Skip to main content
Mathematics LibreTexts

B.1: Definition and Basic Operations

  • Page ID
    91834
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    We'll start with the definition of a complex number and its addition and multiplication rules. You may find the multiplication rule quite mysterious. Don't worry. We'll soon gets lots of practice using it and seeing how useful it is.

    Definition B.1.1
    1. The complex plane is simply the \(xy\)-plane equipped with an addition operation and a multiplication operation. A complex number is nothing more than a point in that \(xy\)-plane. It is conventional to use the notation \(x+iy\)1 to stand for the complex number \((x,y)\text{.}\) In other words, it is conventional to write \(x\) in place of \((x,0)\) and \(i\) in place of \((0,1)\text{.}\)
    2. The first component, \(x\text{,}\) of the complex number \(x+iy\) is called its real part and the second component, \(y\text{,}\) is called its imaginary part, even though there is nothing imaginary 2 about it.
    3. The sum of the complex numbers \(\, z_1=x_1+i y_1 \) and \(\, z_2=x_2 +iy_2 \, \) is defined by

      \[ z_1+z_2 = (x_1+x_2)+i(y_1+y_2) \nonumber \]

      That is, you just separately add the real parts and the imaginary parts.
    4. The product of the complex numbers \(\, z_1=x_1+i y_1 \) and \(\, z_2=x_2 +iy_2 \, \) is defined by

      \[ z_1z_2 = x_1x_2-y_1y_2+i(x_1y_2+x_2y_1) \nonumber \]

      Do not memorise this multiplication rule. We'll see a simple, effective memory aid for it shortly. The heart of that memory aid is the observation that the complex number \(i\) has the special property that

      \[ i^2 = (0+1i)(0+1i) = (0\times 0-1\times 1)+i(0\times 1+1\times 0) = -1 \nonumber \]

    Addition and multiplication of complex numbers obey the familiar addition rules

    \begin{align*} z_1+z_2&=z_2+z_1\\ z_1+(z_2+z_3)&=(z_1+z_2)+z_3 \\ 0+z_1&=z_1 \end{align*}

    and multiplication rules

    \begin{align*} z_1z_2&=z_2z_1 \\ z_1(z_2z_3)&=(z_1z_2)z_3 \\ 1z_1&=z_1 \end{align*}

    and distributive laws

    \begin{align*} z_1(z_2+z_3)&=z_1z_2+z_1z_3 \\ (z_1+z_2)z_3& = z_1z_3+z_2z_3 \end{align*}

    To remember how to multiply complex numbers, you just have to supplement the familiar rules of the real number system with \(i^2=-1\text{.}\) The previous sentence is the memory aid referred to in Definition B.1.1(d).

    Example B.1.2

    If \(z=1+2i\) and \(w=3+4i\text{,}\) then

    \begin{alignat*}{2} z+w&=(1+2i)+(3+4i)&&=4+6i \\ zw&=(1+2i)(3+4i)&&=3+4i+6i+8i^2=3+4i+6i-8\\ &=-5+10i \end{alignat*}

    Definition B.1.3
    1. The negative of any complex number \(z= x+iy\) is defined by

      \[ -z=-x+(-y)i \nonumber \]

      and obviously obeys \(z+(-z)=0\text{.}\)
    2. The reciprocal 3, \(z^{-1}\) or \(\frac{1}{z}\text{,}\) of any complex number \(z= x+iy\text{,}\) other than \(0\text{,}\) is defined by

      \[ \frac{1}{z}z =1 \nonumber \]

      We shall see below that it is given by the formula

      \[ \frac{1}{z}=\frac{x}{x^2+y^2}+\frac{-y}{x^2+y^2}i \nonumber \]

    Example B.1.4

    It is possible to derive the formula for \(\frac{1}{z}\) by observing that

    \[ (a+ib)(x+iy)=[ax-by] + i[ay+bx] \nonumber \]

    equals \(1 = 1+i0\) if and only if

    \begin{align*} ax-by&=1 \\ ay+bx&=0 \end{align*}

    and solving these equations for \(a\) and \(b\text{.}\) We will see a much shorter derivation in Remark B.1.6 below. For now, we'll content ourselves with just verifying that \(\frac{x}{x^2+y^2}+\frac{-y}{x^2+y^2}i\) is the inverse of \(x+iy\) by multiplying out

    \begin{align*} &\left(\frac{x}{x^2+y^2}-\frac{y}{x^2+y^2}i\right)(x+iy)\\ &\hskip0.5in=\frac{x^2}{x^2+y^2}-\frac{xy}{x^2+y^2}i +\frac{xy}{x^2+y^2}i-\frac{y^2}{x^2+y^2}i^2 \\ &\hskip0.5in=\frac{x^2-i^2y^2}{x^2+y^2} =\frac{x^2+y^2}{x^2+y^2}=1 \end{align*}

    Definition B.1.5
    1. The complex conjugate of \(\, z=x+iy\, \) is denoted \(\bar z\) and is defined to be

      \[ \bar z=\overline{x+iy}=x-i y \nonumber \]

      That is, to take the complex conjugate, one replaces every \(i\) by \(-i\) and vice versa.
    2. The distance from \(z=x+iy\) (recall that this is the point \((x,y)\) in the \(xy\)-plane) to \(0\) is denoted \(\, |z|\, \) and is called the absolute value, or modulus, of \(\, z\, \text{.}\) It is given by

      \[ |z| = \sqrt{x^2+y^2} \nonumber \]

      Note that

      \[ z\bar z=(x+iy)(x-iy)=x^2-ixy+ixy+y^2=x^2+y^2 \nonumber \]

      is always a nonnegative real number and that

      \[ |z| = \sqrt{z\,\bar z} \nonumber \]

    Remark B.1.6

    Let \(z=x+iy\) with \(x\) and \(y\) real. Since \(|z|^2=z\,\bar z\text{,}\) we have

    \[ \frac{1}{z} =\frac{1}{z}\frac{\bar z}{\bar z} = \frac{\bar z}{|z|^2} =\frac{x-iy}{x^2+y^2} =\frac{x}{x^2+y^2}+\frac{-y}{x^2+y^2}i \nonumber \]

    which is the formula for \(\frac{1}{z}\) given in Definition B.1.3(b).

    Example B.1.7

    It is easy to divide a complex number by a real number. For example

    \[ \frac{11+2i}{25} = \frac{11}{25}+\frac{2}{25}i \nonumber \]

    In general, the complex conjugate provides us with a trick for rewriting any ratio of complex numbers as a ratio with a real denominator. For example, suppose that we want to find \(\frac{1+2i}{3+4i}\text{.}\) The trick is to multiply by \(1=\frac{3-4i}{3-4i}\text{.}\) The number \(3-4i\) is the complex conjugate of the denominator \(3+4i\text{.}\) Since \((3+4i)(3-4i)=9-12i+12i-16i^2=9+16=25\)

    \[ \frac{1+2i}{3+4i}=\frac{1+2i}{3+4i}\ \frac{3-4i}{3-4i} =\frac{(1+2i)(3-4i)}{25} =\frac{11+2i}{25} = \frac{11}{25}+\frac{2}{25}i \nonumber \]

    Definition B.1.8

    The notations 4 \(\Re z\) and \(\Im z\) stand for the real and imaginary parts of the complex number \(z\text{,}\) respectively. If \(z=x+ iy\) (with \(x\) and \(y\) real) they are defined by

    \[ \Re z=x\qquad \Im z=y \nonumber \]

    Note that both \(\Re z\) and \(\Im z\) are real numbers. Just subbing in \(\bar z=x-iy\text{,}\) you can verify that

    \[ \Re z=\frac{1}{2}(z+\bar z)\qquad \Im z=\frac{1}{2i}(z-\bar z) \nonumber \]

    Lemma B.1.9

    If \(z_1=x_1+iy_1\) and \(z_2=x_2+iy_2\text{,}\) then

    \[ |z_1z_2| = |z_1|\,|z_2| \nonumber \]

    Proof

    Since \(z_1z_2=(x_1+iy_1)(x_2+iy_2)=(x_1x_2-y_1y_2)+i(x_1y_2+x_2y_1)\text{,}\)

    \begin{align*} |z_1z_2| &=\sqrt{(x_1x_2-y_1y_2)^2+(x_1y_2+x_2y_1)^2} \\ &=\sqrt{x_1^2x_2^2-{\color{blue}{2x_1x_2y_1y_2}}+y_1^2y_2^2 +x_1^2y_2^2+{\color{blue}{2x_1y_2x_2y_1}}+x_2^2y_1^2} \\ &=\sqrt{x_1^2x_2^2+y_1^2y_2^2+x_1^2y_2^2+x_2^2y_1^2} =\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)} \\ &=|z_1|\,|z_2| \end{align*}


    This page titled B.1: Definition and Basic Operations is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.