2.3: Division and Angle Measure
The division of the complex number \(z\) by \(w \neq 0\text{,}\) denoted \(\dfrac{z}{w}\text{,}\) is the complex number \(u\) that satisfies the equation \(z = w \cdot u\text{.}\)
For instance, \(\dfrac{1}{i} = -i\) because \(1 = i \cdot (-i)\text{.}\)
In practice, division of complex numbers is not a guessing game, but can be done by multiplying the top and bottom of the quotient by the conjugate of the bottom expression.
We convert the following quotient to Cartesian form:
\( \begin{array} z \dfrac{2+i}{3+2i} &= \dfrac{2+i}{3+2i}\cdot\dfrac{3-2i}{3-2i}\\ &= \dfrac{(6+2)+(-4+3)i}{9+4}\\ &= \dfrac{8-i}{13}\\ &= \dfrac{8}{13} - \dfrac{1}{13}i\text{.} \end{array} \)
Suppose we wish to find \(z/w\) where \(z = re^{i\theta}\) and \(w = se^{i\beta} \neq 0\text{.}\) The reader can check that
\[ \dfrac{1}{w} = \dfrac{1}{s}e^{-i\beta}\text{.} \]
Then we may apply Theorem 2.2.1 to obtain the following result:
So,
\[ \arg\bigg(\frac{z}{w}\bigg)=\arg(z)-\arg(w) \]
where equality is taken modulo \(2\pi\text{.}\)
Thus, when dividing by complex numbers, we can first convert to polar form if it is convenient. For instance,
\[ \dfrac{1+i}{-3 + 3i} =\dfrac{\sqrt{2}e^{i\pi/4}}{\sqrt{18}e^{i3\pi/4}} = \dfrac{1}{3}e^{-i\pi/2} = -\dfrac{1}{3} i\text{.} \]
Angle Measure
Given two rays \(L_1\) and \(L_2\) having common initial point, we let \(\angle(L_1,L_2)\) denote the angle between rays \(L_1\) and \(L_2\), measured from \(L_1\) to \(L_2\text{.}\) We may rotate ray \(L_1\) onto ray \(L_2\) in either a counterclockwise direction or a clockwise direction. We adopt the convention that angles measured counterclockwise are positive, and angles measured clockwise are negative, and admit that angles are only well-defined up to multiples of \(2\pi\text{.}\) Notice that
\[ \angle(L_1,L_2) = - \angle(L_2,L_1)\text{.} \]
To compute \(\angle(L_1,L_2)\) where \(z_0\) is the common initial point of the rays, let \(z_1\) be any point on \(L_1\text{,}\) and \(z_2\) any point on \(L_2\text{.}\) Then
\(\begin{array} \angle(L_1,L_2) & = \arg\bigg(\dfrac{z_2-z_0}{z_1-z_0}\bigg) \notag\\ &= \arg(z_2-z_0)-\arg(z_1-z_0)\text{.} \end{array}\)
Suppose \(L_1\) and \(L_2\) are rays emanating from \(2+2i\text{.}\) Ray \(L_1\) proceeds along the line \(y=x\) and \(L_2\) proceeds along \(y = 3-x/2\) as pictured.
To compute the angle \(\theta\) in the diagram, we choose \(z_1 = 3+3i\) and \(z_2 = 4+i\text{.}\) Then
\[ \angle(L_1,L_2) = \arg(2-i)-\arg(1+i) = -\tan^{-1}(1/2) - \pi/4 \approx -71.6^\circ\text{.} \]
That is, the angle from \(L_1\) to \(L_2\) is \(71.6^{\circ}\) in the clockwise direction.
If \(u,v,\) and \(w\) are three complex numbers, let \(\angle uvw\) denote the angle \(\theta\) from ray \(\overrightarrow{vu}\) to \(\overrightarrow{vw}\text{.}\) In particular,
\[ \angle uvw = \theta = \arg\bigg(\dfrac{w-v}{u-v}\bigg)\text{.} \]
For instance, if \(u = 1\) on the positive real axis, \(v= 0\) is the origin in \(\mathbb{C}\text{,}\) and \(z\) is any point in \(\mathbb{C}\text{,}\) then \(\angle uvz = \arg(z)\text{.}\)
Exercises
Express \(\frac{1}{x+yi}\) in the form \(a + bi\text{.}\)
Express these fractions in Cartesian form or polar form, whichever seems more convenient.
\[ \dfrac{1}{2i},\;\; \dfrac{1}{1+i},\;\; \dfrac{4+i}{1-2i},\;\; \dfrac{2}{3+i}\text{.} \]
Prove that \(\displaystyle|z/w| = |z|/|w|\text{,}\) and that \(\displaystyle\overline{z/w} = \overline{z}/\overline{w}\text{.}\)
Suppose \(z = re^{i\theta}\) and \(w = se^{i\alpha}\) are as shown below. Let \(u = z\cdot w\text{.}\) Prove that \(\Delta 01z\) and \(\Delta 0wu\) are similar triangles.
Determine the angle \(\angle uvw\) where \(u = 2 + i\text{,}\) \(v = 1 + 2i\text{,}\) and \(w = -1 + i\text{.}\)
Suppose \(z\) is a point with positive imaginary component on the unit circle shown below, \(a = 1\) and \(b = -1\text{.}\) Use the angle formula to prove that angle \(\angle b z a = \dfrac{\pi}{2} \text{.}\)