Question

For each of the following two pairs of numbers compute \(z+w, z-w, z w,\) and \(z / w\) (a) \(z=6+8 i\) and \(w=3-4 i \quad\) (b) \(z=8 e^{i \pi / 3}\) and \(w=4 e^{i \pi / 6}\) Notice that for adding and subtracting complex numbers, the form \(x+i y\) is more convenient, but for multiplying and especially dividing, the form \(r e^{i \theta}\) is more convenient. In part (a), a clever trick for finding \(z / w\) without converting to the form \(r e^{i \theta}\) is to multiply top and bottom by \(w^{*}\); try this one both ways.

Short answer

Calculations: (a) 1. \(9+4i\), 2. \(3+12i\), 3. \(50\), 4. \(-\frac{14}{25}+\frac{56}{25}i\); (b) 1. \(6+6\sqrt{3}i\), 2. \(2-2\sqrt{3}+2i\), 3. \(32i\), 4. \(\sqrt{3}+i\).

Step-by-step solution

Addition of Complex Numbers (Part a)

For the given complex numbers \(z = 6 + 8i\) and \(w = 3 - 4i\), we add the real and imaginary parts separately. \[ z + w = (6 + 3) + (8i - 4i) = 9 + 4i \]

Subtraction of Complex Numbers (Part a)

Subtract the real and imaginary parts of the complex numbers: \[ z - w = (6 - 3) + (8i + 4i) = 3 + 12i \]

Multiplication of Complex Numbers (Part a)

Use the distributive property to multiply the complex numbers: \[ z \times w = (6 + 8i)(3 - 4i) \] Expand this using the FOIL method: \[ 6 \cdot 3 + 6 \cdot (-4i) + 8i \cdot 3 + 8i \cdot (-4i) \]This simplifies to: \[ 18 - 24i + 24i - 32i^2 \] Since \(i^2 = -1\), we have:\[ 18 - 32(-1) = 18 + 32 = 50 \] So, \(z \times w = 50 \).

Division of Complex Numbers Using Conjugate (Part a)

Multiply the numerator and the denominator by the conjugate of the denominator (\(w^* = 3 + 4i\)):\[ \frac{z}{w} = \frac{6 + 8i}{3 - 4i} \times \frac{3 + 4i}{3 + 4i} \] Perform the multiplication:Numerator: \[ (6 + 8i)(3 + 4i) = 18 + 24i + 32i - 32 = -14 + 56i \]Denominator: \[ (3 - 4i)(3 + 4i) = 9 + 12i - 12i - 16i^2 = 9 + 16 = 25 \] So, \[ \frac{z}{w} = \frac{-14 + 56i}{25} = -\frac{14}{25} + \frac{56}{25}i \]

Convert Complex Numbers to Polar Form for Multiplication (Part b)

The numbers are given as \( z = 8 e^{i \pi / 3} \) and \( w = 4 e^{i \pi / 6} \). Multiply their magnitudes and add their angles: \[ z \cdot w = 8 \times 4 e^{i (\pi/3 + \pi/6)} = 32 e^{i (\pi/2)} \]Simplifying, this is \( 32i \).

Division of Complex Numbers in Polar Form (Part b)

For division, divide the magnitudes and subtract the angles:\[ \frac{z}{w} = \frac{8}{4} e^{i (\pi/3 - \pi/6)} = 2 e^{i \pi/6} \]Convert back to rectangular form:\[ 2 \left( \cos(\pi/6) + i \sin(\pi/6) \right) = 2 \left( \frac{\sqrt{3}}{2} + i \frac{1}{2} \right) = \sqrt{3} + i \]

Addition of Complex Numbers (Part b)

Convert both numbers to rectangular form:For \( z = 8 e^{i \pi / 3} \):\[ 8 (\cos(\pi/3) + i\sin(\pi/3)) = 8 \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right) = 4 + 4 \sqrt{3}i \] For \( w = 4 e^{i \pi /6} \):\[ 4 \left( \cos(\pi/6) + i \sin(\pi/6) \right) = 4 \left( \frac{\sqrt{3}}{2} + i \frac{1}{2} \right) = 2 \sqrt{3} + 2i \] Add:\[ (4 + 4 \sqrt{3}i) + (2 \sqrt{3} + 2i) = 4 + 2 \sqrt{3} + (4 \sqrt{3} + 2)i = 4 + 2 \sqrt{3} + (6 \sqrt{3}i + 2i) \]

Subtraction of Complex Numbers (Part b)

Using the rectangular form:Subtract \( w \) from \( z \):\[ (4 + 4 \sqrt{3}i) - (2 \sqrt{3} + 2i) = (4 - 2 \sqrt{3}) + (4 \sqrt{3}i - 2i) = 4 - 2 \sqrt{3} + (4 \sqrt{3}i - 2i) \]

Key concepts

Addition of Complex Numbers

Complex numbers are made of two parts, a real part and an imaginary part. Adding complex numbers is as simple as adding the real parts together and the imaginary parts together. For example, if we have two complex numbers \( z = a + bi \) and \( w = c + di \), their sum \( z + w \) is calculated as \((a+c) + (b+d)i\).
This means you handle the real components separately from the imaginary ones. In the provided exercise, the numbers were \( z = 6 + 8i \) and \( w = 3 - 4i \). By adding the real parts: \(6 + 3 = 9\) and the imaginary parts: \(8i - 4i = 4i\), we get the sum: \(9 + 4i\).
Remember, the addition of complex numbers is commutative, which means \(z + w = w + z\). This makes the process quite straightforward and typically easy to understand and apply.

Multiplication of Complex Numbers

Multiplying complex numbers requires a bit more consideration than addition. You can use the distributive property, often employed through the FOIL method, to multiply expressions like \((a + bi)(c + di)\). Here is how it's done:

• Multiply the first terms: \(ac\)
• Multiply the outside terms: \(adi\)
• Multiply the inside terms: \(bci\)
• Multiply the last terms: \(bdi^2\)

Remember that \(i^2 = -1\).
So, when applied, this will eventually simplify into a sum of a real part and an imaginary part. For the case of \( z = 6 + 8i \) and \( w = 3 - 4i \), multiplication gives \(18 - 24i + 24i - 32i^2\). Which simplifies to \(18 + 32 = 50\), since \(-32i^2\) becomes \(+32\). The result here was simply a real number!

Polar Form of Complex Numbers

Complex numbers can also be represented in polar form, which is very useful for multiplication and division. Instead of using \(z = a + bi\), we express the complex number in terms of a magnitude \(r\) and an angle \(\theta\), as \(z = r e^{i\theta}\).
The magnitude \(r\) is the distance from the origin to the point on the complex plane and is calculated by the formula \(r = \sqrt{a^2 + b^2}\). The angle \(\theta\) is the direction measured from the positive real axis and can be found using \(\theta = \tan^{-1}(b/a)\).
In multiplication, you simply multiply the magnitudes and add the angles. For instance, the exercise uses \( z = 8 e^{i\pi/3} \) and \( w = 4 e^{i\pi/6} \). Their product is \(32 e^{i(\pi/2)}\), simplifying directly into \(32i\). Polar form excels in simplifying such operations.

Division using Conjugates

Dividing complex numbers can be done by multiplying the denominator and numerator by the conjugate of the denominator. The conjugate of a complex number \(a + bi\) is \(a - bi\).
Why the conjugate? Multiplying by the conjugate eliminates the imaginary part in the denominator, making the division simpler.
For example, to divide \( z = 6 + 8i \) by \( w = 3 - 4i \): we multiply by \( w^* = 3 + 4i \). This means:\[\frac{z}{w} = \frac{6 + 8i}{3 - 4i} \times \frac{3 + 4i}{3 + 4i}\]You simplify the numerator and denominator separately to get the result.
In this problem, after multiplying, we find the final expression \(-\frac{14}{25} + \frac{56}{25}i\). By applying the conjugate, the division becomes manageable without converting to polar form.