Problem 16 - Advanced Engineering Mathematics

Question

In Problems 15 and 16 , find \(z_{1} z_{2}\) and \(z_{1} / z_{2}\). Write the number in the form \(a+i b\). $$ \begin{aligned} &z_{1}=\sqrt{2}\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right) \\ &z_{2}=\sqrt{3}\left(\cos \frac{\pi}{12}+i \sin \frac{\pi}{12}\right) \end{aligned} $$

Step-by-Step Solution

Answer

\(z_1 z_2 = \frac{\sqrt{6}}{2} + i \frac{\sqrt{18}}{2}\), \(\frac{z_1}{z_2} \approx \frac{1}{2} + i \frac{\sqrt{2}}{2\sqrt{3}}\).

Step 1: Convert to rectangular form

First, convert each complex number from polar form to rectangular form. \( z_1 = \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right) = \sqrt{2} \left( \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right) = 1 + i \). Similarly, \( z_2 = \sqrt{3} \left( \cos \frac{\pi}{12} + i \sin \frac{\pi}{12} \right) = \sqrt{3} \left( \frac{\sqrt{3} + 1}{2\sqrt{2}} + i \frac{\sqrt{3} - 1}{2\sqrt{2}} \right) \). Calculate \( z_2 \) precisely before moving ahead.

Step 2: Calculate magnitude and angle of product

For the product \( z_1 z_2 \), the magnitude is \( |z_1| |z_2| = \sqrt{2} \times \sqrt{3} = \sqrt{6} \). The angle is the sum of angles of \( z_1 \) and \( z_2 \), given by \( \frac{\pi}{4} + \frac{\pi}{12} = \frac{3\pi}{12} + \frac{\pi}{12} = \frac{\pi}{3} \).

Step 3: Calculate the product in rectangular form

The product \( z_1 z_2 = \sqrt{6} \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right) = \sqrt{6} \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right) = \frac{\sqrt{6}}{2} + i \frac{\sqrt{18}}{2} \).

Step 4: Calculate magnitude and angle of quotient

For the quotient \( \frac{z_1}{z_2} \), the magnitude is \( \frac{|z_1|}{|z_2|} = \frac{\sqrt{2}}{\sqrt{3}} = \sqrt{\frac{2}{3}} \). The angle is the difference of the angles of \( z_1 \) and \( z_2 \), given by \( \frac{\pi}{4} - \frac{\pi}{12} = \frac{3\pi}{12} - \frac{\pi}{12} = \frac{\pi}{6} \).

Step 5: Calculate the quotient in rectangular form

The quotient \( \frac{z_1}{z_2} = \sqrt{\frac{2}{3}} \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right) = \sqrt{\frac{2}{3}} \left( \frac{\sqrt{3}}{2} + i \frac{1}{2} \right) = \frac{\sqrt{2}}{\sqrt{3}} \cdot \frac{\sqrt{3}}{2} + i \frac{\sqrt{2}}{2\sqrt{3}} \approx \frac{1}{2} + i \frac{\sqrt{2}}{2\sqrt{3}} \).

Key Concepts

Rectangular and Polar Form

Complex numbers can be expressed in both rectangular and polar forms, each providing different insights into the properties of the number. The rectangular form is written as \(a + ib\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit satisfying \(i^2 = -1\). This form is useful for addition and subtraction.

On the other hand, the polar form expresses a complex number in terms of its magnitude (or modulus) and its angle (or argument). It is written as \(r(\cos \theta + i \sin \theta)\), where \(r\) is the magnitude and \(\theta\) is the argument. The polar form highlights the geometric nature of complex numbers and is especially handy for multiplication and division. The exercise begins by converting complex numbers from polar to rectangular form, which involves calculating \(\cos\) and \(\sin\) for the given angles, then distributing \(r\) to find \(a\) and \(b\).

Knowing how to switch between forms is essential: converting from polar to rectangular helps simplify complex arithmetic, while using polar form can make understanding the geometry of the problem much clearer.

Complex Multiplication

Multiplying two complex numbers is more straightforward when they are in polar form. You simply multiply their magnitudes and add their arguments. Given two complex numbers \(z_1 = r_1(\cos \theta_1 + i \sin \theta_1)\) and \(z_2 = r_2(\cos \theta_2 + i \sin \theta_2)\), their product in polar form will be:

• Magnitude: \(|z_1| \times |z_2| = r_1 \times r_2\)
• Angle: \(\theta_1 + \theta_2\)

The result is still a complex number in polar form. In the exercise, the magnitudes and angle sums are calculated for \(z_1\) and \(z_2\). They are then converted back to rectangular form to express the solution as \(a + ib\).

This method highlights the rotational and scaling properties of complex multiplication: one number rotates and stretches the other in the complex plane.

Complex Division

In polar form, dividing complex numbers involves dividing their magnitudes and subtracting their arguments. Consider two complex numbers, \(z_1 = r_1(\cos \theta_1 + i \sin \theta_1)\) and \(z_2 = r_2(\cos \theta_2 + i \sin \theta_2)\). Their division is:

• Magnitude: \(\frac{|z_1|}{|z_2|} = \frac{r_1}{r_2}\)
• Angle: \(\theta_1 - \theta_2\)

Once you have the result in polar form, you can convert it back to rectangular form. The exercise performs these steps to find the quotient \(\frac{z_1}{z_2}\), particularly focusing on adjusting the magnitudes and angles and finally expressing the answer in the form \(a + ib\).

Complex division often feels like navigating backwards through a geometric transformation: one number is scaled down and rotated in the opposite direction of another.

Trigonometric Form of Complex Numbers

Complex numbers in trigonometric form are intimately related to their polar form, expressed as \(z = r(\cos \theta + i \sin \theta)\). This form is often labeled as the cis form, \(z = r \text{cis} \theta\), where \(\text{cis} \theta\) is shorthand for \(\cos \theta + i \sin \theta\). It allows for an elegant approach to multiplication and division, as demonstrated in this exercise.

The trigonometric or polar form connects complex numbers with vectors and rotations. Vectors are defined by their magnitude and direction, similar to \(r\) and \(\theta\). This representation is particularly handy for understanding the scaling and rotational effects in the plane, especially in operations like multiplication and division which inherently involve scaling (by the magnitudes) and rotation (by the angles).

Thus, having a good grasp of the trigonometric form helps visualize operations on complex numbers as transformations in the complex plane, making the intuitive leap from algebra to geometry.