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Applications of Imaginary Numbers

Date: 10/14/97 at 18:40:14
From: Beatka Zakrzewski
Subject: Imaginary Numbers in the work force (applications)

Dear Dr. Math,

I know that you have already received tens of questions about 
imaginary numbers but I can't seem to find a straight answer to mine. 
Where are imaginary numbers used today in real life, as in the work 
force or other areas that use math?  I have heard that they are used 
in electrical technologies - if so, could you specifically tell me in 
which areas?  

Thank you.      
Sincerely, Beatka Zakrzewski

Date: 10/14/97 at 19:03:06
From: Doctor Tom
Subject: Re: Imaginary Numbers in the work force (applications)

Hi Beatka,

I answered almost exactly the same question for a teacher a couple of 
years ago and saved my reply. I'll send it to you. Unfortunately, as 
you'll see from the description, they are often used with more 
sophisticated math than you know yet, but if you ignore the details 
below and concentrate on the problem descriptions, they may make some 
sense to you.

   I teach Algebra II and Trigonometry in Gilroy High School,
   Gilroy, CA.  The students' ages are 15-17.

   Their question is: Who uses imaginary numbers in the real world?
   The students learn to calculate with i and complex numbers of
   the form a + bi. The book indicates that electrical engineers
   use imaginary numbers, but use j instead of i. What do they
   *do* with the numbers? And does anyone else actually use them?

First off, electrical engineers use "j" because "i" is almost always
used to mean "current" in their equations.  Where most of us would 
write "a + bi", they'd write "a + bj", but it means the same thing.

Electrical engineers often have to solve what are called "differential
equations," which are a bit hard to explain without knowing a bit 
about calculus. Basically, a differential equation relates functions 
to their rates of growth. The solution to a differential equation is 
usually a function, not a number.

As a specific example, suppose you have a snowplow that keeps piling
up more and more snow in front of it so that the farther it goes, the
heavier the load it is pushing, and the heavier the load, the slower 
it goes, and the slower it goes the slower the pile of snow in front 
of it grows.

You can (with a differential equation) relate the amount of snow at a
given time t (call it A(t)) to the velocity of the plow, and the
equations can be solved to give the function A(t) at all times t.

But often, it's easier to solve differential equations in the domain 
of complex numbers because the equations are a lot nicer, but you know 
that the solution you care about is just the real part of the 
solution. It's difficult to give an example without some calculus.

I can, however, show you a nice example that may make it clear that
working in the domain of complex numbers is easier in some cases than
working strictly in the reals. If you're studying the complex numbers
and trigonometry at the same time, in theory you can follow the next
steps, but if you can't, don't worry; just look at the messiness of
the calculation - I'm trying to show how going to the domain of 
complex numbers can drastically simplify a problem.

Try to work out the following:  (cos x + i*sin x) .  Multiply this out
and reduce it to the simplest form.  I get this:

       3           2                         2         3
    cos x + 3*i*cos x * sin x - 3*cos x * sin x - i*sin x.

(It's a nice exercise in algebra, complex numbers, and trigonometry.)

Let's just look at the real part (you can do the same sort of thing
with the imaginary part):

       3                 2
    cos x - 3*cos x * sin x

                   2       2                  2
    = cos x * ( cos x - sin x) - 2*cos x * sin x

                        2       2
Now, since cos(2x) = cos x - sin x, and sin(2x) = 2*sin x * cos x,
we can re-write the equation above as:

     cos x * cos(2x) - sin x * sin(2x).

Now, cos(3x) = cos(x + 2x) = cos x * cos(2x) - sin x * sin(2x), so the
real part is just cos(3x).  With a similar amount of ugly calculation,
we can get that the imaginary part is i*sin(3x).

So the answer is:  cos(3x) + i*sin(3x).

                                     2            3
I assume you understand exponents = x  = x * x,  x  = x * x * x, and
so on.  Well, it turns out that there's a special number called e 
which is equal (approximately) to 2.71828182845... which satisfies the
following equation (in the complex numbers:

    e   = cos(x) + i*sin(x)

So the original problem I stated was to find the cube of the number on
the left:

      ix 3    i(3x)
    (e  )  = e     = cos(3x) + i*sin(3x),

so you're done in a single step.

Another important application of complex numbers to the real world is
in physics. In quantum mechanics, one cannot say with precision where 
a particle is. You can only give an probability distribution of its
position in space. And the only way to calculate the distributions is
using complex variables. Unfortunately, this is even harder to 
explain, but such calculations have to be done for almost any 
calculations about nuclear reactions.

Finally, something that may not be precisely an "application," but one
that you can easily experiment with, is that a certain class of 
complex numbers behave as rotation operators.

For example, draw the usual real and imaginary axes, and plot any 
point on it (say 3 + 5i)  Multiply this number by i, and you get 
(-5 + 3i). If you plot this new point, you'll find that it is the 
original point rotated about the origin by 90 degrees counter-
clockwise.  This works for ANY complex number. Multiply by i, and 
you'll rotate it by 90 degrees.

Now, take any complex number, and multiply it by cos(45) + i*sin(45)
(in degrees), which is about (.707 + .707*i).  This rotates points
clockwise by 45 degrees. And there's nothing special about 45 degrees. 
Multiply any complex number by cos(x) + i*sin(x), and you'll rotate 
the number about the origin by an angle x.

In the same way, adding a fixed complex number is equivalent to a
translation, and multiplying by a real number expands or contracts the
values. By combinations of rotations, translations, and expansions/ 
shrinkages, you can do most 2-dimensional computer graphics operations 
on objects in a plane.

Unfortunately, in 3 dimensions, it's not so easy, and the easy way
involves something called matrix multiplication.

I know none of these is easy to understand, but if you want to do them
eventually, you've got to have a solid foundation in the basic
operations on complex numbers - just as you can't do algebra until
you learn to add and subtract real numbers.

-Doctor Tom,  The Math Forum
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Associated Topics:
High School Calculus

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