Imaginary Impact Slides Imaginary Impact
Multiplying Complex Numbers
ALGEBRA II / UNDERGRADUATE
The Paradox
What happens when you square a number that "doesn't exist"?
\[ i^2 = -1 \]
Multiplying two imaginary things can create something real. This is the fundamental engine of complex arithmetic.
\(i\)
?
?
-1
Algebraic Logic
We treat complex numbers \( a + bi \) as binomials .
Use the FOIL method (First, Outer, Inner, Last).
Distribute every term in the first number to every term in the second.
Crucial Step: Replace any \( i^2 \) with \( -1 \).
The Formula
\[ (a + bi)(c + di) = \] \[ ac + adi + bci + bdi^2 \] \[ \downarrow \] \[ (ac - bd) + (ad + bc)i \]
Walkthrough: \((3 + 2i)(1 - 4i)\)
1. Distribute
\( 3(1) + 3(-4i) + 2i(1) + 2i(-4i) \)
2. Multiply
\( 3 - 12i + 2i - 8i^2 \)
3. Resolve \(i^2\)
\( 3 - 10i - 8(-1) \)
\( 3 - 10i + 8 \)
4. Final Form
\( 11 - 10i \)
Powers of \(i\) Cyclic Nature
First
\(i^1 = i\)
Second
\(i^2 = -1\)
Third
\(i^3 = -i\)
Fourth
\(i^4 = 1\)
When multiplying complex numbers, you will most often encounter \(i^2\). Always replace it immediately to simplify your real component.
Pitfall Patrol
The Sign Swap
The most common error is forgetting that \( - (bi)^2 \) becomes \( +b^2 \).
Incorrect: \( (2i)(2i) = -4i^2 \)
The Standard Form
A result is not finished until it is in \( a + bi \) form.
Don't leave \( 12 - 4i + 7i^2 \).
Product Power Worksheet Product Power
Complex Number Multiplication Workshop
Student:
Date:
Core Strategy: Treat complex numbers as binomials and distribute. Remember that \(i^2 = -1\). Always simplify your result to standard form \(a + bi\).
1 Monomial Foundations
Simplify each expression completely.
\( 3i \cdot 4i \)
\( -2i(5i) \)
\( \sqrt{-4} \cdot \sqrt{-9} \)
\( 2i(-3i)(4i) \)
2 Binomial Distribution (FOIL)
Distribute and simplify into standard form \( a + bi \).
\( (2 + 3i)(1 + 4i) \)
Show Work
\( (5 - 2i)(3 + i) \)
Show Work
\( (-3 + 4i)(2 - 6i) \)
Show Work
3 Powers of Complex Binomials
Expand and simplify: \( (3 + 2i)^2 \)
Expand and simplify: \( (1 - i)^3 \)
Application: AC Circuits
In electrical engineering, impedance \( Z \) is represented as a complex number. The voltage \( V \) is the product of current \( I \) and impedance \( Z \): \( V = I \cdot Z \).
If the current is \( I = (4 + 2i) \) Amperes and the impedance is \( Z = (5 - 3i) \) Ohms, calculate the total voltage \( V \) in standard form.
Product Power Answer Key Product Power
Teacher Answer Key
Lesson 1: Multiplication
1. Monomial Foundations
\( 3i \cdot 4i \)
\( 12i^2 = 12(-1) = \) -12
\( -2i(5i) \)
\( -10i^2 = -10(-1) = \) 10
\( \sqrt{-4} \cdot \sqrt{-9} \)
\( (2i)(3i) = 6i^2 = \) -6
\( 2i(-3i)(4i) \)
\( (-6i^2)(4i) = (6)(4i) = \) 24i
2. Binomial Distribution (FOIL)
\( (2 + 3i)(1 + 4i) \)
\( 2(1) + 2(4i) + 3i(1) + 3i(4i) = 2 + 8i + 3i + 12i^2 \)-10 + 11i
\( (5 - 2i)(3 + i) \)
\( 15 + 5i - 6i - 2i^2 = 15 - i + 2 = \) 17 - i
\( (-3 + 4i)(2 - 6i) \)
\( -6 + 18i + 8i - 24i^2 = -6 + 26i + 24 = \) 18 + 26i
3. Powers of Complex Binomials
\( (3 + 2i)^2 \)
\( (3+2i)(3+2i) = 9 + 6i + 6i + 4i^2 = 9 + 12i - 4 = \) 5 + 12i
\( (1 - i)^3 \)
\( (1-i)(1-i) = 1 - 2i + i^2 = 1 - 2i - 1 = -2i \)-2 - 2i
Application: AC Circuits Solution
\( V = I \cdot Z \)
\( V = (4 + 2i)(5 - 3i) \)
\( V = 20 - 12i + 10i - 6i^2 \)
\( V = 20 - 2i + 6 \)
\( V = 26 - 2i \) Volts
Mirror Image Slides The Mirror Image
The Role of the Complex Conjugate
LESSON 2
Defining the Conjugate
The complex conjugate of a number is its "mirror image" across the real axis.
If the number is:
\( z = a + bi \)
The conjugate is:
\( \bar{z} = a - bi \)
\(z = 3 + 2i\)
\(\bar{z} = 3 - 2i\)
The Neutralizing Property
What happens when you multiply a complex number by its conjugate?
\[ z \cdot \bar{z} = (a + bi)(a - bi) \] \[ = a^2 - abi + abi - (bi)^2 \] \[ = a^2 - b^2(-1) \] \[ \downarrow \] \[ a^2 + b^2 \]
Imaginary Elimination
The imaginary parts always cancel out perfectly, leaving only a real number.
Non-Negativity
Because it's a sum of squares, the product is always real and \(\ge 0\).
The "Rationalization Tool"
Why do we care about conjugates?
Division
We cannot "divide" by an imaginary number easily. The conjugate turns the denominator into a real number.
Norms
It is used to define the absolute value (magnitude) of a complex number.
Physics
In quantum mechanics, the product of a wavefunction and its conjugate gives a probability density.
Conjugate Inquiry Activity Conjugate Inquiry
Discovering the Neutralizing Power of Mirror Images
Investigator:
Date:
Investigation 1: Identifying the Pair
A complex conjugate is formed by changing the sign of the imaginary part. Identify the conjugate \(\bar{z}\) for each given complex number \(z\).
\( z = 3 + 5i \)
\( \bar{z} = \) ________
\( z = -2 - 7i \)
\( \bar{z} = \) ________
\( z = 4i \)
\( \bar{z} = \) ________
\( z = 8 \)
\( \bar{z} = \) ________
\( z = \sqrt{2} - i \)
\( \bar{z} = \) ________
\( z = -1 + 0i \)
\( \bar{z} = \) ________
Investigation 2: The Vanishing Imaginary
Perform the multiplication for each pair. Show all steps and notice what happens to the terms involving \(i\).
\( (4 + 3i)(4 - 3i) \)
\( (1 - 2i)(1 + 2i) \)
\( (-5 + i)(-5 - i) \)
Synthesis & General Rule
1. Observation:
What happened to the "inner" and "outer" terms in every product above?
2. The Rule:
Based on your findings, derive a formula for the product of any complex number and its conjugate:
\( (a + bi)(a - bi) = \)
3. Property Check:
Is the product of a complex number and its conjugate always real? Why or why not?
Conjugate Inquiry Answer Key Conjugate Inquiry
Teacher Solution Guide
Lesson 2: Conjugates
Investigation 1: Identifying the Pair
\( z = 3 + 5i \)
\( \bar{z} = 3 - 5i \)
\( z = -2 - 7i \)
\( \bar{z} = -2 + 7i \)
\( z = 4i \)
\( \bar{z} = -4i \)
\( z = 8 \)
\( \bar{z} = 8 \)
\( z = \sqrt{2} - i \)
\( \bar{z} = \sqrt{2} + i \)
\( z = -1 + 0i \)
\( \bar{z} = -1 \)
Investigation 2: Products
\( (4 + 3i)(4 - 3i) \)=
\( 16 - 12i + 12i - 9i^2 = 16 + 9 = \) 25
\( (1 - 2i)(1 + 2i) \)=
\( 1 + 2i - 2i - 4i^2 = 1 + 4 = \) 5
\( (-5 + i)(-5 - i) \)=
\( 25 + 5i - 5i - i^2 = 25 + 1 = \) 26
Synthesis Solutions
1. Observation:
The "inner" (\(abi\)) and "outer" (\(-abi\)) terms always have opposite signs and equal magnitude, so they cancel out completely (sum to zero).
2. The Rule:
\( (a + bi)(a - bi) = a^2 + b^2 \)
3. Property Check:
Yes. The product \(a^2 + b^2\) consists of two squared real numbers. Since real numbers squared are real (and non-negative), their sum is also real and non-negative. There is no imaginary unit \(i\) left in the expression.
Rationalize This Slides Rationalize This
Complex Division Strategies
LESSON 3
The Roadblock
How do you divide by a binomial that contains \(i\)?
\[ \frac{2 + 3i}{1 - i} = \text{?} \]
There is no simple long-division for complex numbers. Instead, we use a clearing strategy .
The Solution
Rationalize the Denominator
Multiply both the numerator and denominator by the complex conjugate of the denominator.
Since \(\frac{\bar{z}}{\bar{z}} = 1\), we aren't changing the value, just the form.
The 3-Step Algorithm
Step 1
Identify the Conjugate
Look at the denominator and find its conjugate.
\( 3 - 2i \rightarrow 3 + 2i \)
Step 2
Multiply and FOIL
Multiply the entire fraction by \(\frac{\text{Conjugate}}{\text{Conjugate}}\).
Apply FOIL to both top and bottom.
Step 3
Simplify to \( a + bi \)
The bottom will become a real number. Split the fraction.
\( \frac{a}{c} + \frac{b}{c}i \)
Example: \(\frac{2 + i}{3 - i}\)
1
Conjugate of \(3 - i\) is \(3 + i\).
2
\[ \frac{2 + i}{3 - i} \cdot \frac{3 + i}{3 + i} = \frac{6 + 2i + 3i + i^2}{9 + 3i - 3i - i^2} \]
3
\[ \frac{6 + 5i - 1}{9 + 1} = \frac{5 + 5i}{10} \]
Final Answer
\( \frac{1}{2} + \frac{1}{2}i \)
Always split the real and imaginary parts.
Quotient Quest Worksheet Quotient Quest
Complex Division & Rationalization
Student Name:
Date:
Procedural Check
To divide by \( a + bi \), multiply the numerator and denominator by its conjugate \( a - bi \). This transforms the denominator into a real value \( a^2 + b^2 \).
1. Monomial Denominators
Simplify into standard form \( a + bi \).
\( \frac{4}{2i} \)
\( \frac{3 + i}{i} \)
2. Rationalizing with Conjugates
Show step-by-step rationalization.
\( \frac{5}{1 + 2i} \)
Work Area
\( \frac{2 - 3i}{4 + i} \)
Work Area
3. Multi-Step Simplification
Perform the indicated operation and simplify the result to standard form.
\( \frac{(1 + i)(2 - i)}{3 + 2i} \)
Physics Context: Current Phase
In a specific circuit, the voltage is \( V = 10 + 5i \) and the total impedance is \( Z = 2 - i \). The current \( I \) is calculated as \( I = V / Z \). Find the current \( I \) in standard form \( a + bi \).
Quotient Quest Answer Key Quotient Quest
Teacher Answer Key
Lesson 3: Division
1. Monomial Denominators
\( \frac{4}{2i} \)
\( \frac{4}{2i} \cdot \frac{i}{i} = \frac{4i}{2i^2} = \frac{4i}{-2} = \) -2i
\( \frac{3 + i}{i} \)
\( \frac{3 + i}{i} \cdot \frac{i}{i} = \frac{3i + i^2}{i^2} = \frac{3i - 1}{-1} = \) 1 - 3i
2. Rationalizing with Conjugates
\( \frac{5}{1 + 2i} \)
\( \frac{5}{1+2i} \cdot \frac{1-2i}{1-2i} = \frac{5 - 10i}{1^2 + 2^2} = \frac{5 - 10i}{5} = \) 1 - 2i
\( \frac{2 - 3i}{4 + i} \)
\( \frac{2-3i}{4+i} \cdot \frac{4-i}{4-i} = \frac{8 - 2i - 12i + 3i^2}{16 + 1} = \frac{8 - 14i - 3}{17} = \frac{5 - 14i}{17} \)\( \frac{5}{17} - \frac{14}{17}i \)
3. Multi-Step Simplification
\( \frac{(1 + i)(2 - i)}{3 + 2i} \)
Step 1 (Numerator): \( (1+i)(2-i) = 2 - i + 2i - i^2 = 2 + i + 1 = 3 + i \)
Step 2 (Ratio): \( \frac{3+i}{3+2i} \)
Step 3 (Rationalize): \( \frac{3+i}{3+2i} \cdot \frac{3-2i}{3-2i} = \frac{9 - 6i + 3i - 2i^2}{9 + 4} = \frac{9 - 3i + 2}{13} = \frac{11 - 3i}{13} \)
Result: \( \frac{11}{13} - \frac{3}{13}i \)
Physics Solution
\( I = \frac{V}{Z} = \frac{10 + 5i}{2 - i} \)
\( I = \frac{10 + 5i}{2 - i} \cdot \frac{2 + i}{2 + i} = \frac{20 + 10i + 10i + 5i^2}{2^2 + 1^2} \)
\( I = \frac{20 + 20i - 5}{5} = \frac{15 + 20i}{5} \)
\( I = 3 + 4i \) Amps
Magnitude Map Slides Magnitude Map
Modulus & Complex Properties
LESSON 4
What is the "Size" of a Complex Number?
The modulus (or absolute value) is the distance from the origin \((0,0)\) to the number in the complex plane.
For \( z = a + bi \):
\[ |z| = \sqrt{a^2 + b^2} \]
|z|
\(a + bi\)
Distance = \(\sqrt{a^2+b^2}\)
The Crucial Identity
The modulus is deeply linked to the complex conjugate:
\[ |z|^2 = z \cdot \bar{z} \]
"The square of the magnitude is the number times its mirror."
Eliminates square roots in calculations.
Fundamental to complex analysis and proofs.
Rules of Engagement
Multiplication
\[ |z_1 z_2| = |z_1| \cdot |z_2| \]
The magnitude of a product is the product of the magnitudes.
Division
\[ \left| \frac{z_1}{z_2} \right| = \frac{|z_1|}{|z_2|} \]
The magnitude of a quotient is the quotient of the magnitudes (where \(z_2 \neq 0\)).
Triangle Inequality
\[ |z_1 + z_2| \leq |z_1| + |z_2| \]
"The shortest distance between two points is a straight line."
Modulus Mastery Workshop Modulus Mastery
Magnitude, Conjugates, & Arithmetic Proofs
Student:
Date:
1. Magnitude Calculations
Calculate the modulus \( |z| \) for each complex number. Simplify any radicals.
\( z = 3 + 4i \)
\( z = -5 - 12i \)
\( z = 1 + i \)
\( z = 2\sqrt{3} - 2i \)
2. Verifying the Link: \( |z|^2 = z \cdot \bar{z} \)
Choose \( z = 2 + 3i \). Calculate both sides to verify the identity.
Side A: \( |z|^2 \)
Side B: \( z \cdot \bar{z} \)
3. The Product Property: \( |z \cdot w| = |z| \cdot |w| \)
Let \( z = 1 + i \) and \( w = 3 - i \).
A. Calculate the product \( z \cdot w \) first, then find its magnitude \( |z \cdot w| \).
B. Calculate \( |z| \) and \( |w| \) separately, then find their product \( |z| \cdot |w| \).
4. Efficiency Challenge
Without performing the full division (rationalizing), calculate the magnitude of the following quotient:
\[ \left| \frac{10 + 5i}{2 - i} \right| \]
Hint: Use the division property of the modulus
Modulus Mastery Answer Key Modulus Mastery
Teacher Solution Guide
Lesson 4: Modulus
1. Magnitude Calculations
\( z = 3 + 4i \)
\( \sqrt{3^2+4^2} = \sqrt{25} = 5 \)
\( z = -5 - 12i \)
\( \sqrt{(-5)^2+(-12)^2} = \sqrt{169} = 13 \)
\( z = 1 + i \)
\( \sqrt{1^2+1^2} = \sqrt{2} \)
\( z = 2\sqrt{3} - 2i \)
\( \sqrt{(2\sqrt{3})^2+(-2)^2} = \sqrt{12+4} = 4 \)
2. Verifying the Link
Side A: \( |z|^2 \)
\( |z| = \sqrt{2^2+3^2} = \sqrt{13} \)13
Side B: \( z \cdot \bar{z} \)
\( (2+3i)(2-3i) = 4 - 9i^2 \)13
3. Product Property Verification
A. Multiply then Magnitude:
\( z \cdot w = (1+i)(3-i) = 3 - i + 3i - i^2 = 4 + 2i \)\( 2\sqrt{5} \)
B. Individual Magnitudes then Multiply:
\( |z| = \sqrt{1^2+1^2} = \sqrt{2} \)\( 2\sqrt{5} \)
4. Efficiency Solution
\[ \left| \frac{10+5i}{2-i} \right| = \frac{|10+5i|}{|2-i|} \]
\( |10+5i| = \sqrt{100+25} = \sqrt{125} = 5\sqrt{5} \)
\( |2-i| = \sqrt{4+1} = \sqrt{5} \)
Result: \( \frac{5\sqrt{5}}{\sqrt{5}} = \) 5
Operation Overload Slides Operation Overload
Simplifying Rational Expressions
LESSON 5: CULMINATION
The Final Boss
In advanced mathematics and physics, you rarely see simple division. You see compounded expressions .
\[ \frac{(3 + i) + (2 - 5i)}{(1 + 2i)(4 - i)} \]
Our goal is to reduce this chaos into a single, clean \( a + bi \).
Order of Attack
1
Simplify the Numerator (combine like terms or FOIL).
2
Simplify the Denominator into a single complex number.
3
Rationalize the final quotient using the conjugate.
Strategic Efficiency
Hard Mode
Rationalizing a denominator that is still a product of two binomials.
\( \frac{z}{w_1 w_2} \cdot \frac{\overline{w_1 w_2}}{\overline{w_1 w_2}} \)
Lenny's Pro Tip
Efficiency Mode
Always multiply the products in the top and bottom before you try to rationalize.
1. FOIL the bottom \(\rightarrow (c + di)\)
Beyond the Classroom
Why do we master these "Simplification Gauntlets"? Because in Signal Processing and Robotics, transfer functions are complex ratios.
\[ H(s) = \frac{s + 2}{s^2 + 2s + 2} \]
When we evaluate these at specific frequencies (where \(s = j\omega\)), we need the exact skills you've built this week to find phase and magnitude.
Simplification Gauntlet Activity Simplification Gauntlet
The Ultimate Challenge: Mixed Complex Operations
Candidate:
Date:
1 Level 1: Compound Arithmetic
Simplify into standard form \( a + bi \).
\( 2i(3 - i) + (4 + 5i) \)=
\( (1 + i)^2 - (1 - i)^2 \)=
2 Level 2: Rational Mastery
Simplify the fraction completely.
\( \frac{(2 + i)(3 - 2i)}{i} \)
\( \frac{10}{ (1 - i)^2 } \)
3 Level 3: Strategic Rationalization
Reduce the expression below into standard form. Hint: Simplify the denominator completely before rationalizing.
\[ \frac{5 + 5i}{(1 + i)(2 + i)} \]
Show your multi-step process here
Level 4: The Final Gauntlet
This expression appears in the study of electric filter resonance. Simplify it to standard form \( a + bi \).
\[ \frac{1}{1 + i} + \frac{1}{1 - i} \]
Simplification Gauntlet Answer Key Simplification Gauntlet
Teacher Answer Key
Lesson 5: Final Mastery
Level 1: Compound Arithmetic
\( 2i(3 - i) + (4 + 5i) \)
\( (6i - 2i^2) + 4 + 5i = (6i + 2) + 4 + 5i = \) 6 + 11i
\( (1 + i)^2 - (1 - i)^2 \)
\( (1 + 2i + i^2) - (1 - 2i + i^2) = (2i) - (-2i) = \) 4i
Level 2: Rational Mastery
\( \frac{(2 + i)(3 - 2i)}{i} \)
Numerator: \( 6 - 4i + 3i - 2i^2 = 8 - i \)-1 - 8i
\( \frac{10}{ (1 - i)^2 } \)
Denominator: \( 1 - 2i + i^2 = -2i \)5i
Level 3: Strategic Rationalization
\[ \frac{5 + 5i}{(1 + i)(2 + i)} \]
Step 1 (Denominator): \( (1+i)(2+i) = 2 + i + 2i + i^2 = 1 + 3i \)
Step 2 (Expression): \( \frac{5 + 5i}{1 + 3i} \)
Step 3 (Rationalize): \( \frac{5+5i}{1+3i} \cdot \frac{1-3i}{1-3i} = \frac{5 - 15i + 5i - 15i^2}{1^2 + 3^2} = \frac{5 - 10i + 15}{10} \)
Step 4 (Simplify): \( \frac{20 - 10i}{10} = \) 2 - i
Level 4: Final Solution
Method 1 (Conjugates first):
\( \frac{1}{1+i} \cdot \frac{1-i}{1-i} = \frac{1-i}{2} \)
\( \frac{1}{1-i} \cdot \frac{1+i}{1+i} = \frac{1+i}{2} \)
\( \frac{1-i}{2} + \frac{1+i}{2} = \frac{2}{2} = \) 1
Method 2 (Common Denominator):
\( \frac{(1-i) + (1+i)}{(1+i)(1-i)} \)
\( \frac{2}{1^2 + 1^2} = \frac{2}{2} = \) 1
The Answer: 1
