Problem 62 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 57-66: Use trigonometric forms to find \(z_{1} z_{2}\) and \(z_{1} / z_{2}\). $$ z_{1}=2 i, \quad z_{2}=-3 i $$

Step-by-Step Solution

Answer

The product is 6 and the quotient is \(-\frac{2}{3}\).

Step 1: Write the Complex Numbers in Trigonometric Form

The complex numbers are given as \(z_1 = 2i\) and \(z_2 = -3i\). These can be written in the form \(z = r(\cos \theta + i \sin \theta)\), where \(r\) is the modulus and \(\theta\) is the argument.For \(z_1 = 2i\), we can express this in trigonometric form as \((2)(\cos \frac{\pi}{2} + i \sin \frac{\pi}{2})\) because the purely imaginary number \(2i\) is on the positive imaginary axis.For \(z_2 = -3i\), it can be written as \((-3)(\cos -\frac{\pi}{2} + i \sin -\frac{\pi}{2})\) since \(-3i\) is on the negative imaginary axis.

Step 2: Determine the Modulus and Argument for Multiplication

For multiplication of complex numbers in trigonometric form: \[ z_1 z_2 = r_1 r_2 \left( \cos(\theta_1 + \theta_2) + i \sin(\theta_1 + \theta_2) \right) \]The moduli are \(r_1 = 2\) and \(r_2 = 3\). The arguments are \(\theta_1 = \frac{\pi}{2}\) and \(\theta_2 = -\frac{\pi}{2}\).Multiplying these together gives:\[ z_1 z_2 = (2)(3) \left( \cos(\frac{\pi}{2} - \frac{\pi}{2}) + i \sin(\frac{\pi}{2} - \frac{\pi}{2}) \right) \]This simplifies to:\[ 6(\cos 0 + i \sin 0) = 6(1 + 0i) = 6 \]

Step 3: Determine the Modulus and Argument for Division

For division of complex numbers in trigonometric form:\[ \frac{z_1}{z_2} = \frac{r_1}{r_2} \left( \cos(\theta_1 - \theta_2) + i \sin(\theta_1 - \theta_2) \right) \]The modulus division gives \(\frac{r_1}{r_2} = \frac{2}{3}\). The argument difference is \(\theta_1 - \theta_2 = \frac{\pi}{2} - (-\frac{\pi}{2}) = \pi\).Thus:\[ \frac{z_1}{z_2} = \frac{2}{3} \left( \cos \pi + i \sin \pi \right) \]This simplifies to:\[ \frac{2}{3} (-1 + 0i) = -\frac{2}{3} \]

Key Concepts

Modulus of Complex Numbers

In the world of complex numbers, the modulus is a crucial concept that helps us understand the magnitude of these numbers. Think of the modulus as the distance from the origin (0,0) in the complex plane to the point representing the complex number. The complex number can be represented as a point (a, b), where 'a' is the real part and 'b' is the imaginary part. The modulus is the length of the line segment from the origin to this point.

To calculate the modulus of a complex number given by \(z = a + bi\), use the formula:

• \(r = \sqrt{a^2 + b^2}\)

In our example, the complex numbers are \(z_1 = 2i \) and \(z_2 = -3i\). Here, both numbers have no real part, and their imaginary parts are 2 and -3 respectively. This makes it easier because the modulus is simply the absolute value of the imaginary part:

• For \(z_1 = 2i\), the modulus \(r_1 = |2| = 2\).
• For \(z_2 = -3i\), the modulus \(r_2 = |-3| = 3\).

Argument of Complex Numbers

The argument of a complex number provides us with the angle between the positive real axis and the line drawn from the origin to the point in the complex plane. This angle helps us establish the direction of the complex number in its trigonometric form and is crucial for operations like multiplication or division.

For a complex number \(z = a + bi\), the argument \(\theta\) can be calculated using:

• \(\theta = \text{atan2}(b, a)\)

In our exercise:

• For \(z_1 = 2i\), since it lies on the positive imaginary axis, \(\theta_1 = \frac{\pi}{2}\).
• For \(z_2 = -3i\), positioned on the negative imaginary axis, \(\theta_2 = -\frac{\pi}{2}\).

These angles give us the direction for both complex numbers on the imaginary axis.

Multiplication of Complex Numbers in Polar Form

Multiplying complex numbers is seamless when using the trigonometric or polar form. When two complex numbers are given in trigonometric form as \(z_1 = r_1 (\cos \theta_1 + i \sin \theta_1)\) and \(z_2 = r_2 (\cos \theta_2 + i \sin \theta_2)\), we can multiply them easily by multiplying their moduli and adding their arguments.

The formula for multiplication is:

• \(z_1 z_2 = r_1 r_2 [ \cos(\theta_1 + \theta_2) + i \sin(\theta_1 + \theta_2) ]\)

In our example, the moduli are \(r_1 = 2\) and \(r_2 = 3\), and their arguments are \(\theta_1 = \frac{\pi}{2}\) and \(\theta_2 = -\frac{\pi}{2}\). Adding these gives us 0. Therefore:

• Result: \(6\left(\cos(0) + i \sin(0)\right) = 6(1 + 0i) = 6\).

Division of Complex Numbers in Polar Form

The division of complex numbers in polar form is another situation where the trigonometric representation shines. It involves dividing the moduli and subtracting the arguments. For two complex numbers in polar form \(z_1 = r_1 (\cos \theta_1 + i \sin \theta_1)\) and \(z_2 = r_2 (\cos \theta_2 + i \sin \theta_2)\), the division formula becomes:

• \(\frac{z_1}{z_2} = \frac{r_1}{r_2} [ \cos(\theta_1 - \theta_2) + i \sin(\theta_1 - \theta_2) ]\)

For \(z_1\) and \(z_2\), we have:

• Modulus division: \(\frac{2}{3}\)
• Argument subtraction: \(\theta_1 - \theta_2 = \frac{\pi}{2} - (-\frac{\pi}{2}) = \pi\)

This calculation gives us:

• Result: \(\frac{2}{3} (\cos \pi + i \sin \pi) = \frac{2}{3} (-1 + 0i) = -\frac{2}{3}\)

Thus, the division results in \(-\frac{2}{3}\), a real number on the negative real axis.