Problem 44 - Algebra and Trigonometry

Question

Find \(z w\) and \(\frac{z}{w} .\) Write each answer in polar form and in exponential form. \(z=1-i\) \(w=1-\sqrt{3} i\)

Step-by-Step Solution

Answer

In polar form: \( z w = 2 \sqrt{2} \left( \cos (-\frac{7\pi}{12}) + i\sin (-\frac{7\pi}{12}) \right)\), \( z / w = \frac{\sqrt{2}}{2} \left( \cos (\frac{\pi}{12}) + i\sin (\frac{\pi}{12}) \right)\). In exponential form: \( z w = 2 \sqrt{2} e^{i(-\frac{7\pi}{12})}\), \( z / w = \frac{\sqrt{2}}{2} e^{i(\frac{\pi}{12})}\).

Step 1 - Convert Complex Numbers to Polar Form

To convert each complex number to polar form, find the magnitude and the argument.For \(z = 1 - i\):Magnitude: \[|z| = \sqrt{1^2 + (-1)^2} = \sqrt{2}\]Argument: \[\theta_z = \arctan \left( \frac{-1}{1} \right) = -\frac{\pi}{4}\]Therefore, \(z\) in polar form is \[\sqrt{2} \left(\cos\left(-\frac{\pi}{4}\right) + i\sin\left(-\frac{\pi}{4}\right)\right)\]For \(w = 1 - \sqrt{3} i\):Magnitude: \[|w| = \sqrt{1^2 + (\sqrt{3})^2} = 2\]Argument: \[\theta_w = \arctan \left( \frac{-\sqrt{3}}{1} \right) = -\frac{\pi}{3}\]Therefore, \(w\) in polar form is \[2 \left(\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right)\]

Step 2 - Multiply the Complex Numbers

To multiply two complex numbers in polar form, multiply their magnitudes and add their arguments.\(z w\):Magnitude: \[|z w| = |z| |w| = \sqrt{2} \cdot 2 = 2 \sqrt{2}\]Argument: \[\theta_{zw} = \theta_z + \theta_w = -\frac{\pi}{4} + -\frac{\pi}{3} = -\frac{7\pi}{12}\]Therefore, \[z w = 2 \sqrt{2} \left(\cos\left(-\frac{7\pi}{12}\right) + i\sin\left(-\frac{7\pi}{12}\right)\right)\]In exponential form, \[z w = 2 \sqrt{2} e^{i(-\frac{7\pi}{12})}\]

Step 3 - Divide the Complex Numbers

To divide two complex numbers in polar form, divide their magnitudes and subtract their arguments.\(\frac{z}{w}\):Magnitude: \[\left|\frac{z}{w}\right| = \frac{|z|}{|w|} = \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2}\]Argument: \[\theta_{\frac{z}{w}} = \theta_z - \theta_w = -\frac{\pi}{4} - -\frac{\pi}{3} = -\frac{\pi}{4} + \frac{\pi}{3} = \frac{\pi}{12}\]Therefore, \[\frac{z}{w} = \frac{\sqrt{2}}{2} \left(\cos\left(\frac{\pi}{12}\right) + i\sin\left(\frac{\pi}{12}\right)\right)\]In exponential form, \[\frac{z}{w} = \frac{\sqrt{2}}{2} e^{i(\frac{\pi}{12})}\]

Key Concepts

Magnitude of Complex Numbers

The magnitude of a complex number is like its length or distance from the origin on the complex plane. It is denoted by \(|z|\) for a complex number \(z\). If you have a complex number \(z = a + bi\), where \(a\) and \(b\) are real numbers, the magnitude is given by \(|z| = \sqrt{a^2 + b^2}\). Let’s take the given example: for \(z = 1 - i\), you compute its magnitude as:
\(|z| = \sqrt{1^2 + (-1)^2} = \sqrt{2}\)
Similarly, for \(w = 1 - \sqrt{3} i\), you calculate:
\(|w| = \sqrt{1^2 + (\sqrt{3})^2} = 2\).
This length or distance helps in understanding the position of the complex number in the complex plane.

Argument of Complex Numbers

The argument of a complex number is the angle that the line representing the complex number makes with the positive real axis, measured in the counterclockwise direction. It's denoted by \(\theta\) and is usually found using \(\arctan (\frac{b}{a})\). For \(z = 1 - i\), the argument is:
\(\theta_z = \arctan (\frac{-1}{1}) = -\frac{\pi}{4}\).
For \(w = 1- \sqrt{3} i\), the argument is:
\(\theta_w = \arctan (\frac{-\sqrt{3}}{1}) = -\frac{\pi}{3}\).
These angles provide the direction of the complex numbers in the complex plane, which is crucial for their multiplication and division.

Multiplying Complex Numbers

To multiply complex numbers in polar form, multiply their magnitudes and add their arguments. When two complex numbers \(z\) and \(w\) are given in polar form as \(r \cdot (cos \theta + i \sin \theta)\), their product is found by:
\(|z w| = |z||w|\)
\(\theta_{zw} = \theta_z + \theta_w\).
For \(z = \sqrt{2}(cos(-\frac{\pi}{4}) + i \sin(-\frac{\pi}{4}))\) and \(w = 2(cos(-\frac{\pi}{3}) + i \sin(-\frac{\pi}{3}))\),
the product is: \(|z w| = \sqrt{2} \cdot 2 = 2 \sqrt{2}\).
\(\theta_{zw} = -\frac{\pi}{4} + -\frac{\pi}{3} = -\frac{7\pi}{12}\).
Therefore, \((zw = 2\sqrt{2}(cos(-\frac{7\pi}{12}) + i \sin(-\frac{7\pi}{12}))))\), or in exponential form, \( z w = 2 \sqrt{2} e^{i(-\frac{7\pi}{12})}\).

Dividing Complex Numbers

To divide complex numbers in polar form, divide their magnitudes and subtract their arguments. Given \(z\) and \(w\) in polar form, the division is done by:
\(\frac{|z|}{|w|}\) and \(\theta_{\frac{z}{w}} = \theta_z - \theta_w\).
For \(z = \sqrt{2}(cos(-\frac{\pi}{4}) + i \sin(-\frac{\pi}{4}))\) and \(w = 2(cos(-\frac{\pi}{3}) + i \sin(-\frac{\pi}{3}))\),
the division is: \(\frac{|z|}{|w|} = \frac{2}{\sqrt{2}} = \frac{\sqrt{2}}{2}\).
\(\theta_{\frac{z}{w}} = -\frac{\pi}{4} - \-\frac{\pi}{3} = -\frac{\pi}{4} + \frac{\pi}{3} = \frac{\pi}{12}\).
Thus, \( \frac{z}{w} = \frac{\sqrt{2}}{2}(cos(\frac{\pi}{12}) + i \sin(\frac{\pi}{12}))\),
or in exponential form, \(\frac{z}{w} = \frac{\sqrt{2}}{2} e^{i(\frac{\pi}{12})}\).

Exponential Form of Complex Numbers

The exponential form of a complex number is a compact way to express its polar form using Euler's formula \(e^{i \theta} = cos \theta + i \sin \theta\). This form is particularly useful for multiplying and dividing complex numbers. If \(z = r(\text{cos } \theta + i \text{sin } \theta)\), its exponential form is \(z = r e^{i \theta}\).
For the given product \(z w\), where:
\( z w = 2 \sqrt{2} (\text{cos}(-\frac{7\pi}{12}) + i \text{sin}(-\frac{7\pi}{12}))\),
the exponential form is \(2 \sqrt{2} e^{i(-\frac{7\pi}{12})}\).
For the quotient \(\frac{z}{w}\), where: \( \frac{z}{w} = \frac{\sqrt{2}}{2} (\text{cos}(\frac{\pi}{12}) + i \text{sin}(\frac{\pi}{12})\),
the exponential form is \(\frac{\sqrt{2}}{2} e^{i(\frac{\pi}{12})}\).
The exponential form simplifies complex number operations and highlights the significance of the argument and magnitude.