# Complex Numbers with Mathematica: Complex Arithmetic, Methods, and Geometric Interpretations

## Video Lectures

Displaying all 26 video lectures.

Lecture 1Play Video |
The imaginary unit and how to add complex numbersComplex Analysis, Video #1 (Complex Arithmetic, Methods and Geometric Interpretations, Part 1). Topics: the imaginary unit, form of a complex number, addition of complex numbers, entering text and input in Mathematica |

Lecture 2Play Video |
Complex Addition and the Parallelogram Law. Use ListPlot on Mathematica to make the plot.Complex Analysis, Video #2. Complex Arithmetic, Methods and Geometric Interpretations, Part 2. (Complex addition in the complex plane...hint at parallelogram law) Review: imaginary unit, complex numbers (use "Re" and "Im" notation), addition of complex numbers. Plotting complex numbers in the complex plane using ListPlot on Mathematica (Mma) and noting how, when you plot two complex numbers, their sum, and the origin, you always get a parallelogram if the complex numbers don't lie on the same line through the origin (in which case their sum would lie on that line as well). Options used for ListPlot include AxesLabel, PlotRange, PlotStyle, and AspectRatio on Mma. |

Lecture 3Play Video |
Complex Number Addition and the Parallelogram Law. Use of Mathematica to create vectors.Complex Analysis Video #3 (Complex Arithmetic, Part 3). Complex Addition and the Parallelogram Law. Use of Mathematica to create vectors. Review of the Complex Plane and the geometric interpretation of complex addition. Mathematica code to emphasize the parallelogram law, both using line segments and vectors. Use Mathematica commands: Show, Graphics, Line, Arrow along with previous ones. At 1:15, I should have said "seven plus eleven i" (instead of "seven plus five i") |

Lecture 4Play Video |
Complex Number Addition, Parallelogram Law, Triangle Inequality, and Manipulate on MathematicaComplex Analysis Video #4 (Complex Arithmetic, Part 4). Complex Addition and the Parallelogram Law. Intro to the Triangle Inequality and use of "Manipulate" on Mathematica. More detail: Geometric interpretation of complex number addition in terms of vector addition (both in terms of the parallelogram law and in terms of triangles). Emphasize that complex numbers can either be thought of as points or as vectors. Triangle inequality verbally described based on the picture. Introduction to the Mathematica command "Manipulate" to make an animation to illustrate complex number/vector addition when the real part of one of the addends changes. |

Lecture 5Play Video |
Modulus of a Complex Number, Triangle Inequality, Manipulate and Locator on MathematicaComplex Analysis Video #5 (Complex Arithmetic, Part 5). Review the use of "Manipulate" on Mathematica. Introduce the use of "Locator" in Mathematica, all while continuing to illustrate the parallelogram law for complex number and vector addition. Introduce the idea of the modulus of a complex number and illustrate the triangle inequality using the modulus. |

Lecture 6Play Video |
Complex Number Subtraction in terms of Vectors, Manipulate and Locator on MathematicaComplex Analysis, Video #6 (Complex Arithmetic, Part 6). Triangle Inequality and Complex Number Subtraction. Details: Review use of "Locator" within "Manipulate" on Mathematica. Review the Triangle Inequality; write the symbolic form of the triangle inequality. Introduce complex subtraction both arithmetically and geometrically. |

Lecture 7Play Video |
Introduction to Multiplying Complex Numbers and Geometrically Interpreting the ProductComplex Analysis Video #7 (Complex Arithmetic, Methods and Geometric Interpretations, Part 7). Complex Multiplication and Intro to its Geometric Interpretation. Details: Review complex subtraction from previous videos, including details about the diagram. Complex multiplication (2 + 3i)*(-1 + 4i). Make a guess about how to it by using FOIL on binomials...with the caveat that i^2 = -1. This is indeed how to do it. General rule for complex multiplication given. Use of double square bracket notation on Mathematica to pick off the first and second coordinates of a point (representing the real and imaginary parts, respectively, of complex numbers). Plot the product. Describe the geometric interpretation verbally. |

Lecture 8Play Video |
Complex Multiplication in terms of Moduli and Arguments. Use Mathematica to illustrate.Complex Analysis, Video #8 (Complex Arithmetic, Methods and Geometric Interpretations, Part 8). Complex Multiplication and its Geometric Interpretation. Intro to Modulus and Argument. Details: Review complex multiplication. Review visualization using Locator on Mathematica. Verify that |z1*z2| = |z1|*|z2| using Abs on Mathematica (Abs[-14+5i] = Abs[2+3i]*Abs[-1+4i] since 13*17=221=196 + 25). The modulus of the product is the product of the modulii of the factors. The argument of a complex number. Verify that the argument of the product is the sum of the arguments of the factors by creating triangles and using trigonometry. Use the arctangent function on the triangles (via ArcTan on Mma) to confirm this. |

Lecture 9Play Video |
Confirm the Geometry of Complex Number Multiplication with Manipulate and Locator. Principal Value.Complex Analysis Video #9, (Complex Arithmetic, Part 9). Complex Multiplication and its Geometric Interpretation. Modulus and Argument. Intro to Principal Value of the Argument. Details: Review what happens when you multiply two complex numbers. The modulus of the product is the product of the modulii of the factors. The argument of the product is the sum of the arguments of the factors. Write those two facts symbolically (using absolute value symbols for the modulii and "arg" for arguments). Experiment with the Manipulate (Locator) output to test these facts. Briefly mention the "principal value" of the argument (between -pi and pi). Show how to re-do the code using Re and Im in Mathematica). Use the Arg command in Mathematica to find the principal value of the argument. |

Lecture 10Play Video |
Complex Number Reciprocals (Multiplicative Inverses), approached AlgebraicallyComplex Analysis Video #10 (Complex Arithmetic, Part 10). Multiplicative Inverses of Complex Numbers approached Algebraically. Details: Review modulus and argument properties of multiplication. What would it mean for a + b*I to have a multiplicative inverse x + y*i such that (a + b*i)*(x + y*i) = 1? Derive it using a system of equations in the unknowns x & y. Solve via substitution assuming that "a" is not zero. Find a general formula. Have Mathematica check it using Simplify and ComplexExpand. What if a = 0? Still get the same answer if b is nonzero. The only time it doesn't work is if both "a" and "b" equal zero. Find multiplicative inverse of z1 = 5 + 7*i. Use the derived formula to get it. How should we divide z1 = 5 + 7i by z3 = 2 + 3i? Same as z1*(1/z3)...could also do as z1/z3 (the answer is 31/13-i/13). |

Lecture 11Play Video |
Complex Multiplicative Inverses, Complex Division, and Complex ConjugatesComplex Analysis, Video #11 (Complex Arithmetic, Part 11). Complex reciprocal (multiplicative inverse), Complex Division, Intro to Complex Conjugate. Details: Review complex reciprocal & complex division. Introduce complex conjugate as something that is helpful in doing complex number division...the idea (geometrically), notation, and formula. A complex number times its complex conjugate is real and is the square of the modulus of the original complex number. Rewrite modulus and argument principles for multiplication in terms of division. |

Lecture 12Play Video |
Complex Conjugates, Complex Division, and Visualization on Mathematica.Complex Analysis Video #12 (Complex Arithmetic, Part 12). Review of Geometric Interpretation of Complex Multiplication and Division. Complex Conjugates (symbols and use). Details: Review of moduli and argument properties of multiplication and division. Visualize division using Manipulate on Mathematica (starts with (2 + 3i)/(-1 + 4i). Review complex conjugates. Special equations worth remembering: z*zbar = |z|^2; 1/z = zbar/|z|^(2) (show how it comes up in complex division). Example: calculate (5+7i)/(2+3i) by multiplying by (2-3i)/(2-3i). |

Lecture 13Play Video |
Introduction to the Polar Form of a Complex Number and Complex MultiplicationComplex Analysis, Video #13 (Complex Arithmetic, Part 13). Geometry of Complex Multiplication in Polar Coordinates (or "Polar Form"). Look at moduli and argument properties via polar form; multiply (sqrt(8)+sqrt(8)*i)*(sqrt(3)+i) = (2sqrt(6)-2sqrt(2)) + i*(2sqrt(6)+2sqrt(2)); view the product in diagram; think about product in terms of modulii and arguments; calculate modulus and argument of both z1 and z2; write final answer in polar form as 8*cos(5pi/12) + i*8*sin{5pi/12); write polar conversion equations x = r*cos(theta), y = r*sin(theta)...think of in terms of points in the plane. |

Lecture 14Play Video |
Polar Form of Complex Numbers, both with "Cis" & with "e" (Euler's Formula)Complex Analysis Video #14 (Complex Arithmetic, Part 14). Polar Form representations of Complex Multiplication using "cis" and also using the Complex Exponential via Euler's Formula. Details: Polar form. Initially as x + i*y = r cos(theta) + i* r sin(theta); then write as r*cis(theta); moduli and argument rules in terms of polar coordinates: z1*z2 = r1*r2*cis(theta1 + theta2); z1/z2 = r1/r2*cis(theta1-theta2); Euler's formula: e^(I*theta) = cos(theta) + i*sin(theta); also write z1*z2 = r1*r2*e^(i*(theta1 + theta2)); z1/z2 = r1/r2*e^(i*(theta1-theta2)). |

Lecture 15Play Video |
De Moivre's Formula and Trigonometric Identities (mistake at the end...see description below)Complex Analysis, Video #15 (Complex Arithmetic, Part 15). Complex Multiplication in Polar Form. De Moivre's Formula. Application to Derivation of Trigonometric Identity. Details: Review polar form, focusing especially on the exponential form. State De Moivre's formula. Apply De Moivre's formula to derive a couple trigonometric identities. Check answers with "TrigExpand" on Mathematica. Mistake at the end: I forgot to get rid of the "i" in the identity for sin(3*theta). |

Lecture 16Play Video |
De Moivre, Trig Identities, Sine and Cosine in Terms of ExponentialsComplex Analysis, Video 16 (Complex Arithmetic, Part 16). Application of De Moivre's Formula to Deriving Trigonometric Identities. Application of Euler's Formula to derive Exponential Representations of Cosine and Sine Functions. Details: Review De Moivre's formula. Review application to trig identities. Mention mistake in previous video (#15) where I forgot to get rid of the "I" for the sine trigonometric identity. Derive another trigonometric identity by using (cos(theta)+i*sin(theta))^4 = cos(4theta)+i*sin(4theta) and the binomial theorem. Verbal mistake at 2:05: the second term in the binomial expression is NOT squared to get the second term in the expansion. Use Euler's formula in two ways to derive alternative representations of cos(theta) and sin(theta) in terms of exponentials involving the imaginary unit i. |

Lecture 17Play Video |
A Real Integral done using Complex Arithmetic (Euler's Formula)Complex Analysis, Video #17 (Complex Arithmetic, Part 17). Applications of De Moivre's and Euler's Formulas to Trigonometric Identities and Calculation of Integrals. Details: The trig identities derived in previous videos are true, but perhaps not very useful. In this video, we do an application that is useful for (real) integration. Integrate (cos(theta))^3 by using the fact that cos(theta) = (e^(i*theta)+e(-i*theta))/2, the binomial theorem, and then Euler's formula (in your head). Do the integral. Check it graphically on Mathematica (use Mathematica's "Plot" command...as well as the formatting option "PlotStyle"). |

Lecture 18Play Video |
Check the use of Cosine as an Exponential to the Evaluation of an Integral.Complex Analysis, Video #18 (Complex Arithmetic, Part 18). Checking an Application of an Exponential Representation of Cosine to the Evaluation of an Integral. Review the use, from the previous video, of exponential representation of the cosine function and Euler's formula to evaluate the indefinite integral of (cos(theta))^3. Now do the integral using the Pythagorean identity and a substitution. Then use Mathematica commands "ComplexExpand" and "TrigReduce" to check that this is the same answer as from the previous video. |

Lecture 19Play Video |
Powers of Complex Numbers (and an intro to "Table" on Mathematica).Complex Analysis, Video #19 (Complex Arithmetic, Part 19). Powers of Complex Numbers (and an intro to "Table" on Mathematica). First focus on powers of a complex number using Euler's formula. Write formula for z^(n) = r^(n)*e^(i*n*theta) when z = r*e^(i*theta). Example: compute (1+sqrt(3)*i)^(5) by writing it as (2*e^(i*(pi/3)))^(5). Get 16 - 16*sqrt(3)*i. Check it with ComplexExpand on Mathematica. Visualize this using Show, ListPlot, Graphics, and Table on Mathematica. |

Lecture 20Play Video |
Using Mathematica to Visualize Powers of Complex NumbersComplex Analysis, Video #20 (Complex Arithmetic, Part 20). Review polar form and its use in raising a complex number to a power. Modify Mathematica code to include alternate PlotRange and change the AspectRatio to make the angles "true". Review relation between moduli and arguments of products with factors. Use Manipulate and Locator to allow the base number to be moved with a cursor and see how the powers end up moving around as the starting point (base number) is moved around. |

Lecture 21Play Video |
Dynamic Behavior of Powers of Complex Numbers, Intro to Roots and Multi-Valued FunctionsComplex Analysis, Video #21 (Complex Arithmetic, Part 21). Review polar form of positive integer powers. Review use of Manipulate and Locator on Mathematica to animate this process; use Graphics and Circle to create the unit circle and see how the unit circle forms a boundary between two types of behavior of powers of complex numbers (either going to infinity or going to zero). Start with a point on the unit circle (the point 1/2 + sqrt(3)/2*i) and see that the powers generate a periodic sequence of points. Use ComplexExpand to check it. Try it for 3/5 + (4/5)*i and see that it never comes back to itself. Find 1^(1/2) by thinking about polar form. Emphasize that this really represents all possible square roots (it's "multi-valued"). Get +1 and -1. Find 1^(1/3) by thinking about polar form as well. Get 1, e^(i*2pi/3) = -1/2 + i*sqrt(3)/2, and e^(i*4pi/3) = -1/2 - i*sqrt(3)/2. Check with ComplexExpand. |

Lecture 22Play Video |
Deriving and Graphing Complex Roots of UnityComplex Analysis, Video #22 (Complex Arithmetic, Part 22). When you find roots of complex numbers, you get more than one answer (roots are "multi-valued functions"). We're focusing at the moment on finding roots of "unity" (one). The notation 1^(1/3), for instance, represents all possible cube roots of one. We get more than one answer because of the multiplicity of polar representations of complex numbers. Using the polar form of a possible root, we can easily derive the roots. In this video, the main example is to find the 12th roots of unity, 1^(1/12). Find the 12 distinct roots in polar form and then use ListPlot and Table to plot the roots. Also use Graphics and Line along with Table to graph line segments from the origin to the roots. Notice that they are all 30 degrees apart, which makes sense because 360/12 = 30. Also note that we could connect successive roots and get a regular 12-gon (dodecagon). |

Lecture 23Play Video |
Graphing Complex Roots with MathematicaComplex Analysis, Video #23 (Complex Arithmetic, Part 23). Main Topic: Finding the nth roots of a complex number and visualizing them as a regular polygon using Mathematica Review that we found the 12th roots of unity in the last video and visualized them with ListPlot. Add lines between the vertices in this video to create a regular dodecagon (12-sided regular polygon). Discuss how to define the general nth roots of an arbitrary complex number, given in polar form (n is a positive integer). Write a proposed nth root in polar form and consider what the possibilities for the modulus and argument would be. Use the radical notation to indicate non-negative real nth roots of non-negative real numbers. The ambiguity in the argument of the original number gives n possibilities for the nth root...every nonzero complex number has exactly n nth roots. The nth root function is therefore multi-valued. Finish the video by making an animation using Manipulate and Locator to graph the nth roots of an arbitrary complex number (and Locator will allow us to move the starting point around)...initially accidentally raise the modulus to the 12th power, but then I catch the mistake and fix it (should raise it to the 1/12th power). |

Lecture 24Play Video |
More on Visualizing Complex Roots with MathematicaComplex Analysis, Video #24 (Complex Arithmetic, Part 24). Main Topic: finding the nth roots of a complex number and visualizing them to see that they form a regular polygon in the complex plane. Review animation from previous video to visualize the set of 12th roots of any nonzero complex number. Add an extra animation parameter that allows us to visualize mth roots for integer values of m from 2 to 12. Write a formula for the set of nth roots, first in exponential polar form, then in trigonometric polar form. Find the 5th roots of z = -1 + sqrt(3)*i. The modulus of z is 2 and the principal value of the argument is 2*pi/3. The modulus of the 5th roots is the non-negative (real) 5th root of 2. The arguments are 2*pi/15, 8*pi/15, 14*pi/15, 20*pi/15, and 26*pi/15. |

Lecture 25Play Video |
Introduction to Basic Topology of the Complex Plane (Define an Open Disk)Complex Analysis, Video #25 (Complex Arithmetic, Part 25) Main Topics: numerically check 5th roots from previous video. Define an open disk and visualize it using RegionPlot on Mathematica. Do a quick numerical check of the 5th roots of -1 + sqrt(3)*I from the previous video (use "N" on Mathematica). Use the percent sign % to refer back to the preceeding output (as a list) and raise every number in the list to the 5th power. Definition of an "open disk" of radius positive radius "rho", centered at a given complex number z0. Use set-builder notation to define, as well as the modulus of the difference z - z0, which represents the distance between the points representing z and z0 in the complex plane. Also write in terms of rectangular coordinates x and y by using the distance formula (Pythagorean Theorem). To be an open disk, we must use a strict inequality so that we do NOT include the boundary points of the disk. Use RegionPlot to graph a disk of radius 2 centered at the point 3 + 4*i. RegionPlot graphs it as if the boundary circle is included, but it is not. We can use Show, Graphics, Dashed, and Circle to emphasize that the boundary is not included. |

Lecture 26Play Video |
Open Sets in the Complex Plane and illustrating the definition with MathematicaComplex Analysis, Video #26 (Complex Arithmetic, Part 26. Main Topics: definition of an open disk ("circular neighborhood"), interior point, and open set. Example: the open right half plane is an open set. Visualize why it is with Mathematica. Review the definition of an open disk of radius "rho" centered at z0 and the set-builder notation used to define it. It's also called a circular neighborhood (I spelled it wrong in the video) of radius "rho" centered at z0. Define "interior point" of a set S. Discuss the visual interpretation of an interior point. Define a set S to be open if every point of S is an interior point of S. Use RegionPlot to illustrate for the (open) "right half plane". Show how to find a radius that proves that 2+i, 1+i, 0.5+i, and 0.1+i are interior point of the right half plane. Use Manipulate and Locator on Mma to make an animation that shows how the radius of the disk needs to decrease as we approach the boundary (in order to prove points are interior points). If we included the imaginary axis to create the closed right half plane, it would no longer be an open set because the points on that boundary would not be interior points of the set. |