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Complex Numbers

Tags
math

Are a class of numbers with both real and imaginary parts. (eg. \(z = x + ji\)).

The Cartesian form of all complex numbers is: \[ z = a + bi \] where \( Re(z) = a \) is the real part, \( Im(z) = b \) is the imaginary part and \( i \) is the imaginary constant.

Arithmetic Operations

For two complex numbers \( z_1 = 2 + i \) and \( z_2 = 3 - 2i \).

  • Addition \( z_1 + z_2 = 5 - i \)
  • Subtraction \( z_1 - z_2 = -1 + 3i \)
  • Multiplication

    \begin{align*} z_1z_2 &= (2+i)(3-2i) \\

         &= 6 - 4i + 3i - 2i^2 \\\\
         &= 6 - 4i + 3i - 2(-1) \\\\
         &= 8 - i

    \end{align*}

  • Reciprocals using the complex conjugate \( z_1^{*} = 2 - i \)

    \begin{align*} \frac{1}{z_1} &= \frac{1}{2 + i} \\

                &= \frac{1}{2+i} \frac{z\_1^{\*}}{z\_1^{\*}} \\\\
                &= \frac{1}{2+i} \frac{2-i}{2-i} \\\\
                &= \frac{2-i}{5}

    \end{align*}

  • Division using the complex conjugate of the denominator \( z_2^{*} = 3 + 2i \).

    \begin{align*} \frac{z_1}{z_2} &= \frac{2 + i}{3 - 2i} \\

                  &= \frac{2 + i}{3 - 2i} \frac{z\_2^{\*}}{z\_2^{\*}} \\\\
                  &= \frac{2 + i}{3 - 2i} \frac{3 + 2i}{3 + 2i} \\\\
                  &= \frac{4 + 5i}{13}

    \end{align*}

  • Length Refers to the length of the complex number when plotted on an Argand diagram, and is equivalent to its euclidean distance. \[ |z_1| = \sqrt{2^2 + 1^2} = \sqrt{5} \] There's also some general facts about the relation between the sum and product of two complex number lengths.

    \begin{align*} |z_1 z_2| &= |z_1| \times |z_2| \
    |\frac{z_1}{z_2}| &= \frac{|z_1|}{|z_2|} \end{align*}

Complex Conjugate

Of a complex number \( z_1 = a + bi \) is \( z_1^{*} = a - bi \).

Note: The pair of a complex number and its conjugate is known as the Complex Conjugate Pair.

For any complex conjugate pair \( Z, Z^{*} \), both \( Z Z^{*} \) and \( Z + Z{*} \) cancels out the imaginary component and leaves only a real value solution.

Note: If the roots \( a \) & \( b \) of a quadratic equation are complex, \( a \) & \( b \) will always form a complex conjugate pair. For example consider the formula \( x^2 - 4x + 29 = 0 \), we can refactor this into \( (x - 2)^2 + 25 = 0 \) which leads to \( x - 2 = \pm \sqrt{-25} \). There's always a positive and negative \( i \) term leading to the two being complex conjugates of each other.

Completing a Quadratic Factor

With this property we can also work backwards from the knowledge that a quadratic equation has a complex factor to complete a quadratic factor. For example \( 3 - 2i \) is a root of the formula \( 5z^3 + az^2 + bz + 52 = 0 \), what are the values of \( a \) and \( b \)?

First we know the complex conjugate \( 3 + 2i \) is also a factor. We can expand this into a quadratic

\begin{align*} x = 3 \pm 2i (x - 3)^2 = -4 x^2 - 6x + 13 = 0 \end{align*}

and then pair the factors with the cube and \( x^0 \) term to get: \[ (x^2 - 6x + 13)(5x + 4) = 5x^3 + az^2 + bz + 52 \] Notice that the only term that expands to get \( x^3 \) is the \( x^2 \times x \) meaning we get \( 5x \) in the second factor. Similarly \( 13 \times 4 = 52 \).

Now to find \( a, b \) we can simply expand this final factor to get \( a = -26, b = 41 \).

Argand Diagram

Is a 2D visualisation of the Cartesian form of a complex number with the Imaginary component along the \( y \) axis and real part along the \( x \) axis.

\begin{figure}
  \begin{tikzpicture}
    \begin{scope}[thick,font=\scriptsize]
      \draw [stealth-stealth] (-4,0) -- (4,0);
      \draw [stealth-stealth] (0,-4) -- (0,4);

      \draw [-stealth] (0,-0.25) -- (1.5,-0.25);
      \draw [-stealth] (1.5,-0.25) -- (1.5,-2);

      \foreach \n in {-3,...,-1,1,2,...,3}{%
	\draw (\n,-3pt) -- (\n,3pt)   node [above] {$\n$};
	\draw (-3pt,\n) -- (3pt,\n)   node [right] {%
	  \ifnum\n=1
	  \textcolor{red}{$i$}
	  \else
	  \ifnum\n=-1
	  \textcolor{red}{$-i$}
	  \else
	  $\n i$
	  \fi
	  \fi
	};
      }
    \end{scope}
  \end{tikzpicture}
  \caption{Argand diagram for the complex number $z = 1.5 + 2i$.}
\end{figure}