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Imaginary Numbers

Let's think about the idea of a square root. The square root of a number is another number which, when multiplied by itself is the first number. For example, 3 is the square root of 9 because . Let's consider this a little further: a positive number muliplied by itself is a positive number (for example, ... 4 is positive and 16 is also positive). A negative number multiplied by itself is also positive (i.e. ).

Now, in the first case, the square root of 16 is 4 because . (Some people would be really picky and they'll tell you that 16 has two roots: 4 and -4. Those people are slightly geeky, but technically correct.) There's just one small snag - what if you were asked for the square root of a negative number? There is no such thing as a number which, when multiplied by itself results in a negative number. So asking for the square root of -16 doesn't make sense. In fact, if you try to do this on your calculator, it'll probably tell you that it gets an error instead of producing an answer.

Mathematicians as a general rule don't like loose ends - they aren't the type of people who leave things lying around... and having something as simple as the square root of a negative number lying around unanswered got on their nerves so they had a bunch of committee meetings and decided to do something about it. Their answer was to invent a new number called i (for imaginary) although some people call it j just to screw everyone up.^2.3 Generally speaking, mathematicians use i and physicists and engineers use j so we'll stick with j - we'll see why in a later chapter.)

``What is j?'' I hear you cry. Well, j is the square root of -1. Of course, there is no number that is the square root of -1, but since that answer is inadequate, j will do the trick, so we just define it with the equation

$\begin{displaymath} \sqrt{-1} = j \end{displaymath}$

(2.16)

and therefore

$\begin{displaymath} j^{2} = -1 \end{displaymath}$

(2.17)

Now, remember that j * j = -1. This is useful for any square root of any negative number, you just calculate the square root of the number pretending that it was positive, and then stick an j after it. So, since the square root of 16, abbreviated $\sqrt{16} = 4$ and $\sqrt{-1} = j$ , then $\sqrt{-16} = j4$ . Let's do a couple:

$\begin{displaymath} \sqrt{-9} = j3 \end{displaymath}$

(2.18)

$\begin{displaymath} \sqrt{-4} = j2 \end{displaymath}$

(2.19)

Another way to think of this is $\sqrt{-a} = \sqrt{-1 * a} = \sqrt{-1}*\sqrt{a} = j \sqrt{a}$ so:

$\begin{displaymath} \sqrt{-9} = \sqrt{-1} * \sqrt{9}= j * \sqrt{9} = j3 \end{displaymath}$

(2.20)

Of course, this also means that

$\displaystyle j3 * j3$	$\textstyle =$	$\displaystyle (jj) (3*3)$	(2.21)
	$\textstyle =$	$\displaystyle -1 * 9$	(2.22)
	$\textstyle =$	$\displaystyle -9$	(2.23)

Next: Complex numbers Up: Complex Numbers Previous: Real Numbers Contents Index

Geoff Martin 2006-10-15

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