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Complex Numbers: Graphing Functions with Imaginary Numbers

Date: 01/21/2003 at 22:31:22
From: Carolyn
Subject: Graphing functions with imaginary numbers

My pre-calculus class was graphing the function  f(x)=x^(2/3) on our 
graphing calculators, and my teacher asked us for the domain of the 
function. We all agreed that it could include all real numbers, and I 
asked if  it could also include imaginary numbers, so we tried to 
figure it out, using 5i for x, but we couldn't get the same answer if 
we switched techniques. How can you tell if a graph would include 
imaginary numbers? If not, wouldn't it have spaces, even if they were 
only very small?

Also, we had trouble with the imaginary number that I picked to 
substitute in the function. The calculators we used (TI-83 or TI-83 
plus) came up with different answers using different methods. We can't 
figure out why. When we simplified the problem f(x)=x^(2/3), with 
x=5i, we got  the cube root of -25. If we typed this in on our 
calculators, we came up with -2.924017738. If, however, you enter 
(5i)^2, and then the cube root of  that answer, you come up with 
1.462008869+2.532273642i. Is this the same answer?

Date: 01/22/2003 at 00:22:33
From: Doctor Tom
Subject: Re: Graphing functions with imaginary numbers

Hi Carolyn,

You have asked a very interesting (and difficult) question. Much of 
the field of complex analysis and Riemann surfaces is devoted to 
providing a complete answer.

I'll try to show you what the problem is with some simple examples.

If I ask you what the square root of 25 is, you'll probably say 5, 
although (-5)(-5) = 25, so in a sense, -5 is just as good an answer. 
We agree that when you use that square root symbol over a number we 
mean the positive square root since we run into positive numbers more 
often than negative ones, but it could have been defined to be the 
negative square root and everything would work out fine.

But what does it mean to take the square root of a negative number?  
sqrt(-1) = i, but it's also equal to -i, and there is no reasonable 
way to assign an ordering to the complex numbers like you can the 
reals so you can't say that i is positive and -i is negative.  In 
fact, every complex number except zero has two complex square roots, 
where by "complex number" I mean the entire set of complex numbers 
that includes as a small subset the reals.

Cube roots are worse -- every complex number except zero has three 
cube roots (and four fourth roots, five fifth roots, et cetera).  The 
three cube roots of 1 are 1, -1/2 + sqrt(3)i/2 and -1/2 - sqrt(3)i/2.  
Cube either of the last two to check that their cubes are, in fact, 1.

1 has four fourth roots: 1, -1, i, -i.

These are called "multiple-valued functions," and in complex analysis, 
you can't get away from them without running into trouble.  But by 
definition, a function has only one value for each input, so the 
problem is "fixed" by defining the solutions to lie on a weird surface 
(possibly different for each function) on which the function behaves 
locally like a single-valued function.

For example, you would expect that if the square root of 1 is 1, then 
the square root of a number close to 1, like 1.0001 + .000324i, would 
also be close to 1, and on the surface (called a Riemann surface), it 
is.  The other square root of 1.0001 + .000324i is, of course, close 
to -1, as you'd expect.

We say that the Riemann surface for the square root function has two 
"sheets," since there are two answers everywhere except at the origin, 
and you can "glue" the sheets together in your imagination (but not 
literally) such that one passes through the other.  For example, if 
the square root of 1 is 1, and you pick numbers near it, those square 
roots will also be close to 1, and numbers close to that new number 
will be close to the new square root, and so on.  But if you start at 
1, and trace the square roots as you trace a path of numbers around 
the origin 0, each time picking the square root that is near the last 
one, when you get back to 1, you'll find your square roots are close 
to -1.  In one more loop, you'll be back to 1.

With cube roots, there are three sheets, and each time around the loop 
you'll get to a different cube root of 1 until three times around when 
you return to the original value.  Fourth roots would require four 
loops, et cetera.

On some functions you would never get back, like a spiral staircase 
that goes on forever.  The complex logarithm function is like that.  
For other functions, the value you get depends on what path you took, 
like the function square_root(x(1-x)).

So in short, you've asked a very difficult, but very interesting 
question that's usually not covered completely until you take a full-
year course in complex analysis in college.

The good news is that among the elementary functions, the only time 
you have this sort of bad news is with the roots (square roots, cube 
roots, ...), with the logarithm, and with functions that involve those 
functions.  Also with things like arcsine, arctangent, ..., since 
they, too, are multiple valued.  What is the arcsine of zero?  There 
are an infinite number of answers, really: 0, pi, 2pi, 3pi, -pi and so 

Functions where there is no problem and a reasonable complex function 
occurs includes polynomials, quotients of polynomials except where the 
denominator goes to zero, and even then you can say it "goes to 
infinity" in a well-controlled way, e^x, sin, cos, tan, sec, csc, cot, 
and combinations of those.

Since you haven't even gotten to calculus yet, you'd probably have a 
tough time looking at a book on Riemann surfaces, but I hope this at 
least gives you a flavor of what's going on. It's a difficult but 
wonderful subject!

- Doctor Tom, The Math Forum 
Associated Topics:
High School Imaginary/Complex Numbers

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