Problem 50 - Algebra and Trigonometry

Question

\(49-56\) me product \(z_{1} z_{2}\) and the quotient \(z_{1} / z_{2}\) . Express your answer in polar form. $$ z_{1}=\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}, \quad z_{2}=\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4} $$

Step-by-Step Solution

Answer

Product: \( -1 \), Quotient: \( -i \) in polar form.

Step 1: Identify Polar Forms

Determine the polar form for each complex number given. \( z_1 = \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \) implies an angle of \( \theta_1 = \frac{\pi}{4} \) and magnitude \( |z_1| = 1 \). Similarly, \( z_2 = \cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4} \) denotes \( \theta_2 = \frac{3\pi}{4} \) and magnitude \( |z_2| = 1 \). Thus, in polar form, \( z_1 = 1(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4}) \) and \( z_2 = 1(\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4}) \).

Step 2: Multiply the Complex Numbers

To find the product \( z_1 z_2 \), multiply their magnitudes and add their angles: \( |z_1 z_2| = |z_1| \cdot |z_2| = 1 \cdot 1 = 1 \) and \( \theta_{z_1 z_2} = \theta_1 + \theta_2 = \frac{\pi}{4} + \frac{3\pi}{4} = \pi \). Therefore, \( z_1 z_2 = 1(\cos \pi + i \sin \pi) = -1 + 0i = -1 \).

Step 3: Divide the Complex Numbers

To determine the quotient \( \frac{z_1}{z_2} \), divide the magnitudes and subtract the angles: \( \left| \frac{z_1}{z_2} \right| = \frac{|z_1|}{|z_2|} = \frac{1}{1} = 1 \) and \( \theta_{\frac{z_1}{z_2}} = \theta_1 - \theta_2 = \frac{\pi}{4} - \frac{3\pi}{4} = -\frac{\pi}{2} \). Thus, \( \frac{z_1}{z_2} = 1\left( \cos\left(-\frac{\pi}{2}\right) + i \sin\left(-\frac{\pi}{2}\right) \right) = 0 - i \).

Step 4: Express Final Answers in Polar Form

The product of the two complex numbers is \( z_1 z_2 = -1 \), which in polar form is \( 1(\cos \pi + i \sin \pi) \). The quotient is \( \frac{z_1}{z_2} = -i \), which in polar form translates to \( 1(\cos\left(-\frac{\pi}{2}\right) + i \sin\left(-\frac{\pi}{2}\right)) \).

Key Concepts

Polar Form of Complex Numbers

Complex numbers can be represented in more than one way. The polar form is particularly elegant and useful in performing operations like multiplication and division. Instead of expressing a complex number as a sum of its real and imaginary parts, the polar form uses a radius (or magnitude) and an angle. This gives it a form of \( r(\cos \theta + i \sin \theta) \) or alternatively \( r e^{i\theta} \).

Here, \( r \) is the magnitude of the complex number. It represents the distance from the origin to the point in the complex plane. On the other hand, \( \theta \) (theta) denotes the angle made with the positive real axis. This angle is typically measured in radians.

In our example, \( z_1 \) has a polar form of \( 1(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4}) \) and \( z_2 \) is represented by \( 1(\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4}) \). This form highlights both the direction of the vector representation of the complex number and its magnitude.

Product of Complex Numbers

When dealing with the product of two complex numbers in polar form, it becomes extraordinarily simple. The key is to multiply their magnitudes and add their angles. This operation reflects how angles add when two vectors are combined directionally.

Mathematically, if you have two complex numbers \( z_1 = r_1(\cos \theta_1 + i \sin \theta_1) \) and \( z_2 = r_2(\cos \theta_2 + i \sin \theta_2) \), the product \( z_1 z_2 \) becomes:

\[ r_1 r_2 \left( \cos(\theta_1 + \theta_2) + i \sin(\theta_1 + \theta_2) \right) \]

In our specific instance, \( z_1 z_2 = 1(\cos \pi + i \sin \pi) = -1 \). Here, since both magnitudes are 1, the multiplication effect mainly influences the angle, resulting in a rotation on the complex plane without stretching.

Quotient of Complex Numbers

Dividing one complex number by another becomes equally straightforward using the polar form, as you work with their magnitudes and angles. For the quotient, you divide their magnitudes and subtract their angles. Consider two complex numbers \( z_1 = r_1(\cos \theta_1 + i \sin \theta_1) \) and \( z_2 = r_2(\cos \theta_2 + i \sin \theta_2) \). The division \( \frac{z_1}{z_2} \) is computed as:

\[\frac{r_1}{r_2} \left( \cos(\theta_1 - \theta_2) + i \sin(\theta_1 - \theta_2) \right)\]

In the given problem, this results in the computation \( \frac{z_1}{z_2} = 1(\cos\left(-\frac{\pi}{2}\right) + i \sin\left(-\frac{\pi}{2}\right)) = -i \). Notice how the negative angle translates into a clockwise rotation, leading to our final result.