To understand this section, you'll want to know your basic trigonometry, how to use complex numbers, and you'll need to be familiar with polar coordinates. Everything else is pretty straightforward.
The complex plane defined
We often use a set of Cartesian-like coordinates to graphically represent complex numbers like
$$z = x + i y$$
with a real part and an imaginary part. Its major elements are illustrated here →.
The horizontal and vertical axes of the graph (which are perpendicular, or "orthogonal") are called the real axis and the imaginary axis, respectively. The are usually labeled Re and Im, as shown. The plane defined with these coordinates is called the complex plane.
Notice that the same transformations between the the Cartesian and polar representations of complex numbers apply as for polar coordinates. It's just simple trigonometry. Also notice in the figure above that the two complex numbers graphed are actually complex conjugates, which we'll generally label z and z*, where (*) stands for complex conjugate. On the complex plane, complex conjugates are reflections of one-another across the real axis.
The polar form of a complex number is:
$$z = r\cdot cos(\phi) + i r\cdot sin(\phi)$$
In this section, I will refer to the polar angle as φ, as is commonly done when working on the complex plane.
We will use the term complex number to refer to a number like $z = x + iy$ and a complex vector to be its representation on the complex plane. r is commonly called the modulus of a complex number and φ the angle or the argument.
When we work on the complex plane, we do little to no arithmetic with the imaginary number, i. It just labels the imaginary component of the complex vector; it's like a sign saying "this is the y-coordinate".
Example 1
Converting a complex number, $z = x + iy,$ to polar form
Convert z = 2 + 2i to polar form and plot it on a polar graph.
In this function x = 2 and y (the coordinate along the imaginary axis) is also 2. Remember, the pattern is z = x ± iy. From those coordinates, we can calculate r:
$$r = \sqrt{2^2 + 2^2} = 2 \sqrt{2}$$
and polar angle, θ:
$$\theta = tan^{-1}(1) = 45˚$$
Now $x = r·cos(\theta)$ and $y = r·sin(\theta),$ so finally,
$$z = 2 \sqrt{2} cos(45˚) + 2i \sqrt{2} sin(45˚)$$
Here is a graphical depiction of that coordinate, a vector on the complex plane:
Example 2
Converting a complex number, $z = x + iy,$ to polar form
Problem: Convert z = -2 + 2i to polar form and plot it on a polar graph.
This location on the complex plane isn't too different from the first example. The polar radius is the same:
$$r = \sqrt{2^2 + (-2)^2} = 2 \sqrt{2}$$
The angle is different, and calculating it is a cautionary tale: Recall that the range of the inverse tangent function is (-π/2, π/2), so it is incapable of directly giving us the angle we need in the upper left quadrant. We have to add π/2 or 90˚ to get the correct angle.
$$ \require{cancel} \begin{align} \theta = tan^{-1}(-1) &= \cancel{-45˚} \\[5pt] &= 135˚ \end{align}$$
Using $x = r·cos(\theta)$ and $y = r·sin(\theta),$ we have
$$z = 2 \sqrt{2} cos(135˚) + 2i \sqrt{2} sin(135˚)$$
Why use the complex plane?
The complex plane allows us to visualize complex numbers geometrically. We can treat them as we do vectors in physics, applying all of the rules of trigonometry to use and manipulate them. On the complex plane, addition of two complex numbers is just normal vector addition—see below. Notice that each of the starting vectors and the sum has a real part and an imaginary part.
Addition of complex vectors on the complex plane
$$ \begin{align} z_1 + z_2 &= x_1 + x_2 + i(y_1 + y_2) \\[6pt] &= r_1 cos(\phi_1) + r_2 cos(\phi_2) \\[5pt] &\phantom{00} + i(r_1 sin(\phi_1) + r_2 sin(\phi_2)) \end{align}$$
Complex vector multiplication
Multiplication of two complex numbers is also easy to visualize on the complex plane, and it's even simpler to perform algebraically using the polar representation.
Consider two complex numbers, z1 = 1 + i and z2 = 2 + i. These are shown on the right. We know that
$$ \begin{align} z_1 z_2 &= (1 + i)(2 + i) \\[5pt] &= 2 + i + 2i - 1 \\[5pt] &= 1 + 3i \end{align}$$
The graph shows what the multiplication looks like on the complex plane. Multiplication of two complex vectors results in a vector with a different length, and rotated by the sum of the arguments, φ1 and φ2.
The dashed gray line is the sum of the two vectors, just to emphasize that the two are not the same.
In polar coordinates, the multiplation looks like this:
(Narrow screens: scroll wide equations R ↔ L)
$$ \begin{align} z_1 z_2 &= [r_1 cos(\phi_1) + ir_1 sin(\phi_1)]\cdot[r_2 cos(\phi_2) + ir_2 sin(\phi_2)] \\[5pt] &= r_1r_2 cos(\phi_1)cos(\phi_2) + ir_1r_2cos(\phi_1)sin(\phi_2) + ir_1r_2 sin(\phi_1)cos(\phi_2) - r_1r_2 sin(\phi_1)sin(\phi_2) \\[5pt] &= r_1r_2 [cos(\phi_1)cos(\phi_2) - sin(\phi_1)sin(\phi_2)] + ir_1r_2[cos(\phi_1)sin(\phi_2) + sin(\phi_1)cos(\phi_2)] \\[5pt] &= r_1r_2 cos(\phi_1 + \phi_2) + ir_1r_2 sin(\phi_1 + \phi_2) \end{align}$$
To do the last step, we made use of the trigonometric sum and difference formulae:
Trig. sum and difference formulas
$$ \begin{align} cos(a \pm b) &= cos(a) \, cos(b) \mp sin(a) \, sin(b) \\[5pt] sin(a \pm b) &= cos(a) \, sin(b) \pm sin(a) \, cos(b) \end{align}$$
Vector multiplication in the complex plane
To multiply complex vectors in polar coordinates, multiply the moduli and add the angles:
$$z_1 z_2 = r_1 \, r_2, cos(\phi_1 + \phi_2) + i\, r_1 \, r_2 sin(\phi_1 + \phi_2)$$
A similar formula can be derived for complex vector division in the polar representation. You should derive this one yourself as an exercise.
Vector division in the complex plane
To divide complex vectors in polar coordinates, divide the moduli in the same order as the division and subtract the angles (also in that order):
$$\frac{z_1}{z_2} = \frac{r_1}{r_2} \, cos(\phi_1 - \phi_2) + i \, \frac{r_1}{r_2} \, sin(\phi_1 - \phi_2)$$
Multiplication of complex conjugates
You will recall that the complex conjugate of a complex number $z = x + iy$ is $z^* = x - iy.$ The product $z·z^*$ is $(x + iy)(x - iy) = x^2 + y^2,$ a real number.
The product of a complex number and its complex conjugate is a real number: $z \cdot z^* \in \mathbb{R}.$
Because the product of complex conjugates is real, it must lie on the real axis, with no imaginary part. Often, when it is understood (for example in quantum mechanics) that we're dealing with complex numbers, the expression z2 is taken to mean z·z*.
Now let's take a look at a full example of multiplication of two complex numbers using the polar representation:
Example 3
Calculate the product of $z_1 = 2 + i$ and $z_2 = 3 - 5i$
The first thing we need to do to multiply these complex vectors is to convert them to their polar forms. The modulus of z1 is
$$r_1 = \sqrt{2^2 + 1^2} = \sqrt{5}$$
and the rotation angle is
$$\theta_1 = tan^{-1} \left( \frac{1}{2} \right) = 26.6˚$$
So the polar form of z1 is
$$z_1 = \sqrt{5} \, cos(26.6) + i \sqrt{5} \, sin(26.6)$$
Now the modulus of z2 is
$$r_2 = \sqrt{3^2 + 5^2} = \sqrt{34}$$
and its angle:
$$\theta_2 = tan^{-1} \left( \frac{-5}{3} \right) = -59.0˚$$
So z2 in polar form is:
$$z_2 = \sqrt{34} \, cos(-59.0) + i \sqrt{34} \, sin(-59.0)$$
Now it's just a matter of multiplying the angles and adding the moduli:
$$ \begin{align} z_1 z_2 &= \sqrt{5 \cdot 34} \, cos(26.6 - 59.0) \\[4pt] &\phantom{000}+i \, \sqrt{5\cdot 34} \, sin(26.6 - 59.0) \\[5pt] &= 13(0.844) + i\, (13)(-0.536) \\[5pt] &= 11 - 7i \end{align}$$
Isn't this overly complicated?
You can confirm that our answer in the example above, $z_1 z_2 = 11 - 7\, i,$ is what you'd get by straight multiplication of the complex numbers in $a + bi$ form. So now you're asking, "why convert to polar form? It seems so complicated." But remember, when we multiply many numbers, that's a lot of binomials and a lot of instances of multiplication of the imaginary number i, and that's a lot of opportunities to make errors. Simply multiplying the moduli and adding the angles is much easier.
Finding powers of complex numbers: De Moivre's formula
To raise a number to a power, especially a large one, it is much more convenient to work in the polar representation and use De Moivre's formula:
De Moivre's formula
In polar form, the complex number $z = x + i\, y$ raised to the nth power, $z^n,$ is
$$[r \, cos(\phi) + r \, sin(\phi)]^n = r^n \, cos(n\phi) + i \, r^n \, sin(n\phi)$$
Proof of De Moivre's formula
To prove De Moivre's formula, begin with the multiplication rule, but multiply one complex number, z, by itself:
$$zz = r\cdot r\cdot cos(\phi + \phi) + ir\cdot r \cdot sin(\phi + \phi)$$
or
$$z^2 = r^2 cos(2\phi) + ir^2 sin(2\phi)$$
Then if we multiply z2 by z, we get:
$$z^3 = r^3 cos(3\phi) + ir^3 sin(3\phi)$$
It is easy to see that the pattern will keep repeating ad infinitum (to infinity) to give:
$$z^n = r^n cos(n \phi) + ir^n sin(n\phi)$$
This kind of reasoning, building a case that must extend in the same direction to some logical conclusion, is called induction or inductive reasoning.
Example 4
Given that $z = 2 + 3i,$ find $z^7$
First (we're getting used to this), we convert to polar coordinates on the complex plane. The modulus is:
$$r = \sqrt{2^2 + 3^2} = \sqrt{3}$$
The angle is
$$\theta = tan^{-1} \left( \frac{3}{2} \right) = 56.3˚$$
The complex vector in polar coordinates is then:
$$z = \sqrt{13} \, cos(56.3) + i\sqrt{13} \, sin(56.3)$$
Now we just apply De Moivre's formula to calculate the power:
$$z^7 = (\sqrt{13})^7 cos(7\cdot 56.3) + i(\sqrt{13})^7 sin(7\cdot 56.3)$$
$$ \begin{align} &= 7921 \cdot 0.828 + i\, 7291 \cdot 0.561 \\[4pt] &= 6558 + 4444 \, i \end{align}$$
Roots of complex numbers
We can use De Moivre's formula, which was an extension of our multiplication rule, to derive a formula for finding the nth roots of a complex number, where n is an integer. We simply change n to 1/n:
$$z^{\frac{1}{n}} = r^{\frac{1}{n}}\left[ cos \left( \frac{\phi}{n} \right) + i\, sin \left( \frac{\phi}{n} \right) \right]$$
That seems fine, but it's not quite finished. The fundamental theorem of algebra tells us that every complex number has n complex nth roots, and this formula would just give us one. To find the others, we have to recognize the cyclic nature of the sine and cosine functions . . .
... they repeat themselves every 2π rad. That means if an angle φ satisfies a trigonometric equation, so must φ + 2π, φ + 4π, and so on.
In our power formula this just didn't matter because φ + 2π is the same angle as φ (we would call these degenerate solutions). But in the root equation above, division by n makes φ + 0π, φ + 2π, φ + 4π, and so on yield different solutions. It looks like this:
$$a_k = r^{\frac{1}{n}} \left[ cos \left( \frac{\phi + 2k\pi}{n} \right) + i\, sin \left( \frac{\phi + 2k\pi}{n} \right) \right]$$
Here, k is an integer that starts at zero and counts up. When k is zero, we just have our initial formula for a single root at the beginning of this section.
When k = n, we have this situation:
So our index variable k is only useful up to n-1: k = 0, 1, 2 ... n-1. A couple of examples will make this all a little more clear.
nth roots of complex numbers
There are n nth roots of a complex number, which we label ak. They are given by:
$$a_k = r^{\frac{1}{n}} \left[ cos \left( \frac{\phi + 2\pi k}{n} \right) + i\, sin \left( \frac{\phi + 2\pi k}{n} \right) \right]$$
where $k = 0, 1, 2, \dots, n-1$
Example 5
Calculate the five fifth roots of -5. In other words, solve for all roots of $x^5 = -5.$
The angle of rotation is $\theta = \pi \; \text{rad},$ and $r = 5.$ Note that the length of the vector to our point on the complex plane is always a postive number.
Now our complex number in polar form is
$$z = 5 \, cos(\pi) + 5i \, sin(\pi).$$
Now we can calculate our roots. They are
$$ \begin{align} a_0 &= (5)^{\frac{1}{5}} \left[ cos\left( \frac{\pi}{5} \right) + i \, sin\left( \frac{\pi}{5} \right) \right] \\[4pt] &= 1.38 (0.81 + 0.59 i) \\[4pt] &= 1.12 + 0.81 i \end{align}$$
$$ \begin{align} a_1 &= (5)^{\frac{1}{5}} \left[ cos\left( \frac{3\pi}{5} \right) + i \, sin\left( \frac{3\pi}{5} \right) \right] \\[4pt] &= 1.38 (-0.31 + 0.95 i) \\[4pt] &= -0.43 + 1.31 i \end{align}$$
$$ \begin{align} a_2 &= (5)^{\frac{1}{5}} \left[ cos( \pi ) + i \, sin( \pi ) \right] \\[4pt] &= 1.38 (-1 + 0 i) \\[4pt] &= -1.38 \end{align}$$
$$ \begin{align} a_3 &= (5)^{\frac{1}{5}} \left[ cos\left( \frac{7\pi}{5} \right) + i \, sin\left( \frac{7\pi}{5} \right) \right] \\[4pt] &= 1.38 (-0.31 - 0.95 i) \\[4pt] &= -0.43 - 1.31 i \end{align}$$
$$ \begin{align} a_4 &= (5)^{\frac{1}{5}} \left[ cos\left( \frac{\pi}{5} \right) + i \, sin\left( \frac{\pi}{5} \right) \right] \\[4pt] &= 1.38 (0.81 - 0.59 i) \\[4pt] &= 1.12 - 0.81 i \end{align}$$
So our five fifth roots are $1.12 ± 0.81i, \; -0.43 ± 1.31 i,$ and $-1.38.$ Notice that the four roots with imaginary parts come in complex-conjugate pairs. Each is plotted in the graph below. This is always the case for the full complement of roots of a complex number: they are evenly distributed around the circle of radius r.
Practice problems
1. Sketch the complex vector on the complex plane and find its polar representation, $z = r·cos(\phi) + i\, r\, sin(\phi).$
| a. |
$4i$ Solution
$$ r = 4, \; \; \theta = \frac{\pi}{2} \\[5pt] \begin{align} z &= 4i \\[4pt] &= 4\left[( cos \frac{\pi}{2} + i sin \frac{\pi}{2} \right] \\[4pt] &= 4i sin\left(\frac{\pi}{2}\right) \\[4pt] &= 4i \end{align}$$ |
| b. |
$2$ Solution
$$ r = 2, \; \; \theta = 0 \\[5pt] \begin{align} z &= 2 \\[4pt] &= 2(cos(0) + i sin(0)) \\[4pt] &= 2 cos(0)\\[4pt] &= 2 \end{align}$$ |
| c. |
$3 + i$ Solution
The coordinates correspond to a 30-60-90 triangle, making the angle $\theta = \frac{\pi}{6}$, and $$ \begin{align} r^2 &= 1^2 + (\sqrt{3})^2 \\[4pt] &= 4 \\[4pt] r &= 2 \end{align}$$ $$ \begin{align} z &= \sqrt{3} + i \\[4pt] &= 2 \left( cos \frac{\pi}{6} + i sin \frac{\pi}{6} \right) \end{align}$$ |
| d. |
$7 - 3i$ Solution
$$ r = \sqrt{49 + 9} = \sqrt{58} \\[5pt] \theta = tan^{-1} \left( \frac{-3}{7}\right) = -23.2˚ \\[5pt] \begin{align} z &= \sqrt{58}(cos(-23.2) + i sin(-23.2)) \\[5pt] &= \sqrt{58}(cos(23.2) - i sin(23.2)) \end{align}$$ The last step took advantage of the even/odd properties of the sine and cosine functions: cos(-x) = cos(x) and sin(-x) = -sin(x). |
2. Sketch the complex number, z, and also 2z, -z and z/2 on the same complex plane:
| a. |
$z = 1 + i$ Solution
|
| b. |
$z = -1 + i\sqrt{3}$ Solution
|
3. Sketch z1, z2, z1 + z2 and z1·z2 on the same complex plane:
| a. |
$z_1 = 2 - i, \: \: z_2 = 2 + i$ Solution
$$ \begin{align} r &= \sqrt{2^2 + 1^2} = \sqrt{5} \\[4pt] \theta &= tan^{-1} \frac{1}{2} = 26.5˚ \\[4pt] z_1 &= \sqrt{5} (cos(-26.5) + i sin(-26.5)) \\[4pt] z_2 &= \sqrt{5} (cos(26.5) + i sin(26.5)) \\[4pt] z_1 z_2 &= \sqrt{5} \sqrt{5} (cos(0) + i sin(0))\\[4pt] &= 5 \end{align}$$ also: $$ \begin{align} z_1 z_2 &= (2 - i)(2 + i) \\[4pt] &= 4 + 2i - 2i + 1 = 5 \end{align}$$ $$\text{and } \; z_1 + z_2 = 4$$ |
| b. |
$z_1 = -1 + i, \: \: z_2 = 2 - 3i$ Solution
$$ \begin{align} z_1 &= -i + 1 \rightarrow r_1 = \sqrt{2} \\[4pt] \theta_1 &= tan^{-1}\frac{5\pi}{4} = -45˚ = 315˚ \end{align}$$ $$ \begin{align} z_2 &= 2 - 3i \rightarrow r_2 = \sqrt{13} \\[4pt] \theta_2 &= tan^{-1}\frac{3}{2} = -56.3˚ = 304˚ \end{align}$$ Now the sum: $$ \begin{align} z_1 + z_2 &= 3 - 4i \rightarrow r = 5, \\[4pt] \theta &= tan^{-1}\frac{4}{3} = -53.1˚ = 307˚ \\[4pt] &= 5(cos(307˚) + i sin(307˚)) \end{align}$$ ... and the product: $$ \begin{align} z_1z_2 &= (2 - 3i)(1 - i) \\[4pt] &= 2 - 2i - 3i - 3 = -1 -5i \\[4pt] &= \sqrt{2}\sqrt{13} (cos(315+304) + \\[4pt] &\; \; \; \; i sin(315+304)) \\[4pt] &= \sqrt{26}(cos(259) + i sin(259)) \end{align}$$ In the last step, the angle was 315 + 304 = 619˚, but subtracting one complete circle of 360˚ gets us to 259˚. |
4. Write the complex number in polar form with the angle φ between 0 and 2π
| a. |
$1 + i$ Solution$$1 + i = \sqrt{2}\left(cos \frac{\pi}{4} + i sin \frac{\pi}{4}\right)$$ |
| b. |
$1 + i \sqrt{3}$ Solution$$ \begin{align} 1 &+ \sqrt{3}i \rightarrow \\[4pt] r &= \sqrt{1^2 + (\sqrt{3})^2} \\[4pt] &= \sqrt{4} = 2 \\[4pt] \theta &= tan^{-1} \frac{\sqrt{3}}{1} = \frac{\pi}{3} \\[4pt] &\text{so then}\\[4pt] z &= 2 \left( cos \frac{\pi}{3} + i sin \frac{\pi}{3} \right) \end{align}$$ |
| c. |
$8i$ Solution$$ 8i \rightarrow \theta = \frac{\pi}{2}, \; r = 8 \\[5pt] \begin{align} z &= 8\left(cos \frac{\pi}{2} + i\, sin \frac{\pi}{2}\right) \\[4pt] &= 8 i sin \frac{\pi}{2} = 8 i \end{align}$$ |
| d. |
$2(i - 1)$ Solution$$2i - 2 \rightarrow r = \sqrt{4 + 4} = 2 \sqrt{2}$$ $$\theta = tan^{-1}(-1) = -\frac{\pi}{4} = \frac{7\pi}{4}$$ $$z = 2 \sqrt{2} \left[ cos \frac{7\pi}{4} + i\, sin \frac{7\pi}{4} \right]$$ |
5. Find the product, $z_1 \cdot z_2$ and the quotient, $z_1/z_2$:
| a. |
$z_1 = cos(\pi) + i sin(\pi),$ SolutionProduct: z1z2 $$(1)(1)\left[ cos \left( \pi + \frac{\pi}{3} \right) + i\,sin \left( \pi + \frac{\pi}{3} \right) \right] \\[4pt] = cos\left( \frac{4\pi}{3} \right) + i\, sin \left( \frac{4\pi}{3} \right)$$ Quotient: z1/z2 $$\frac{1}{1}\left[ cos \left( \pi - \frac{\pi}{3} \right) + i\,sin \left( \pi - \frac{\pi}{3} \right) \right]\\[4pt] = cos\left( \frac{2\pi}{3} \right) + i\, sin \left( \frac{2\pi}{3} \right)$$ |
| b. |
$z_1 = cos(\pi/4) + i sin(\pi/4),$ SolutionProduct: z1z2 $$cos \left( \frac{\pi}{4} + \frac{3\pi}{4} \right) + i\,sin \left( \frac{\pi}{4} + \frac{3\pi}{4} \right)\\[4pt] = cos(\pi) + i\, sin(\pi) = cos(\pi)$$ Quotient: z1/z2 $$cos \left( \frac{\pi}{4} - \frac{3\pi}{4} \right) + i\,sin \left( \frac{\pi}{4} - \frac{3\pi}{4} \right)\\[4pt] = cos\left( \frac{-\pi}{2} \right) + i\, sin\left( \frac{-\pi}{2} \right)\\[4pt] = cos\left( \frac{3\pi}{2} \right) + i\, sin\left( \frac{3\pi}{2} \right)$$ |
6. Find the indicated power using De Moivre's formula:
| a. |
$(1 + i)^20$ Solution$$1 + i = \sqrt{2}\left( cos \frac{\pi}{4} + i\, sin \frac{\pi}{4} \right)$$ $$(1 + i)^{20} =\\[5pt] \begin{align} &= (\sqrt{2})^{20} \left( cos \frac{20\pi}{4} + i\, sin \frac{20\pi}{4} \right) \\[4pt] &= 2^{10}(cos(5\pi) + i\, sin(5\pi)) \\[4pt] &= 1024(cos(\pi) + i\, sin(\pi)) \\[4pt] &= -1024 \end{align}$$ |
| b. |
$(1 - i\sqrt{3})^5$ Solution$$ \begin{align} r &= \sqrt{1 + 3} = 2 \\ \theta &= tan^{-1}(-\sqrt{3}) = \frac{-\pi}{3} = \frac{5\pi}{3} \end{align}$$ $$ \begin{align} z &= 2 \left( cos \frac{5\pi}{3} + i\, sin \frac{5\pi}{3} \right) \\[4pt] z^5 &= 2^5\left( cos \frac{25\pi}{3} + i\, sin \frac{25\pi}{3} \right) \end{align}$$ $\frac{25\pi}{3} = \text{ 4 round trips} + \frac{\pi}{3}$ $$ \begin{align} z^5 &= 2^5\left( cos \frac{\pi}{3} + i\, sin \frac{\pi}{3} \right) \\[4pt] &= 32\left( \frac{1}{2} + \frac{i\sqrt{3}}{2} \right) \\[4pt] &= 16(1 + i\sqrt{13}) \end{align}$$ |
| c. |
$(2\sqrt{3} + 2i)^5$ Solution$$ \begin{align} r &= \sqrt{(2 \sqrt{3})^2 + 2^2} = \sqrt{16} = 4 \\[4pt] \theta &= tan^{-1}\left( \frac{2}{2\sqrt{3}} \right) = \frac{\pi}{6} = \frac{5\pi}{3} \end{align}$$ $$ \begin{align} z &= 4 \left( cos \frac{\pi}{6} + i\, sin \frac{\pi}{6} \right) \\[4pt] z^5 &= 4^5\left( cos \frac{5\pi}{6} + i\, sin \frac{5\pi}{6} \right) \\[4pt] &= 4^4 \left( \frac{\sqrt{3}}{2} + \frac{1}{2} \right) \\[4pt] &= 4^4 \cdot 2(i - \sqrt{3}) \\[4pt] &= 512 (i - \sqrt{3}) \end{align}$$ |
7. Find the indicated roots and graph them in the complex plane:
| a. |
The square roots of $4\sqrt{3} + 4i$ |
| b. |
The cube roots of $4\sqrt{3} + 4i$ |
| c. |
The eighth roots of 1 |
| d. |
The eighth roots of -1 |
8. Convert these complex numbers to polar form: $x + iy \; \rightarrow \; r[cos(\phi) + i·sin(\phi)]$:
| a. |
$-2 + 2i \sqrt{3}$ Solution
$$ \begin{align} r &= \sqrt{(-2)^2 + (2\sqrt{3})^2} \\ &= \sqrt{4 + 4.3} \approx \sqrt{16} \approx 4 \end{align}$$ $$ \begin{align} \theta &= tan^{-1}\left( \frac{2 \sqrt{2}}{-2} \right) \\ &= tan^{-1} (-\sqrt{3}) = -60˚ \rightarrow 120˚ \end{align}$$ (Convert to the positive angle.) $$ \begin{align} z &= 4(cos(120˚) + i\, sin(120˚)) \\ \\ &= 4 cos \left( \frac{2\pi}{3} \right) + 4i \, sin\left( \frac{2\pi}{3} \right) \end{align}$$ |
| b. |
$\sqrt{3} + i$ Solution
$$ \begin{align} r &= \sqrt{1^2 + (\sqrt{3})^2} \\ &= \sqrt{4} = 2 \end{align}$$ $$ \begin{align} \theta &= tan^{-1}\left( \frac{1}{\sqrt{3}} \right) \\ &= 30˚ = \frac{\pi}{6} \; rad \end{align}$$ $$ \begin{align} z &= 2(cos(30˚) + 2i\, sin(30˚)) \\ \\ &= 2 \left[ cos \left( \frac{\pi}{6} \right) + 4i \, sin\left( \frac{\pi}{6} \right)\right] \end{align}$$ |
| c. |
$4 - 7i$ Solution
$$ \begin{align} r &= \sqrt{4^2 + (-7)^2} \\ &= \sqrt{16 + 49} = \sqrt{65} \end{align}$$ $$ \begin{align} \theta &= tan^{-1}\left( \frac{-7}{\sqrt{4}} \right) \\ &= -60.2˚ \end{align}$$ $$z = \sqrt{65}\left[ cos(60.2˚) + 2i\, sin(60.2˚)) \right]$$ In the last step we used the even/odd properties of the sine and cosine functions: cos(-x) = cos(x) and sin(-x) = -sin(x) |
| d. |
$3 - 3i$ Solution
$$ \begin{align} r &= \sqrt{3^2 + (-3)^2} \\ &= \sqrt{18} = 3\sqrt{2} \end{align}$$ $$ \begin{align} \theta &= tan^{-1}\left( \frac{-3}{\sqrt{3}} \right) \\ &= tan^{-1}(-1) = -45˚ = -\frac{\pi}{4} \; rad \end{align}$$ $$ \begin{align} z &= 3 \sqrt{2}\left[ cos \left( \frac{\pi}{4} \right) - i\, sin \left( \frac{\pi}{4} \right)\right] \\ \\ &= 3 \sqrt{2}\left[ cos \left( -\frac{\pi}{4} \right) + i\, sin \left( -\frac{\pi}{4} \right)\right] \\ \\ &= 3 \sqrt{2}\left[ cos(-45˚) + i\, sin (-45˚)\right] \\ \\ &= 3 \sqrt{2}\left[ cos(45˚) - i\, sin (45˚)\right] \end{align}$$ All of those solutions are equivalent. In addition to changing units, we used the even/odd properties of the sine and cosine functions: cos(-x) = cos(x) and sin(-x) = -sin(x). |
9. Convert these complex numbers in polar form to standard form: $r[cos(\phi) + i·sin(\phi)] \; \rightarrow \; x + iy$:
| a. |
$3 \, [cos(30˚) - i sin(30˚)]$ |
| b. |
$\sqrt{2} \, [cos(7\pi/6) + i sin(7\pi/6)]$ |
| c. |
$8 \, [cos(210˚) - i sin(210˚)]$ |
| d. |
$7 \, [cos(\pi/12) + i sin(\pi/12)]$ |
10. Find the product of $z_1$ and $z_2,$ and convert the answer to standard form: $x + iy$:
| a. |
$z_1 = 7[cos(25˚) + i sin(25˚)],$ Solution$$ \begin{align} z_1 &= 7(cos(25) + i\, sin(25)) \\ z_2 &= 2(cos(130) + i \, sin(130)) \\ z_1z_2 &= 7\cdot 2 (cos(155) + i\,sin(155)) \\ &= \bf -6.52 + 3.05 i \end{align}$$ |
| b. |
$z_1 = \sqrt{3}\, [cos(3\pi/4) + i sin(3\pi/4)],$ Solution$$ \begin{align} z_1 &= \sqrt{3}\left(cos \frac{2\pi}{4} _ i\, sin \frac{3\pi}{4}\right) \\ z_2 &= \frac{1}{3}\left( cos \frac{\pi}{6} + i \, sin\frac{\pi}{6}\right) \\ z_1z_2 &= \frac{\sqrt{3}}{3} \left( cos \frac{11\pi}{12} + i \, sin \frac{11\pi}{12} \right) \\ &= \bf -0.558 + 0.149 i \end{align}$$ |
11. Find the quotient and leave it in polar form:
| a. |
$$\frac{2(cos(30˚) + i\, sin(30˚))}{3(cos(60˚) + i\, sin(60˚))}$$ Solution$$\frac{2[cos(30) + i\, sin(30)]}{3[cos(60) + i\, sin(60)]}$$ $$ \begin{align} &= \frac{2}{3}[cos(30 - 60) + i\, sin(30 - 60)] \\ &= \frac{2}{3}[cos(30) - i \, sin(30)] \end{align}$$ We used the even-odd properties of sine and cosine: cos(-x) = cos(x), and sin(-x) = -sin(x). Make sure you know those; they're very handy. |
| b. |
$$\frac{cos(\pi/2) + i\, sin(\pi/2)}{cos(\pi/4) + i\, sin(\pi/4)}$$ Solution$$\frac{cos\frac{\pi}{2} + i\, sin\frac{\pi}{2}}{cos\frac{\pi}{4} + i\, sin\frac{\pi}{4}}$$ $$ \begin{align} &= cos\left( \frac{\pi}{2} - \frac{\pi}{4} \right) + i \, sin \left( \frac{\pi}{2} - \frac{\pi}{4} \right) \\ &= cos\frac{\pi}{4} + i\, sin \frac{\pi}{4} \end{align}$$ |
12. Find the indicated power of the complex number and express your answer in $z = x + iy$ form:
| a. |
$(3 + 4i)^{18}$ Solution$$ \begin{align} r &= \sqrt{3^2 + 4^2} = \sqrt{25} = 5 \\ \\ \theta &= tan^{-1} \left( \frac{4}{3} \right) = 53.13˚ \\ \\ z^{18} &= 5^{18}[ cos(18\cdot 53.13) \\ &\; \; \; + i\,sin(18\cdot 52.13)] \\ \\ &= 5^{18}[-0.554 - i\, 0.832] \end{align}$$ I won't bother to multiply the 518. It's a huge number! (Oh, what the heck, it's 3,814,697,265,625) |
| b. |
$(1 - i\sqrt{3})^3$ Solution$$ \begin{align} r &= \sqrt{1^2 + 3} = sqrt{4} = 2 \\ \\ \theta &= tan^{-1} (-\sqrt{3}) = -60˚\\ \\ z^{3} &= 2^3[ cos(3\cdot-60) + i\,sin(3\cdot -60˚)] \\ \\ &= 8[cos(-180) + i\, sin(-180)] \\ \\ &= 8[cos(180) - i\, sin(180)] \\ \\ &= -8 - 0 = \bf -8 \end{align}$$ We used the even-odd properties of sine and cosine: cos(-x) = cos(x), and sin(-x) = -sin(x). Make sure you know those; they're very handy. Notice that this cube is a real number. |
| c. |
$[1/2 + i(\sqrt{3}/2)]^3$ Solution$$ \begin{align} r &= \sqrt{\left( \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2} \\ \\ &\; \; \; = \sqrt{\frac{1}{4} + \frac{4}{4}} = 1 \\ \\ \theta &= tan^{-1} (\sqrt{3}) = 60˚\\ \\ z^{3} &= 1^3[ cos(3\cdot 60) + i\,sin(3\cdot 60˚)] \\ \\ &= cos(180) + i\, sin(180) \\ \\ &= -1 + 0 = \bf -1 \end{align}$$ |
13. Find the indicated roots of the complex number:
| a. |
$(-2 + 2i)^{1/4}$ SolutionSolution |
| c. |
$(2i)^{1/5}$ SolutionSolution |
14. Solve: $x^5 = 32$; find all roots.
Solution
Solution
Problem 7a solution
Find $(4 \sqrt{3} + 4i)^{1/2}$
$$r = \sqrt{(4 \sqrt{3})^2 + 4^2} = \sqrt{48 + 16} = 8$$
Using the triangle above, we find θ:
$$\theta = tan^{-1} \left( \frac{4}{4\sqrt{3}}\right) = \frac{\pi}{6}$$
So $z = 8 \left( cos \frac{\pi}{6} + i\, sin \frac{\pi}{6} \right)$
Now the roots are:
$$ \begin{align} a_0 &= 8^{1/2} \left( cos \frac{\pi/6}{2} + i\, sin \frac{\pi/6}{2} \right) \\ \\ &= 8^{1/2} \left( cos \frac{\pi}{12} + i\, sin \frac{\pi}{12} \right) \\ \\ a_1 &= 8^{1/2} \left( cos \frac{\pi/6 + 2\pi}{2} + i\, sin \frac{\pi/6+ 2\pi}{2} \right) \\ \\ &= 8^{1/2} \left( cos \frac{13\pi}{12} + i\, sin \frac{13\pi}{12} \right) \end{align}$$
Now to go a bit further, recall your trig. sum formulas:
$$ \begin{align} cos \frac{\pi}{12} &= sin \left( \frac{\pi}{3} - \frac{\pi}{4} \right) \\ &= cos \frac{\pi}{3} cos \frac{\pi}{4} + sin \frac{\pi}{3} sin\frac{\pi}{4} \\ &= \frac{1 + \sqrt{3}}{2 \sqrt{2}} \end{align}$$
$$ \begin{align} sin \frac{\pi}{12} &= sin \left( \frac{\pi}{3} - \frac{\pi}{4} \right) \\ &= sin \frac{\pi}{3} cos \frac{\pi}{4} - cos \frac{\pi}{3} sin\frac{\pi}{4} \\ &= \frac{1 - \sqrt{3}}{2 \sqrt{2}} \end{align}$$
Using these we can reduce ao and a1 to
$$a_0 = 1 + \sqrt{3} + i(1 - \sqrt{3})$$
$$a_1 = -1 - \sqrt{3} - i(1 - \sqrt{3})$$
Problem 7b solution
From problem 7a we have $z = 8 \left( cos \frac{\pi}{6} + i\, sin \frac{\pi}{6} \right)$, where $8^{1/3} = 2$. The roots are:
$$ \begin{align} a_0 &= 2 \left( cos \frac{\pi/6}{3} + i\, sin \frac{\pi/6}{3} \right) \\ \\ &= 2 \left( cos \frac{\pi}{12} + i\, sin \frac{\pi}{12} \right) \\ \\ a_1 &= 2 \left( cos \frac{\pi/6 + 2\pi}{3} + i\, sin \frac{\pi/6+ 2\pi}{3} \right) \\ \\ &= 2 \left( cos \frac{13\pi}{12} + i\, sin \frac{13\pi}{12} \right) \\ \\ a_2 &= 2 \left( cos \frac{\pi/6 + 4\pi}{3} + i\, sin \frac{\pi/6+ 4\pi}{3} \right) \\ \\ &= 2 \left( cos \frac{25\pi}{18} + i\, sin \frac{25\pi}{18} \right) \end{align}$$
Here is the graphical representation of the three solutions. Notice that they are evenly-distributed around a circle with an equal angle between the complex vectors.
Problem 9a solution
Problem 9b solution
Induction
When we reason inductively, or use induction to reason, we extend a pattern. The sun has risen every day of my life, so I'm reasonably confident it will rise again tomorrow. That's induction. To extend a series like 2, 4, 6, 8, x, by concluding that x = 10 is also induction.
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