{-1}(1/\sqrt{2})\).
Note that in polar form, we always express angles in radians, not degrees.
Hence, we can write \(z\) as \[z = 2\sqrt{2} e^{i\frac{\pi}{4}}\nonumber\]
Notice that the standard and polar forms are completely equivalent. That is not only can we transform a complex number from standard form to its polar form, we can also take a complex number in polar form and convert it back to standard form.
Example \(\PageIndex{3}\): Polar to Standard Form
Let \(z = 2 e^{ 2\pi i/3}\). Write \(z\) in the standard form \[z = a+bi\nonumber \]
Solution
Let \(z = 2 e^{2\pi i/3}\) be the polar form of a complex number. Recall that \(e^{i\theta} = \cos \theta + i \sin \theta\). Therefore using standard values of \(\sin\) and \(\cos\) we get: \[\begin{aligned} z = 2 e^{i 2\pi/3} &= 2 (\cos (2\pi/3)+i\sin (2\pi/3))\\ &= 2 \left ( -\frac{1}{2} + i \frac{\sqrt{3}}{2} \right) \\ &=-1 + \sqrt{3}i \end{aligned}\] which is the standard form of this complex number.
You can always verify your answer by converting it back to polar form and ensuring you reach the original answer.