The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Square Roots in Complex Numbers

Date: 11/06/97 at 04:35:21
From: Tim Jurd
Subject: Square Roots in Complex

Dear Dr. Math,

My students and I were wondering why in the complex number system 
every number has two square roots when in the real number system we 
teach that the square root of any positive number is by definition 

The method I use for finding the sqrt z, z belongs to C, is to let 
(a+ib)^2 = z and find real values of a,b which solve this equation.
Please enlighten us.

Tim Jurd

Date: 11/06/97 at 10:57:02
From: Doctor Pete
Subject: Re: Square Roots in Complex


The short and simple answer to your question is that the square root 
of a positive real number is in fact not a unique number. For 

     Sqrt[4] = {-2, 2}.

But because there is a "natural" way to choose one of these values, 
namely, the positive one, we assign the square root this *principal* 
value. That is, 2 is the principal value of the square root of 4.  
However, -2 is also a root, because the equation x^2 = 4 has two 
solutions. This explains why the quadratic formula for the roots of 
ax^2 + bx + c = 0 is written

         -b +- Sqrt[b^2 - 4ac]
     x = --------------------- 

because +- emphasizes the fact that the square root may take on two 

Now, as you have seen, with the complex numbers this choice of 
"positive" over "negative" no longer seems so clear. For instance, 
what is the square root of 3-4i ?  Clearly,

     (2-i)^2 = 4 - 2(2i) + (-i)^2 = 3-4i


    (-2+i)^2 = (-2)^2 - 2(2i) + i^2 = 3-4i

There are two values, but which one is "positive" and which one is
"negative?"  One could say that the principal root should be the one 
with positive real part, but this doesn't always work. For example, 
what are the cube roots of -1?  There are three (verify for yourself):

     (-1)^3 = [(1+Sqrt[3]i)/2]^3 = [(1-Sqrt[3]i)/2]^3 = -1.

Now, note that there are *two* roots which have positive real part, 
yet the most commonly taken root is in fact -1, because it is a pure 
real number. So we need more powerful concepts than just "positive/
negative." But a full description of this, which we call the theory of 
complex analysis (specifically, branch cuts of complex analytic 
functions), is perhaps beyond the scope of a simple explanation.  

The underlying idea behind all this is that in the complex numbers, 
one should be accustomed to thinking of functions as maps of sets to 
sets, not numbers to numbers. For instance, the function f(z) = z^2 is 
a map from the complex numbers to the complex numbers, but it is not 
one-to-one. Rather, it is TWO-to-one, because 

     f(z) = z^2 = (-z)^2 = f(-z)

So there are two values, z and -z, in the domain, which map to the 
same value z^2 in the range. Conversely, the inverse function 
g(z) = Sqrt[z] must be a one-to-two mapping - that is, for one value 
of z in the domain, g maps this to *two* values in the range. While 
this may seem strange or unnecessary, it is neither....  And it is 
easy to show that the functions f(z) = z^k, g(z) = z^(1/k) where k is 
a positive integer, are k-to-one and one-to-k functions in the complex 

One last tidbit:  e^(i*t) = Cos[t]+i*Sin[t].  This is Euler's formula, 
and from it we can see that the function 

     f(z) = e^z = Exp[z] 

is an infinity-to-one function, and therefore its inverse 

     g(z) = Ln[z] = Log[z] 

is a one-to-infinity function.

-Doctor Pete,  The Math Forum
 Check out our web site!   
Associated Topics:
College Imaginary/Complex Numbers
High School Imaginary/Complex Numbers
High School Square & Cube Roots
Middle School Square Roots

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.