###### Example2.3.1Dividing complex numbers in Cartesian form

We convert the following quotient to Cartesian form:

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The division of the complex number \(z\) by \(w \neq 0\text{,}\) denoted \(\frac{z}{w}\text{,}\) is the complex number \(u\) that satisfies the equation \(z = w \cdot u\text{.}\)

For instance, \(\frac{1}{i} = -i\) 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{align*}
\frac{2+i}{3+2i} \amp = \frac{2+i}{3+2i}\cdot\frac{3-2i}{3-2i}\\
\amp = \frac{(6+2)+(-4+3)i}{9+4}\\
\amp = \frac{8-i}{13}\\
\amp = \frac{8}{13} - \frac{1}{13}i.
\end{align*}

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

\begin{equation*}
\frac{1}{w} =
\frac{1}{s}e^{-i\beta}.
\end{equation*}

Then we may apply Theorem 2.2.4 to obtain the following result:

\begin{align*}
\frac{z}{w} \amp = z\cdot\frac{1}{w}\\
\amp = re^{i\theta}\cdot \frac{1}{s}e^{-i\beta}\\
\amp = \frac{r}{s}e^{i(\theta-\beta)}.
\end{align*}

So,

\begin{equation*}
\arg\bigg(\frac{z}{w}\bigg)=\arg(z)-\arg(w)
\end{equation*}

where equality is taken modulo \(2\pi\text{.}\)

Thus, when dividinig by complex numbers, we can first convert to polar form if it is convenient. For instance,

\begin{equation*}
\frac{1+i}{-3 + 3i} =\frac{\sqrt{2}e^{i\pi/4}}{\sqrt{18}e^{i3\pi/4}}
= \frac{1}{3}e^{-i\pi/2}
= -\frac{1}{3} i.
\end{equation*}

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

\begin{equation*}
\angle(L_1,L_2) = - \angle(L_2,L_1).
\end{equation*}

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{align*}
\angle(L_1,L_2) \amp = \arg\bigg(\frac{z_2-z_0}{z_1-z_0}\bigg) \notag\\
\amp = \arg(z_2-z_0)-\arg(z_1-z_0).
\end{align*}

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

\begin{equation*}
\angle(L_1,L_2) = \arg(2-i)-\arg(1+i) = -\tan^{-1}(1/2) -
\pi/4 \approx -71.6^\circ.
\end{equation*}

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,

\begin{equation*}
\angle uvw = \theta = \arg\bigg(\frac{w-v}{u-v}\bigg).
\end{equation*}

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 \(\angle uvz = \arg(z)\text{.}\)

Express \(\frac{1}{x+yi}\) in the form \(a + bi\text{.}\)

Express these fractions in Cartesian form or polar form, whichever seems more convenient.

\begin{equation*}
\frac{1}{2i},~~ \frac{1}{1+i},~~ \frac{4+i}{1-2i},~~ \frac{2}{3+i}.
\end{equation*}

Prove that \(\displaystyle|z/w| = |z|/|w|\text{,}\) and that \(\displaystyle\overline{z/w} = \overline{z}/\overline{w}.\)

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 \(\angle b z a = \pi/2.\)