Page Description Equation
3 complex number \[ z=a+bi \]
3 complex addition \[ \left(a+bi\right)+\left(c+di\right)=\left(a+c\right)+\left(b+d\right)i \]
3 complex subtraction \[ \left(a+bi\right)-\left(c+di\right)=\left(a-c\right)+\left(b-d\right)i \]
3 complex multiplication \[ \left(a+bi\right)\left(c+di\right)=\left(ac-bd\right)+\left(bc+ad\right)i \]
3 complex division where $z_2\neq0.$ Note. $c^2+d^2\neq0\Leftrightarrow z_2\neq0$ \[ \frac{a+bi}{c+di} =\frac{ac+bd}{c^2+d^2}+\frac{bc-ad}{c^2+d^2}i \]
3 real part of complex product \[ \Re z_1z_2=ac-bd \]
3 imaginary part of complex product \[ \Im z_1z_2=bc+ad \]
3 real part of complex quotient where $z_2\ne0.$ \[ \Re{\frac{z_1}{z_2}}=\frac{ac+bd}{c^2+d^2} \]
3 imaginary part of complex quotient where $z_2\ne0.$ \[ \Im{\frac{z_1}{z_2}}=\frac{bc-ad}{c^2+d^2} \]
3 additive inverse \[ -z=\left(-a,-b\right) \]
3 additive identity \[ 0=\left(0,0\right) \]
3 multiplicative inverse \[ \frac{1}{z}=\left(\frac{a}{a^2+b^2},\frac{-b}{a^2+b^2}\right) \]
3 multiplicative identity \[ 1 = (1, 0) \]
5 For any $n\in\Z$ and $k=n\pmod{4},$ $0\le k\le1,$ the given equation follows. This hypothesis is equivalent to the statement that $n=4k_1+k$ for some $k_1\in\Z$ and $k=0,1,2,3.$ \[ i^n=i^k=\left\{ \begin{matrix} 1k &=0\\ ik &=1\\ -1k &=2\\ -ik &=3 \end{matrix} \right. \]
3 Commutative Law Addition \[ \]
3 Commutative Law Multiplication \[ \]
3 Associative Law Addition \[ \]
3 Associative Law Multiplication \[ \]
3 Distributive Law \[ \]
4 real part. where $z=a+bi.$ Note that $\Re z$ is a real number. \[ a=\Re z \]
4 imaginary part. where $z=a+bi.$ Note that $\Im z$ is a real number. \[ b=\Im z \]
4 complex variable in terms of $\Re$ and $\Im.$ \[ z=\Re z+i\Im z \]
4 equality of complex numbers \[ z_1=z_2\\ \text{iff}\\ \Re z_1=\Re z_2\\ \text{and}\\ \Im z_1=\Im z_2 \]
4 pure imaginary number \[ z=bi \]
8 point form of a complex number $z$ \[ a+bi=\left(a,b\right) \]
8 modulus. absolute value $z=(a, b)$ \[ \abs{z}=\sqrt{a^2+b^2} \]
9 modulus of $i$ \[ \abs{i}=1 \]
9 modulus of complex difference is the distance between corresponding points $z_1 = (a_1, b_1)$ and $z_2 = (a_2, b_2)$ \[ \left|z_1-z_2\right|=\sqrt{\left(a_1-a_2\right)^2+\left(b_1-b_2\right)^2} \]
10 complex conjugate of $z.$ The symbol $z*$ is also used \[ z=a-bi\Leftrightarrow z=a+bi \]
10 conjugate of a sum is a sum of conjugates \[ \overline{z_1+z_2}=\overline{z_1}+\overline{z_2} \]
10 conjugate of a difference is a difference of conjugates \[ \overline{z_1-z_2}=\overline{z_1}-\overline{z_2} \]
11 conjugate of a product is a product of conjugates \[ \overline{z_1z_2}=\overline{z_1}\ \overline{z_2} \]
11 conjugate of a quotient is a quotient of conjugates \[ \overline{\left(\frac{z_1}{z_2}\right)}=\frac{\overline{z_1}}{\overline{z_2}} \]
11 conjugate is the inverse of conjugate \[ \overline{(\overline{z})}=z \]
11 modulus of $z$ and $\overline{z}$. $z$ and $\overline{z}$ are equidistant from the origin. \[ \abs{z}=\abs{\overline{z}} \]
11 product of a complex number with its conjugate \[ z\overline{z}=\abs{z}^2 \]
11 modulus in terms of the product of a complex number with its conjugate \[ \sqrt{\left(z\overline{z}\right)}=\left(z\overline{z}\right)^{1/2}=z \]
11 This implies that the sum of a complex number and its conjugate $z+\overline{z}$ is real since $2\Re z$ is real. \[ \Re{z}=\frac{z+\overline{z}}{2} \]
11 This implies that the difference of a complex number and its conjugate $z-z$ is pure imaginary since i 2 Im z is pure imaginary. \[ \Im{z}=\frac{z-\overline{z}}{2i} \]
12 rationalizing the denominator \[ \frac{z_1}{z_2}=\frac{z_1\overline{z_2}}{z_2\overline{z_2}}=\frac{z_1\overline{z_2}}{\abs{z_2}^2} \]
9 equation of circle with radius $r$ and center $z_0$ \[ \abs{z-z_0}=r \]
9 equation of circle with radius $r$ and center at the origin \[ \abs{z}=r \]
15 triangle inequality \[ \abs{z_1+z_2}\le\abs{z_1}+\abs{z_2} \]
15 complex number as vector. vector and complex addition are the same. \[ z_1+z_2=\vect{z}_1+\vect{z}_2 \]
14 norm vs. modulus. The length (norm) of vector $\vect{z}$ is equal to the modulus of the complex number $z$. \[ \norm{z}=\abs{z} \]
17 argument (phase) of $z.$ $\theta=$ angle in radians measured from the $x$-axis to the vector $\vect{z}.$ \[ \begin{align*} \arg z &= \left\{ \theta: \cos\theta=\frac{x}{\abs{z}}, \sin\theta=\frac{y}{\abs{z}} \right\}\\ &= \left\{ \theta+2k\pi: \cos\theta=\frac{x}{\abs{z}}, \sin\theta=\frac{y}{\abs{z}},k\in\Z \right\}\\ &= \theta+2k\pi\quad(k=0,\pm1,\pm2,\ldots) \end{align*} \]
18 $\tau$ branch of $\arg z.$ A half-open interval of length $2\pi.$ Every such interval necessarily contains exactly one value of $\arg z.$ \[ \left(\tau,\tau+2\pi\right] \]
18 $\tau$ branch of $\arg z$ \[ \mathrm{arg}_\tau z=\theta\Leftrightarrow\theta\in\left(\tau,\tau+2\pi\right] \]
18 principle value of the argument $\arg z$ where $\theta=\left(-\pi,\pi\right]$ \[ \Arg z=\mathrm{arg}_{-\pi}z=\theta \]
18 cis function \[ \cis\theta=\cos{\theta}+i\sin{\theta} \]
18 polar coordinates for $z$ \[ \left(r,\theta\right) = \left(\abs{z}, \arg z\right) \]
18 polar form of a complex number. Follows, since $z=x+iy$ $=r\cos{\theta}+ir\sin{\theta}$ $=r\left(\cos{\theta}+i\sin{\theta}\right)$ $=r\cis\theta$ \[ z=r\left(\cos{\theta}+\sin{\theta}\right) =r\cis\theta =\abs{z}\cis\arg z \]
20 geometric interpretation of the product $z_1z_2.$ Vector $z_1z_2$ has length equal to the product of the lengths of the vectors $z_1$ and $z_2$ and has angle equal to the sum of the angles of the vectors $z_1$ and $z_2.$ \[ z_1z_2=r_1r_2\mathrm{cis}\theta_1+\theta_2\\ \abs{z_1z_2}=\abs{z_1}\abs{z_2}\\ \arg z_1z_2=\arg z_1+\arg z_2 \]
20 geometric interpretation of the quotient $\frac{z_1}{z_2}.$ Vector $\frac{z_1}{z_2}$ has length equal to the quotient of the lengths of the vectors $z_1$ and $z_2$ and has angle equal to the difference of the angles of the vectors $z_1$ and $z_2.$ \[ \frac{z_1}{z_2}=r_1r_2\cis\theta_1-\theta_2\\ \abslr{\frac{z_1}{z_2}}=\frac{\abs{z_1}}{\abs{z_2}}\\ \arg\frac{z_1}{z_2}=\arg{z_1}-\arg z_2 \]
21 orthogonal complex numbers. Vectors $\vect{z}_1$ and $\vect{z}_2$ are orthogonal iff this equation holds. The equation is equivalent in vector notation to ${\vect{z}}_1\cdot{\vect{z}}_0=0$ \[ \arg\frac{z_1}{z_2}=\arg z_1-\arg z_2=\pm\frac{\pi}{2} \]
27 Euler's equation \[ e^{iy}=\cos{y}+i\sin{y}=\cis y \]
27 complex exponential function. $z=x+iy$ \[ \begin{align*} e^z &=e^{x+iy}=e^xe^{iy}\\ &=e^x\cis ye^x\left(\cos{y}+i\sin{y}\right) \end{align*} \]
31 complex exponential function. It is periodic with period $2\pi i$ \[ e^z=e^{z+2\pi i} \]
27 product of exponentials \[ e^{z_1}e^{z_2}=e^{z_1+z_2} \]
27 quotient of exponentials \[ \frac{e^{z_1}}{e^{z_2}}=e^{z_1-z_2} \]
29 power rule for exponentials \[ \left(e^z\right)^n=e^{nz} \]
28 polar form complex number. $r=\abs{z}$ and $\theta=\arg z.$ \[ z=re^{i\theta}=r\left(cos{\theta}+sin{\theta}\right) \]
29 polar form of complex conjugate \[ z=re-iθ \]
31 modulus of the complex exponential function \[ \left|e^{x+iy}\right|=e^x \]
11
28		 complex cosine function \[ \cos{\theta} =\Re{e^{i\theta}} =\frac{e^{i\theta}+e^{-i\theta}}{2} \]
21
18		 complex sine function \[ \sin{\theta} =\Im{e^{i\theta}} =\frac{e^{i\theta}-e^{-i\theta}}{2i} \]
29 product of complex numbers, geometric interpretation \[ z_1z_2=r_1r_2e^{i\left(\theta_1+\theta_2\right)} \]
29 quotient of complex numbers, geometric interpretation \[ \frac{z_1}{z_2}=\frac{r_1}{r_2}e^{i\left(\theta_1-\theta_2\right)} \]
29 De Moivre's Theorem. \[ \left(\cos{\theta}+i\sin{\theta}\right)^n =\cos{n}\theta+i\sin{n}\theta \]
31 argument of the complex exponential function. $k=0,\pm1,\pm2,\ldots$ \[ \arg{e^{x+iy}}=y+2k\pi \]
33 De Moivre's theorem extended \[ z^n=r^ne^{in\theta} =r^n\left(\cos{n}\theta+i\sin{n}\theta\right) \]
34 mth roots of unity $k=0,\ldots,m-1$ \[ 1^{1/m}=e^{i2k\pi/m} =\cos{\frac{2k\pi}{m}}+i\sin{\frac{2k\pi}{m}} \]
34 complete set of $m$th roots of unity. $k=1$ from preceding equation. \[ \begin{multline*} 1,\omega_m,\omega_m^2,\ldots,\omega_m^{m-1} \omega_m =e^{i2\pi/m}\\ =\cos{\frac{2\pi}{m}}+i\sin{\frac{2\pi}{m}} \end{multline*} \]
36 mth root of a complex number. $k=0,\ldots,m-1$ \[ z^{1/m}=\sqrt[m]{\abs{z}}e^{i\left(\theta+2k\pi\right)/m} \]
36 quadratic formula. $a,b,c$ are complex constants, $\sqrt{b^2-4ac}$ is one of the values of $\left(b^2-4ac\right)^{1/2},$ and $a\neq0.$ \[ z=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \]

Notes

Page Notes
4 There is no natural ordering of $\C.$
The addition field properties for complex numbers follow from their real counterparts. The multiplication properties do not so readily follow and are demonstrated from the definition of multiplication for complex numbers.
The definition of multiplication for complex numbers implies that $i^2=–i^2=–1$ (cf. solution to exercise 1.1.1). As a result, we see that if we expand the product $\left(a+bi\right)\left(c+di\right)$ in the same manner that we would a product of real binomials, the result is consistent with that specified by complex multiplication. An important consequence of this consistency is that we can apply the binomial theorem to expand the complex number $\left(a+bi\right)^n.$
real axis. The $x$-axis is called the real axis.
imaginary axis. The $y$-axis is called the imaginary axis.
Argand diagram, z-plane, complex plane. A plot of a complex variables using real and imaginary axes.
15 real and imaginary vectors. Every vector parallel to the real axis is a real number, every vector parallel to the imaginary axis is a pure imaginary number
34 There are exactly $m$ distinct roots of a complex number $z.$
39 open disk, circular neighborhood of $z_0.$ The set of all points that satisfy the inequality $\abs{z-z_0}\lt\rho$ where $\rho$ is a positive real number, is called an open disk or circular neighborhood of $z_0.$
39 open unit disk. the neighborhood $\abs{z} \lt 1$ is the open unit disk.
39 interior point of $S.$ A point $z_0$ which lies in a set $S$ is called an interior point of $S$ if there is some circular neighborhood of $z_0$ that is completely contained in $S.$
39 open set. If every point of a set $S$ is an interior point of $S,$ we say that $S$ is an open set.
39 polygonal path.
40 An open set $S$ is said to be connected if every pair of points $z_1,z_2\in S$ can be joined by a polygonal path that lies entirely in $S.$
40 An open connected set is a domain.

Procedure for Finding Complex Roots

According to the fundamental theorem of algebra, the polynomial equation \begin{equation} \label{eq_ZetaPol} \zeta^n=z \end{equation} in $\zeta$ has $n$ roots in the complex field. That is, the complex number $z$ has exactly $n$ $n$th roots in the complex field.

Since an $n$th root of the complex number $z$ is itself a complex number, we can denote such a root in the general polar form \begin{equation} \label{eq_PolarRoot} \zeta=\rho e^{i\varphi} \end{equation} where, as usual for complex numbers, $\rho=\abs{\zeta}$ and $\varphi\in\arg{\zeta}.$

If we number the $n$ roots, say from $k=0,\ldots,n–1$, then we can denote the $k$th $n$th root of $z$ by \begin{equation} \label{eq_KthRoot} \zeta_k=\rho e^{i\varphi_k} \end{equation}

We denote the set of $n$ $n$th roots of $z$ by $z^{1/n}$ which can be expressed as follows: \begin{equation} \label{eq_NthRoots} z^{1/n}=\left\{\zeta_k=\rho e^{i\varphi_k}:\zeta_k^n=z,k=0,1,\ldots n-1\right\} \end{equation}

However, this equation does not tell us how to find $z^{1/n}$ when $z$ is known. That is, it does not tell us how to solve equation \eqref{eq_ZetaPol} for $\zeta.$ We obtain such a formula for $z^{1/n}$ from page 36. \begin{equation} \label{eq_SlnNthRoots} z^{1/n}=\left\{\sqrt[n]{\abs{z}}e^{i\left(\theta+2k\pi\right)/n}:\theta\in\arg{z},k=0,1,\ldots n-1\right\} \end{equation}

This theorem provides the value of $\zeta_k$ (the $k$th $n$th root of $z$) in terms of $z$ by revealing the values of $\rho$ and $\phi_k$ in terms of $z.$

In particular, from \eqref{eq_NthRoots} and \eqref{eq_SlnNthRoots} we see that: \begin{equation} \label{eq_RhoPhi} \rho=\sqrt[n]{\abs{z}} \text{ and } \varphi_k =\frac{\theta}{n} +\frac{2k\pi}{n}\left(\theta\in\arg{z}\right) \end{equation}

Another form of \eqref{eq_SlnNthRoots}, then, is: \begin{equation} \label{eq_SlnNthRootsRev} z^{1/n}= \left\{ \zeta_k=\rho e^{i\varphi_k}:\rho=\sqrt[n]{\abs{z}}, \varphi_k=\frac{\theta}{n}+\frac{2k\pi}{n}, \theta\in\arg{z},k=0,1,\ldots n-1 \right\} \end{equation}

Note that $\rho$ and $\theta$ do not depend on $k$ so need to be computed only once. Also, it follows from equation \eqref{eq_RhoPhi} that two consecutive $n$th roots of $z$ are $2\pi/n$ radians apart. \[ \varphi_k-\varphi_{k-1}=\frac{\theta}{n}+\frac{2k\pi}{n}-\frac{\theta}{n}-\frac{2k\pi}{n}+\frac{2\pi}{n}=\frac{2\pi}{n} \]

Thus, adding $\phi_{k–1}$ to both sides of this last equation gives us the following recursive expression for $\phi k:$ \begin{equation} \label{eq_PhiRec} \varphi_k=\varphi_{k-1}+\frac{2\pi}{n} \end{equation} This equation means we can compute the first root we want using equation \eqref{eq_RhoPhi} , and then recursively add $2\pi/n$ to obtain subsequent roots by equation \eqref{eq_PhiRec}.

Thus, we can find the $\phi_k$ for each $n$th root as follows: \[ \begin{align*} \phi_0&=\frac{\theta}{n}\\ \phi_1&=\phi_0+\frac{2\pi}{n}\\ \phi_2&=\phi_1+\frac{2\pi}{n}\\ \vdots\\ \phi_{n-1}&=\phi_{n-2}+\frac{2\pi}{n}\\ \end{align*} \]

It also may be helpful to recall from algebra the conjugate pair theorem and two of its corollaries (below), which apply when solving polynomial equations with real coefficients over the complex field. Equation \eqref{eq_ZetaPol} is such a polynomial, so these theorems can be used to simplify finding the $n$th roots of a complex number $z.$

Conjugate Pair Theorem. If $P$ is a polynomial function of degree $n\ge1$ with real coefficients, and if the imaginary number $a+ib$ is a zero of $P,$ then its conjugate $a–ib$ is also a zero of $P.$

Corollary 1. A polynomial function with real coefficients of odd degree must have at least one real zero. In fact, such a polynomial function must have an odd number of real zeros.

Corollary 2. A polynomial function with real coefficients of even degree must have either an even number of real zeros or no real zero at all.

Here is a general procedure for finding the $k$th $n$th root of a complex number $z,k=0,\ldots,n–1,$ that summarizes all of these details. Let $z=x+iy,r=\abs{z},\theta=\Arg z,$ and compute the following:

  1. \( \quad\abs{z}\\ \)
  2. \( \quad\rho=\sqrt[n]{\abs{z}}\\ \)
  3. \( \quad\theta=\Arg z\\ \)
  4. \( \quad\varphi_k=\frac{\theta}{n} +\frac{2k\pi}{n},\ (k=0,\ldots,n–1) \text{ or } \varphi_k=\varphi_{k-1}+\frac{2\pi}{n},\ (k=1,\ldots,n–1)\\ \)
  5. \( \quad\zeta_k=\sqrt[n]{\abs{z}}e^{i\varphi_k},\ (k=0,\ldots,n–1) \)

By raising the general element of the set $z^{1/n}$ defined in equation \eqref{eq_SlnNthRoots} to the $n$th power, and noting that the exponential function is periodic with period $2\pi i$, we see that it does indeed satisfy equation \eqref{eq_ZetaPol} as a root. \[ \left[\sqrt[n]{\abs{z}}e^{i\left(\theta+2k\pi\right)/n}\right]^n =\abs{z}e^{i\left(\theta+2k\pi\right)} =\abs{z}e^{i\theta}=z \]

However, how do we know that these are all the $n$ roots, and that they are non-congruent roots? This question is not really answered in the chapter and equation \eqref{eq_SlnNthRoots} is not proved to be $z^{1/n}$ but mostly just defined as such. It remains to be proved that indeed the set in equation \eqref{eq_SlnNthRoots} is equal to $z^{1/n}.$ See the footnote on page 34, however, and refer to sections 11, 12 and 14 in Modern Algebra, by Durbin, to figure it out.

See also page 36. "Equivalently, we can form these roots by taking any single one such as given in (3) and multiplying by the $m$th roots of unity."