Precalculus Lesson 9: Introduction to Complex Numbers and the Complex Plane

Introduction to Complex Numbers and the Complex Plane

• AoPS Online is closed May 24-26th.

Motivation

• A simple quadratic equation with no real solutions is presented as motivation:
  $x^2 + 1 = 0$
• The equation has no real solutions because the square of any real number is non-negative.
• Instead of giving up, a new kind of number is introduced to solve this problem.
• Let's use the symbol $i$ to denote this new "number".

The Imaginary Unit

• $i$ is defined such that $i^2 = -1$.
• Then:
  • $i^2 = -1$
  • $i^3 = i \cdot i^2 = -i$
  • $i^4 = i(i^3) = i(-i) = -i^2 = 1$
• It is possible to figure out any power of $i$ in this manner.

Complex Numbers

• Complex numbers are defined as all numbers of the form $a + bi$, where $a$ and $b$ are real numbers.
  • $a$ is the real part of $z$, written as $Re(z)$.
  • $b$ is the imaginary part of $z$, written as $Im(z)$.
  • Both $a$ and $b$ are real numbers.
• Example: $\pi = \pi + 0i$, so $\pi$ is a complex number.
• Quick Quiz: What is $Im(3+4i)$? Notice that the imaginary part of $3 + 4i$ is $4$, not $4i$.
• Two complex numbers are equal when their real parts are equal to each other and their imaginary parts are equal to each other.
• Complex numbers with a zero imaginary part are the real numbers.
• Complex numbers with a zero real part are called purely imaginary.
• We add, subtract, and multiply complex numbers just like you might expect -- to add and subtract, we gather like terms, and to multiply, we use the distributive property.

The Complex Plane

• To graph complex numbers, we use the two-dimensional cousin of the real line: the complex plane.
• The number $a + bi$ corresponds to the point $(a, b)$ on the complex plane.
• The horizontal axis is called the real axis, representing the real part of the complex number.
• The vertical axis is called the imaginary axis, representing the imaginary part of the complex number.

Example

• Given a complex plane with a grid where each square has side lengths of 1 unit, we can identify complex numbers.
• If $z$ is at the point (2, 4) then $z = 2 + 4i$.
• If $w$ is at the point (3, -1) then $w = 3 - i$.
• If you want a single variable denoting a complex number, it is customary to use the letter $z$, if a second complex variable is needed, $w$.

Complex Number Arithmetic

• We add and subtract in the usual way, by gathering terms.
• If $z = 2 + 4i$ and $w = 3 - i$, then:
  • $z + w = (2 + 4i) + (3 - i) = 5 + 3i$
  • $z - w = (2 + 4i) - (3 - i) = -1 + 5i$

Geometric Interpretation of Addition and Subtraction

• Addition and subtraction have interesting geometric interpretations.
• To get from $w$ to $z + w$, we go right 2 and up 4. And what do we do to get from $0$ to $z$?
• To get from $0$ to $z$, we also go right 2 and up 4.
• The line segment from $0$ to $z$ is a translation of the line segment from $w$ to $z + w$, meaning they have the same length and direction.
• To add $z$ to $w$, we take the line segment from $0$ to $z$ and translate the endpoint at 0 over to $w$. The end of this shifted line segment will be $z + w$. This is called head to tail addition of complex numbers.
• Complex number addition is commutative: $z + w = w + z$.
• The shape formed by $0$, $z$, $z + w$, and $w$ is a parallelogram.
  • Both pairs of opposite sides are equal in length and parallel.
• $z - w$ is the complex number we need to add to $w$ to get $z$.
• If we shift the line segment from $w$ to $z$ to start at $0$, we get $z - w$.
• From the diagram, we can see that to go from $w$ to $z$, we need to go 1 unit left and 5 units up. This means that $z - w = -1 + 5i$.
• Head to tail addition lets us draw sums and differences of complex numbers without doing any calculations.
• Completing the parallelogram: the complex number $z<em>4$ is the unique point in the plane such that $0$, $z</em>1$, and $z_2$ form a parallelogram. (Assuming that those first three points are not collinear.)
• It can be surprisingly easy to incorrectly visualize differences of complex numbers. The best way to avoid this issue is to always check what you find using the addition equation.

Multiplication and Division

• The obvious next candidates are multiplication and division.
• If $z = 2 + 4i$ and $w = 3 - i$, we can calculate $zw$ and $z/w$.

Multiplication

• Make sure that you don't just multiply the real parts by the real parts and the imaginary parts by the imaginary parts! Think of what you'd do to multiply $(a + b)$ and $(c + d)$.
• $(2 + 4i)(3 - i) = 6 - 2i + 12i - 4i^2 = 10 + 10i$

Division

• Division is a bit trickier.
  $\frac{2 + 4i}{3 - i}$
• When we have messy denominators, we multiply top and bottom by something (the same thing, of course) to rationalize the denominator.
• In this case, we are going to "realize" the denominator. We're going to multiply numerator and denominator by something that makes the denominator a real number.
• Multiply numerator and denominator by $3 + i$:
  $\frac{2 + 4i}{3 - i} = \frac{(2 + 4i)(3 + i)}{(3 - i)(3 + i)} = \frac{6 + 12i + 2i + 4i^2}{9 - i^2} = \frac{2 + 14i}{10} = \frac{1}{5} + \frac{7}{5}i$
• We've figured out geometric interpretations of addition and subtraction. We might wonder about whether multiplication and division have geometric interpretations, too.
• It turns out that they do. However, we won't learn about them today. That'll have to wait until the next couple of weeks. Stay tuned!
• (As a little teaser: our work with polar coordinates last week will prove very useful for those operations!)

Complex Conjugates

• What would we multiply the numerator and denominator by when "realizing" the denominator of $\frac{c+di}{a+bi}$?
• Multiply the numerator and denominator by $a - bi$.
• If $z = a + bi$, then we define $\overline{z} = a - bi$. We call $\overline{z}$ the conjugate of $z$.
• In LaTeX you can type $\overline{z}$ to get $\overline{z}$. If you're using plain text, you can use conj(z) to mean the conjugate of $z$.
• We usually read $\overline{z}$ as "z bar".
• If $z = 2 + 4i$ and $w = 3 - i$, then $\overline{z} = 2 - 4i$ and $\overline{w} = 3 + i$.
• Conjugates are reflections of each other in the real axis.
• $\overline{\overline{z}} = z$ In fact, this always works! For any complex number $z$ we have $\overline{\overline{z}} = z$. The conjugate of a conjugate is the original number.
• The conjugate of a sum also works out nicely. Say that $z = a + bi$, $w = c + di$, where $a$, $b$, $c$, $d$ are real numbers.
  • Then $z + w = (a + c) + (b + d)i$, so $\overline{z + w} = (a + c) - (b + d)i$
  • Conjugating $z$ and $w$ gives $\overline{z} = a - bi, \overline{w} = c - di$, so $\overline{z} + \overline{w} = (a + c) - (b + d)i$.
  • Therefore, for any complex numbers, $\overline{z + w} = \overline{z} + \overline{w}$. Or in other words, the conjugate of a sum is the sum of the conjugates.
• Now let's think about multiplication and look at $\overline{zw}$, the conjugate of $zw$.
• We might guess that $\overline{zw} = \overline{z} \cdot \overline{w}$, that is, the conjugate of a product is the product of the conjugates.
• Let's test that out! This time, the calculations with $a, b, c, d$ get ugly fast, so before we dive into that, let's try an example and see what happens. After all, if it doesn't work for our example, it will definitely not work in general!
• Let $z = 2 + i$ and $w = 1 + 3i$. What's $zw$?
  • $zw = (2 + i)(1 + 3i) = 2 + i + 6i + 3i^2 = -1 + 7i$
• From here, it's simple to see that the conjugate of $zw$ is $\overline{zw} = -1 - 7i$
• Now let's check whether $\overline{zw}$ is indeed $\overline{z} \cdot \overline{w}$. What do we get when we multiply $\overline{z}$ and $\overline{w}$?
• $\overline{z} \cdot \overline{w} = (2 - i)(1 - 3i) = 2 - i - 6i + 3i^2 = -1 - 7i$ which is precisely $\overline{zw}$!
• That means that for this one example, the product of the conjugates equal to the conjugate of the products! This, of course, is not a proof that this works in general, but it's certainly suggestive.
• Let's quickly see the actual proof. If $z = a + bi$ and $w = c + di$, then $zw = (a + bi)(c + di) = ac + adi + bci + bdi^2 = (ac - bd) + (ad + bc)i$.
• Thus we see that $\overline{zw} = (ac - bd) - (ad + bc)i$.
• $\overline{z} = a - bi$ and $\overline{w} = c - di$.
  $\overline{z} \cdot \overline{w} = (a - bi)(c - di) = (ac - bd) - (ad + bc)i$
• Therefore, for any complex numbers $\overline{zw} = \overline{z} \cdot \overline{w}$. Or in other words, the conjugate of a product is always the product of the conjugates.
• To recap, we've now observed that in general:
  • $\overline{z + w} = \overline{z} + \overline{w}$
  • $\overline{zw} = \overline{z} \cdot \overline{w}$
• I will sometimes refer to these properties by saying that "conjugation plays nicely with addition and multiplication."
• If you want a more conventional way to say this, we can also say that conjugation commutes with addition and multiplication.
• Conjugation also plays nicely with subtraction and division, since subtraction is just addition by an inverse, and division is just multiplication by an inverse:
  • $\overline{z - w} = \overline{z} - \overline{w}$
  • $\overline{(\frac{z}{w})} = \frac{\overline{z}}{\overline{w}}$

Complex Number Algebra

• It'll be helpful for us to think about how to solve equations involving complex numbers.
• When we solve equations in the real numbers, we perform the same operations on both sides of an equation to manipulate the equation. The same ideas apply to complex numbers.
• If we have an equation, and we add the same complex number to both sides, the equation is still true.
• Our equation is still true if we subtract the same complex number from both sides, or multiply both sides by the same complex number, or divide both sides by the same nonzero complex number.

Example 1

• Solve the equation:
  $z(1 + i) + i = 3 - i$
• Subtract $i$ from both sides:
  $z(1 + i) = 3 - 2i$
• Multiply both sides by $\frac{1}{1 + i}$:
  $z = \frac{(1 + i)(3 - 2i)}{1 + i} = (1 + i)(3 - 2i) = 3 + 3i - 2i - 2i^2 = 5 + i$

Example 2

• Find all solutions to $z^2 = i$.
• Write $z = a + bi$ and substitute it into the equation. We get $(a + bi)^2 = i$
• Expanding the left side, we get:
  $(a + bi)^2 = a^2 - b^2 + 2abi$
  $(a^2 - b^2) + (2ab)i = 0 + 1i$
• We know that the two sides are equal when the real and imaginary parts are equal!
  • $a^2 - b^2 = 0$
  • $2ab = 1$
• The first equation tells us $a = \pm b$.
• Since the second equation tells us that the product $ab$ is positive, $a$ and $b$ must be the same sign!
• Therefore, $a = -b$ is impossible, which means that $a = b$.
• substituting into $2ab = 1$ we find that $2a^2 = 1$.
  • $a = \pm \sqrt{\frac{1}{2}} = \pm \frac{\sqrt{2}}{2}$
• Since $b = a$, our answers $z$ are:
  • $z = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i$
  • $z = - \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i$
• If these numbers look familiar to you — say, from the work we did with terminal points on the unit circle — that is not a coincidence! We'll see why in our next class.

Example 3

• Two complex numbers $z$ and $w$ satisfy:
  • $z + wz^3 = 7 + 3i$
  • $\overline{z} - \overline{wz^3} = 3 - i$
• What is $z$?
• One approach is to write $z = a + bi$ and $w = c + di$ and write out our equations in terms of $a, b, c, d$. That looks really ugly with terms.
• The left sides of the equations look very similar. Maybe we can combine them somehow to make things cancel out. Let's see. If we add the two equations, we get:
  $z + wz^3 + \overline{z} - \overline{wz^3} = 10 + 2i$
• We could take the conjugates of both sides of one of our equations! Taking the conjugate of both sides of the second equation we get:
  $\overline{\overline{z} - \overline{wz^3}} = \overline{3 - i}$
• The right side is just $3 + i$.
• Since $\overline{z - w} = \overline{z} - \overline{w}$, we can turn the left side into:
  $\overline{\overline{z} - \overline{wz^3}} = \overline{\overline{z}} - \overline{\overline{wz^3}}$
• Similarly, since $\overline{\overline{z}} = z$, we can turn this into:
  $\overline{\overline{z}} - \overline{\overline{wz^3}} = z - wz^3$
• Finally, taking the conjugate of both sides of the second equation turned it into:
  $z - wz^3 = 3 + i$
• Now we can just add this to the first equation and the $wz^3$ terms will cancel out! We get:
  $2z = 10 + 4i$
  $z = 5 + 2i$

The Magnitude

• Our last important function of the day is the magnitude or absolute value of a complex number, which we write as $|z|$. This is the distance of the complex number from the origin.
• This is a generalization of the familiar absolute value of real numbers. For a real number $x$, the absolute value $|x|$ is the distance from $x$ to 0 on the number line, which we can also think of as its distance to the origin in the complex plane.
• Both in LaTeX and in plain text, you can use vertical bars to denote the magnitude. That is, we write $|z|$ in LaTeX and |z| in plain text.

Example

• Calculate $|z|$ and $|w|$ for the $z$ and $w$ in the picture below:
• By definition, $|z|$ is the distance of $z$ from the origin. Using Pythagoras,
  • If $z = 2 + 4i$, then $|z| = \sqrt{2^2 + 4^2} = \sqrt{20} = 2\sqrt{5}$.
• What is $|w|$ for the $w$ above?
  • $w = \sqrt{3^2 + 1^2} = \sqrt{10}$
• In general, again using Pythagoras, $|a + bi| = \sqrt{a^2 + b^2}$.

Algebraic Interpretation

• How could we factor the expression $a^2 + b^2$?
• We can write $a^2 + b^2 = a^2 - (-b^2) = a^2 - (bi)^2$ since $(bi)^2 = -b^2$.
• Factoring this as a difference of squares, we get:
  $a^2 + b^2 = (a + bi)(a - bi)$
• Since $a + bi = z$ and $a - bi = \overline{z}$, this tells us that $|z|^2 = z \cdot \overline{z}$.
• This is an important fact!

Magnitude of Sums

• Let's figure out whether there are nice expressions for $|z + w|$ and $|zw|$ in terms of $|z|$ and $|w|$.
• Let's start with $|z + w|$. We'll start by trying this with some actual numbers, just to get a sense.
• Let's use $z = 2 - 2i$ and $w = 2 + 3i$. Adding them and keeping things organized, we see that $z + w = 4 + i$.
• We know that the magnitude of a complex number is its distance from the origin. So what are $|z|, |w|, |z + w|$?
• Calculating the distances using Pythagoras, we have that:
  • $|z| = \sqrt{8} = 2\sqrt{2}$
  • $|w| = \sqrt{13}$
  • $|z + w| = \sqrt{17}$
• $\sqrt{17} \neq \sqrt{8} + \sqrt{13}$
• In particular, $|z + w| \neq |z| + |w|$ in this case.
• Note that unlike before, this is actually a proof that $|z + w| \neq |z| + |w|$ since one counterexample shows that this property can't possibly hold in general!
• There turns out to be a good geometric reason that this doesn't work.
• Which line segments above have length $|z|$?
  • the blue ones do: the ones from 0 to $z$ and from $w$ to $z + w$
• Similarly, the line segments from 0 to and from to have length ; they are color-coded in red in the diagram above.
• The green line segment, the diagonal of the parallelogram that goes from 0 to, has length ||.
• Since these are the sides of a triangle, we don't usually have = + although we do have ≤ + due to the triangle inequality.
• So we now know that is not equal to + but it's possible that there's a different relationship!
• The next question is whether knowing and is enough to figure out
• Notice that and form three sides of a triangle in the complex plane. This question is equivalent to asking, "is knowing two side lengths in a triangle sufficient to figure out the third side length?"
• Not at all! We'd need at least one angle in order to figure out the length of the third side.
• If we fix and there are many possible values for :
• Hence, we generally don't have an equation that relates to and (We do have the triangle inequality, though, which has its uses.)
• When does the triangle inequality become an equality? In other words, under what circumstances do we have = +?
• Right, it's when and "point in the same direction". This is equivalent to saying that and are collinear and in the same quadrant
• So that's addition and magnitude! It doesn't play nicely in the way we might want it to, which is very useful to know.

Magnitude of Products

• Now let's think about whether has a nice relationship to and
• Let's use and again. Then what's
• That's right,
• Since we're calculating magnitudes, let's use the complex plane.
• Putting it all together,
• putting it all together, that is, the magnitude of the product is the product of the magnitudes.
• Letting and, we have that
• Therefore, That's pretty neat!
• Now let's also demonstrate another method.
• Perhaps we can use that somehow.
• What nice relationship between|| and did we just discover?
• That's right, we have that
• Did we figure out about earlier this class?
• We learned that conjugation plays nicely with multiplication. So and therefore We get: |zw|^2 = ((z z )(w w) = |z|^2|W|^2
• We now know that magnitudes do not play nice with addition, but do play nice with multiplication.
• Since division undoes multiplication, magnitudes do play nice with division: we have that

Complex Number Algebra, Part 2

• We've already seen earlier that just like with real number equations, an equation with complex numbers stays true if we do the same thing to both sides.

*We also learned that we could take the conjugate of both sides, because doing the same thing to both sides keeps the two sides equal. We also discovered that this can be useful when the original equation involves conjugation.

The problem below feels tricky at first, but becomes very simple if we take the magnitude of both sides!

Summary

• Today we introduced complex numbers and defined the rules of arithmetic for them: addition, subtraction, multiplication, and division. We figured out geometric interpretations of addition and subtraction, but we haven't yet figured out geometric interpretations of multiplication and division. That will come in later classes.
• We also defined two very useful functions on complex numbers: the conjugate and the magnitude.
  • Magnitudes, on the other hand, play nice with multiplication and division, but not addition or subtraction.
  • Along the way, we talked a lot about the utility of working with complex numbers without breaking them up into real and imaginary parts.
  • We also realized if we have an equation in complex numbers, we can conjugate both sides, and we can also take the magnitude of both sides.
• Next week we'll link complex numbers to trigonometry and polar coordinates.
• Today's topics are pretty foundational for the next few weeks. So if you don't get the hang of them, you'll get pretty lost.