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Complex Numbers for Dummies

Thursday, April 29, 2010

Today I decided to open a new chapter entitled "Mathematics". This new chapter will be mainly about Algebraic Coding Theory and Algebraic Geometry. If you are a mathematician, then this is not what your are looking for because I will try to avoid all the unnecessary mathematical notations for a basic understanding. However, if you are just interested in learn some curious facts about mathematics, then, probably this is the right place for you.


We all know that $(-2)^2 = 4$ and $2^2 = 4$, but what happens when we want to take the square root of a negative number? Until now, we simply left it as "undefined", since we had no numbers which were negative after squared. Therefore, we couldn't "go backwards" by taking the square root.


Now, you can take the square root of a negative number, but it involves using a new number to do it. This new number was invented around the time of the Reformation. Don’t be surprised! If you think about it, aren't all numbers inventions? It's not like numbers grow on trees! They live in our heads. Why not invent a new one, as long as it works with what we already have?


Anyway, this new number is called $i$ (stands for "imaginary") and is defined to be:


$i = \sqrt{-1}$


Now, we have a number $i$ that has the property that $i^2 = -1$. Using this, we can now find the square roots of negative numbers in terms of real numbers and $i$. Therefore:


$i^2 = (\sqrt{-1})^2 = -1$


Now, you may think you can do this:


$i^2 = (\sqrt{-1})^2 = \sqrt{(-1)^2} = \sqrt{1} =1$


But you can't! You already have two numbers that square to 1; namely -1 and +1. And $i$ already squares to -1. So it's not reasonable that $i$ would also square to 1. This points out an important detail: When dealing with imaginaries, you gain something (the ability to deal with negatives inside square roots), but you also lose something (some of the flexibility and convenient rules you used to have when dealing with square roots). In particular, you must always do the imaginary part first.


Examples


  • Simplify $\sqrt{-18}$
    $\sqrt{-18} = \sqrt{9\cdot 2 \cdot (-1)} = \sqrt{9}\;\sqrt{2}\;\sqrt{-1} = 3\sqrt{2}i$
  • Simplify $-\sqrt{-6}$
    $-\sqrt{-6} = -\sqrt{6 \cdot (-1)} = -\sqrt{6}\;\sqrt{-1} = -\sqrt{6}i$
  • Simplify $i^9$
    $i^9 = i^2\;i^2\;i^2\;i^2\;i = (-1)(-1)(-1)(-1)i = i$

    Note that we can't simplify more then this.


  • Simplify $\sqrt{-49}$
    $\sqrt{-49} = \sqrt{-1}1 \sqrt{49}= \pm 7i$

Now let's analyze that the pattern of powers, signs, 1's, and $i$'s is a cycle:


$\begin{eqnarray}
i^1&=&i\\
i^2&=&-1\\
i^3&=&-i\\
i^4&=&i\\
i^5&=&i^1=i\\
i^6&=&i^2=-1\\
i^7&=&i^3=-i\\
i^8&=&i^4=1\\
\end{eqnarray}$

In other words, to calculate any high power of $i$, you can convert it to a lower power by taking the closest multiple of 4 that's no bigger than the exponent and subtracting this multiple from the exponent. For example, a common trick question on tests is something along the lines of "Simplify $i^{99}$", the idea being that you'll try to multiply $i$ ninety-nine times and you'll run out of time, and the teachers will get a good giggle at your expense in the faculty lounge. Here's how the shortcut works:


$i^{99} = i^{96+3} = i^{(4\cdot 24)+3} = i^3 = -i$

That is, $i^{99} = i^3$, because you can just lop off the $i^{96}$. (Ninety-six is a multiple of four, so $i^{96}$ is just 1, which you can ignore.) In other words, you can divide the exponent by 4 (using long division), discard the answer, and use only the remainder. This will give you the part of the exponent that you care above. Here are a few more examples:


  • Simplify $i^{120}$
    $i^{120} = i^{4 \cdot 30} = i^{4\cdot 30 + 0} = i^0 = 1$
  • Simplify $i^{64,002}$
    $i^{64,002} = i^{64,000 + 2} = i^{4 \cdot 16,000 + 2} = i^2 = -1$

Now you've seen how imaginaries work; it's time to move on to complex numbers. "Complex" numbers have two parts, a "real" part (being any "real" number that you're used to dealing with) and an "imaginary" part (being any number with an $i$ in it). The "standard" format for complex numbers is $a + bi$; that is, real-part first and $i$-part last.


Operations on Complex Numbers


Unlike real numbers, complex numbers can produce negative numbers when squared; because of this, all polynomials have complex roots even though some of them may lack real roots.


Complex numbers can always be reduced to the form $a + bi$. If there are any terms with higher powers of $i$, you can factor out $i^2$ as many times as you need.


Furthermore, arithmetic on complex numbers obeys the same laws of algebra real numbers do.


$\begin{eqnarray}
(a + bi) + (c + di) &=& (a + c) + (b + d)i\\
(a + bi) - (c + di) &=& (a - c) + (b - d)i\\
(a + bi) * (c + di) &=& ac + adi + bci + bdi2\\
&=& (ac - bd) + (ad + bc)i
\end{eqnarray}$

Examples


  • Simplify $(2 + 3i)(1 - i)$
    $(2 + 3i)(1 - i) = 2+3i-2i-3i^2 = 2+i-3(-1) = 5+i$
  • Simplify $2i + 3i$
    $2i + 3i = (2 + 3)i = 5i$
  • Simplify $16i - 5i$
    $16i - 5i = (16 - 5)i = 11i$
  • Multiply and simplify $(3i)(4i)$
    $(3i)(4i) = (3\cdot 4)(i\cdot i) = (12)(i^2) = (12)(-1) = -12$
  • Multiply and simplify $(i)(2i)(-3i)$
    $\begin{eqnarray}
    (i)(2i)(-3i) &=& (2 \cdot -3)(i \cdot i \cdot i)\\
    &=& (-6)(i^2 \cdot i)\\
    &=& (-6)(-1 \cdot i) = (-6)(-i) = 6i\\
    \end{eqnarray}$

    Note this last problem. Within it, you can see that $i^3 = -i$, because $i^2 = -1$.


  • Simplify $(2 + 3i) + (1 - 6i)$
    $(2 + 3i) + (1 - 6i) = (2 + 1) + (3i - 6i) = 3 + (-3i) = 3 - 3i$
  • Simplify $(5 - 2i) - (-4 - i)$
    $\begin{eqnarray}
    (5 - 2i) - (-4 - i) &=& (5 - 2i) - 1(-4 - i) = 5 - 2i - 1(-4) - 1(-i)\\
    & = &5 - 2i + 4 + i = (5 + 4) + (-2i + i)\\
    & = & (9) + (-1i) = 9 - i
    \end{eqnarray}$

    You may find helpful to insert the "1" in front of the second set of parentheses so you can better keep track of the "minus" being multiplied through the parentheses.


Adding and multiplying complexes isn't too bad. It's when you work with fractions that things turn ugly. Most of the reason for this ugliness is actually arbitrary. Remember back in elementary school, when you first learned fractions? Your teacher would get her panties in a wad if you used "improper" fractions. For instance, you couldn't say "$\frac{3}{2}$"; you had to convert it to "$1 + \frac{1}{2}$". But now that you're in algebra, nobody cares, and you've probably noticed that "improper" fractions are often more useful than "mixed" numbers. The issue with complex numbers is that your professor will get his boxers in a bunch if you leave imaginaries in the denominator. So how do you handle this?


Suppose you have the following exercises:


  • Simplify $\frac{3}{2i}$

    This is pretty "simple", but they want you to get rid of that $i$ underneath, in the denominator ? To do this, you will use the fact that $i^2 = -1$. If you multiply the fraction, top and bottom, by $i$, then the $i$ underneath will vanish in a puff of negativity


    $\frac{3}{2i} = \frac{3}{2i} \cdot \frac{i}{i} = \frac{3i}{2i^2} = \frac{3i}{2\cdot (-1)} = \frac{3i}{-2} = -\frac{3i}{2} = -\frac{3}{2}i$

    So the answer is $-\frac{3}{2}i$.


  • Simplify $\frac{3}{2+i}$

    If you multiply this fraction, top and bottom, by $i$, I'll get:


    $\frac{3}{2+i} = \frac{3}{2+i} \cdot \frac{i}{i} = \frac{3i}{2i+i^2} = \frac{3i}{2i-1} = \frac{3i}{-1 + 2i}$

    Since you still have an $i$ underneath, this didn't help much. So how do you handle this simplification? You use something called "conjugates". The conjugate of a complex number $z=a + bi$ is the same number, but with the opposite sign in the middle: $\bar{z}=a - bi$. When you multiply conjugates, you are, in effect, multiplying to create something in the pattern of a difference of squares:


    $\begin{eqnarray}
    (a+bi)(a-bi) &=& a^2 -abi+abi-(bi)^2\\
    &=& a^2 - b^2(i^2)\\
    &=& a^2-b^2(-1)\\
    &=&a^2+b^2
    \end{eqnarray}$

    Note that the $i$'s disappeared, and the final result was a sum of squares. This is what the conjugate is for, and here's how it is used:

    $\begin{eqnarray}
    \frac{3}{2+i}&=&\frac{3}{2+i}\cdot \frac{2-i}{2-i} \\
    &=&\frac{3(2-i)}{(2+i)(2-i)}\\
    &=&\frac{6-3i}{4-2i+2i-i^2}\\
    &=&\frac{6-3i}{4-(-1)}\\
    &=&\frac{6-3i}{4+1}\\
    &=&\frac{6-3i}{5}\\
    &=&\frac{6}{5}-\frac{3}{5}i
    \end{eqnarray}$

    So the answer is $\frac{6}{5}-\frac{3}{5}i$


    In the last step, note how the fraction was split into two pieces. This is because, technically speaking, a complex number is in two parts, the real part and the $i$ part. They aren't supposed to "share" the denominator. To be sure your answer is completely correct, split the complex-valued fraction into its two separate terms.


Conjugation


As I explained before, the complex conjugate of the complex number $z = x + yi$ is defined to be $x − yi$, written as \bar{z} or z^*. You can imagine $\bar{z}$ to be the "reflection" of $z$ about the real axis. Therefore, both $z+\bar{z}$ and $z\cdot\bar{z}$ are real numbers.

We also have the square of the absolute value obtained by multiplying a complex number by its conjugate:

  • $|z|^2 = z\cdot\bar{z}$
  • $|z|=|\bar{z}|$
  • $z^{-1} = \frac{\bar{z}}{|z|^{2}}$ if $z$ is non-zero.

The real and imaginary parts of a complex number can also be extracted using the conjugate:

  • $\bar{z}=z$ if and only if $z$ is real
  • $\bar{z}=-z$ if and only if $z$ is purely imaginary
  • Re$\{z\} = \frac{1}{2}(z+\bar{z})$
  • Im$\{z\} = \frac{1}{2i}(z-\bar{z})$

Complex Numbers and The Quadratic Formula


You'll probably only use complexes in the context of solving quadratics for their zeroes. However there are many other practical uses for complexes, but for now you'll have to wait my next post on this topic.


Remember that the Quadratic Formula solves $ax^2 + bx + c = 0$ for the values of $x$. Also remember that this means that you are trying to find the x-intercepts of the graph. When the Formula gives you a negative inside the square root, you can now simplify that zero by using complex numbers. The answer you come up with is a valid "zero" or "root" or "solution" for $ax^2 + bx + c = 0$, because, if you plug it back into the quadratic, you'll get zero after you simplify. But you cannot graph a complex number on the x,y-plane. So this "solution to the equation" is not an x-intercept.


As an aside, you can graph complexes, but not in the x,y-plane. You need the "complex" plane. For the complex plane, the x-axis is where you plot the real part, and the y-axis is where you graph the imaginary part. For instance, you would plot the complex number $3 - 2i$ in the position $(x,yi) = (3,2)$


This leads to an interesting fact: When you learned about regular ("real") numbers, you also learned about their order (this is what you show on the number line). But x,y-points don't come in any particular order. You can't say that one point "comes after" another point in the same way that you can say that one number comes after another number. For instance, you can't say that (4, 5) "comes after" (4, 3) in the way that you can say that 5 comes after 3. Pretty much all you can do is compare "size", and, for complex numbers, "size" means "how far from the origin". To do this, you use the Distance Formula, and compare which complexes are closer to or further from the origin. This "size" concept is called "the modulus". For instance, looking at our complex number plotted above, its modulus is computed by using the Distance Formula:


$|3-2i| = \sqrt{3^2+2^2} = \sqrt{9+4} = \sqrt{13} \approx 3.61$


Note that all points at this distance from the origin have the same modulus. All the points on the circle with radius $\sqrt{13}$ are viewed as being complex numbers having the same "size" as $3 - 2i$.


Complex Number in Phasor Form


A complex number $z=x+iy$ can be written in "phasor" form $z=|z|(cos\; \theta + i\; sin\; \theta)=|z|e^{i\theta}$


Here, $|z|$ is known as the complex modulus (or sometimes the complex norm) and $\theta$ is known as the complex argument or phase. The plot above shows what is known as an Argand diagram of the point z, where the dashed circle represents the complex modulus $|z|$ of $z$ and the angle $\theta$ represents its complex argument. Historically, the geometric representation of a complex number as simply a point in the plane was important because it made the whole idea of a complex number more acceptable. In particular, "imaginary" numbers became accepted partly through their visualization.



Matrix representation of complex numbers


While usually not useful, alternative representations of the complex field can give some insight into its nature. One particularly elegant representation interprets each complex number as a $2\times 2$ matrix with real entries which stretches and rotates the points of the plane. Every such matrix has the form


$\begin{pmatrix}a & -b\\b & a\end{pmatrix}$

where a and b are real numbers. The sum and product of two such matrices is again of this form, and the product operation on matrices of this form is commutative. Every non-zero matrix of this form is invertible, and its inverse is again of this form.

Therefore, the matrices of this form are a field, isomorphic to the field of complex numbers. Every such matrix can be written as

$\begin{pmatrix}a & -b\\b & a\end{pmatrix} =a\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}+b\begin{pmatrix}0 & -1\\1 & 0\end{pmatrix}$

which suggests that we should identify the real number 1 with the identity matrix

$\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}$

and the imaginary unit $i$ with

$\begin{pmatrix}0 & -1\\1 & 0\end{pmatrix}$

a counter-clockwise rotation by 90 degrees. Note that the square of this latter matrix is indeed equal to the 2x2 matrix that represents -1.

The square of the absolute value of a complex number expressed as a matrix is equal to the determinant of that matrix.

$|z|^2 = det\; \left( \begin{array}{c c}a & -b\\b & a\end{array}\right) = (a^2) - ((-b)(b)) = a^2 + b^2$

References


Stapel, Elizabeth. "Complex Numbers & The Quadratic Formula." Purplemath. Available from http://www.purplemath.com/modules/complex3.htm. Accessed 28 April 2010


Complex Numbers. Wikipedia. Available from http://en.wikipedia.org/wiki/Complex_number. Accessed 28 April 2010



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