9.4: Polar Form of Complex Numbers

Writing Complex Numbers in Polar Form

Given a complex number expressed as \(z = x + y \, i\), we can use the material from the previous section to visualize \(z\) as the point \(\left( x,y \right)\) on the Cartesian plane.¹ We then observe that we can rewrite both \(x\) and \(y\) in trigonometric form:\[\begin{array}{rcl} x & = & r \cos\left( \theta \right) \\ y & = & r \sin\left( \theta \right) \\ r & = & \sqrt{x^2+y^2} \end{array} \nonumber \]where \(\theta\) is the angle in standard position with terminal side being the line segment connecting the origin to the point \(\left( x,y \right)\). We review these relationships in Figure \(\PageIndex{1}\).

Thus, we can rewrite the complex number, \(z = x + y \, i\), in the form\[z = r \cos\left( \theta \right) + \left( r \sin\left( \theta \right) \right) i = r \left( \cos\left( \theta \right) + i \, \sin\left( \theta \right) \right). \nonumber \]This form is known as the polar form of the complex number.

The notation \(z = r \, \mathrm{cis}\left( \theta \right)\) is read as "cosine plus i sine of theta."

As we did in the previous section, we use the term modulus to represent the "size" of a complex number, or the distance from the origin to the point \(\left(x,y\right)\) on the complex plane. The modulus, therefore, is the same as \(r\), the radius in polar form.

It's very tempting to evaluate the trigonometric functions at \(\frac{3\pi}{4}\) in Example \(\PageIndex{2}\). I would not blame you for wanting to do that because that's what we have been doing this entire course; however, in doing so, you would be reverting back to the standard (rectangular) form of the complex number (thereby undoing all of your work to get the number in polar form).

Converting a Complex Number from Polar to Rectangular Form

Converting a complex number from polar form to rectangular form is a matter of evaluating what is given and using the distributive property. In other words, given\[z = r\left(\cos\left( \theta \right)+i \, \sin\left( \theta \right)\right), \nonumber \]we first evaluate the trigonometric functions \(\cos\left( \theta \right)\) and \(\sin\left( \theta \right)\), then multiply through by \(r\).

Finding Products of Complex Numbers in Polar Form

Now that we can convert complex numbers to polar form we will learn how to perform operations on complex numbers in polar form. For the rest of this section, we will work with formulas developed by French mathematician Abraham de Moivre (1667-1754). These formulas have made working with products, quotients, powers, and roots of complex numbers much simpler than they appear. The rules are based on multiplying the moduli and adding the arguments.

Notice that the product calls for multiplying the moduli and adding the angles. The proof of this theorem is left as a homework exercise.

Finding Quotients of Complex Numbers in Polar Form

The quotient of two complex numbers in polar form follows the same patterns set for products.

The proof of this theorem is left as a homework exercise.

Finding Powers of Complex Numbers in Polar Form

Finding powers of complex numbers is greatly simplified when they are written in polar form. For example, let \(z = 1 + i\). This is a fairly innocuous complex number; however, let's try to compute \(z^5 = \left( 1 + i \right)^5\).\[\begin{array}{rcl}
z^5 & = & \left( 1 + i \right)^5 \\[6pt] & = & \left( 1 + i \right)^2 \left( 1 + i \right)^2 \left( 1 + i \right) \\[6pt] & = & \left( 1 + 2i + i^2 \right) \left( 1 + 2i + i^2 \right) \left( 1 + i \right) \\[6pt] & = & \left( 1 + 2i - 1 \right) \left( 1 + 2i - 1 \right) \left( 1 + i \right) \\[6pt] & = & \left( 2i \right) \left( 2i \right) \left( 1 + i \right) \\[6pt] & = & 4i^2 \left( 1 + i \right) \\[6pt] & = & -4 \left( 1 + i \right) \\[6pt] & = & -4 - 4i \\[6pt] \end{array} \nonumber \]This is only the fifth power of a somewhat simple complex number. Imagine working with the tenth power of something like \(z = -\sqrt{3} + i\). If our complex number is written in polar form, however, something fantastic happens.

(Note: The following derivation of why De Moivre's Theorem works requires a proof method called Mathematical Induction. If you have already taken College Algebra, then you have worked with, and performed, proofs using Mathematical Induction; however, since College Algebra is not a prerequisite for this course, you might not have seen it. As such, I provide the proof only for those who are interested.)

De Moivre's Theorem states that, for a positive integer \(n\), \(z^n\) is found by raising the modulus to the \(n^{th}\) power and multiplying the argument by \(n\). It is the standard method used in modern Mathematics.

Finding Roots of Complex Numbers in Polar Form

DeMoivre's Theorem also works for fractional values of \(n\), so we can compute roots of complex numbers. For example, by applying the theorem with \(n=\frac{1}{2}\), we see that one of the square roots of \(z=r(\cos\left( \theta \right)+i \sin\left( \theta \right))\) is\[z^{1 / 2} = r^{1 / 2}\left(\cos \left(\dfrac{\theta}{2}\right)+i \, \sin \left(\dfrac{\theta}{2}\right)\right). \nonumber \]Now, every number, whether real or complex, has two square roots. To find the other root, remember that we can add a multiple of \(2 \pi\) to the argument of \(z\), that is, we can also write the polar form of \(z\) as\[z=r(\cos (\theta+2 \pi)+i \, \sin (\theta+2 \pi)).\nonumber\]

The second square root of \(z\) is thus\[r^{1 / 2}\left(\cos \left(\dfrac{\theta}{2}+\pi\right)+i \, \sin \left(\dfrac{\theta}{2}+\pi\right)\right). \nonumber \]As an example, consider \(w=-8+8 i \sqrt{3}\), whose polar form is \(16\left(\cos \left(\frac{2 \pi}{3}\right)+i \,\sin\left( \frac{2 \pi}{3}\right)\right)\), or by adding \(2 \pi\) to the argument, \(16\left(\cos \left(\frac{8 \pi}{3}\right)+i \, \sin \left(\frac{8 \pi}{3}\right)\right)\). The two square roots of \(w\) are\[\begin{array}{rclcrcl}
z_1 & = & 16^{1 / 2}\left(\cos \left(\dfrac{\pi}{3}\right) + i \, \sin \left(\dfrac{\pi}{3}\right)\right) & \quad & z_2 & = & 16^{1 / 2}\left(\cos\left( \dfrac{4 \pi}{3}\right) + i \, \sin \left(\dfrac{4 \pi}{3}\right)\right) \\[6pt] & = & 4\left(\dfrac{1}{2}+i \, \dfrac{\sqrt{3}}{2}\right) & \quad & & = & 4\left(\dfrac{-1}{2}+i \dfrac{-\sqrt{3}}{2}\right) \\[6pt] & = & 2+2 i \sqrt{3} & \quad & & = & -2-2 i \sqrt{3} \\[6pt] \end{array} \nonumber \]You can verify that both of these numbers are square roots of \(w=-8+8 i \sqrt{3}\), for instance,\[\begin{array}{rcl}
(2+2 i \sqrt{3})^2 & = & 4+2(2)(2 i \sqrt{3})+12 i^2 \\[6pt] & = & -8+8 i \sqrt{3} \\[6pt] \end{array} \nonumber \]The graphs of \(w\) and its two square roots are shown below.

It is not hard to show that every number has three complex cube roots, four complex fourth roots, and so on.

In general, to find the \(n^{th}\) root of a complex number in polar form, we use the \(n^{th}\) Root Theorem, or we can use De Moivre's Theorem as we did in our work above. The following theorem is presented without proof.