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2.1: Algebra with Complex Numbers

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    106806
  • The imaginary unit \(i\) is defined as the square root of -1: \(i= \sqrt{-1}\). If \(a\) and \(b\) are real numbers, then the number \(c= a+ib\) is said to be complex. The real number \(a\) is the real part of the complex number \(c\), and the real number \(b\) is its imaginary part. If \(a=0\), then the number is pure imaginary. All the rules of ordinary arithmetic apply with complex numbers, you just need to remember that \(i^2=-1\) For example, if \(z_1= 2+3i\) and \(z_2=1-4i\):

    • \(z_1+ z_2=3-i\)
    • \(z_1- z_2=1+7i\)
    • \(\dfrac{1}{2} z_1+z_2=2-\dfrac{5}{2} i\)
    • \(z_1z_2=(2+3i)(1-4i)=2-12 i^2-5i=14-5i\) (remember that \(i^2 = -1\)!)
    • \(z_1^2=(2+3i)(2+3i)=4+9 i^2+12i=-5+12i\)

    In order to divide complex numbers we will introduce the concept of complex conjugate.

    Complex Conjugate

    The complex conjugate of a complex number is defined as the number that has the same real part and an imaginary part which is the negative of the original number. It is usually denoted with a star: If \(z = x + iy\), then \(z^∗ = x − iy\)

    For example, the complex conjugate of \(2-3i\) is \(2+3i\). Notice that the product \(zz^*\) is always real:

    \[\label{complexeq:eq1}(x+iy)(x-iy)=x^2-ixy+ixy+y^2=x^2+y^2.\]

    We’ll use this result in a minute. For now, let’s see how the complex conjugate allows us to divide complex numbers with an example:

    Example \(\PageIndex{1}\): Complex Division

    Given \(z_1= 2+3i\) and \(z_2=1-4i\) obtain \(z_1/z_2\)

    Solution

    \[\dfrac{z_1}{z_2}=\dfrac{2+3i}{1-4i} \nonumber \]

    Multiply the numerator and denominator by the complex conjugate of the denominator:

    \[\dfrac{z_1}{z_2}=\dfrac{2+3i}{1-4i}\dfrac{1+4i}{1+4i} \nonumber\]

    This “trick” ensures that the denominator is a real number, since \(zz^*\) is always real. In this case,

    \[\begin{align*} (1-4i)(1+4i) &=1-4i+4i-16i^2\\[4pt] &=17. \end{align*}\]

    The numerator is

    \[\begin{align*} (2+3i)(1+4i) &=2+3i+8i+12i^2 \\[4pt] &=-10+11i \end{align*}\]

    Therefore,

    \[\begin{align*}\dfrac{z_1}{z_2} &=\dfrac{2+3i}{1-4i} \\[4pt] &=\displaystyle{\color{Maroon}-\dfrac{10}{17}+\dfrac{11}{17}i} \end{align*}\]

    Example \(\PageIndex{2}\)

    Calculate \((2-i)^3\) and express your result in cartesian form (\(a + bi\))

    Solution

    \[ \begin{align*} (2-i)^3 &= (2-i)(2-i)(2-i) \\[4pt] (2-i)(2-i) &=4-4i+i^2 \\[4pt] &=4-4i-1 \\[4pt] &=3-4i \\[4pt] (2-i)(2-i)(2-i) &=(3-4i)(2-i) \\[4pt] &=6-3i-8i+4i^2 \\[4pt] &=6-11i+4(-1) \\[4pt] &=\displaystyle{\color{Maroon}2-11i} \end{align*}\]

    The concept of complex conjugate is also useful to calculate the real and imaginary part of a complex number. Given \(z = x+iy\) and \(z^*=x-iy\), it is easy to see that \(z+z^*=2x\) and \(z-z^*=2iy\). Therefore:

    \[\label{eq2} Re(z)=\dfrac{z+z^*}{2}\]

    and

    \[Im(z)=\dfrac{z-z^*}{2i}\]

    You may wonder what is so hard about finding the real and imaginary parts of a complex number by visual inspection. It is certainly not a problem if the number is expressed as \(a+bi\), but it can be more difficult when dealing with more complicated expressions.

    algebra with complex numbers

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