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### Square Roots in Complex Numbers

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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
POSITIVE.

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.

Tim Jurd
```

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

Hi,

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
instance,

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 = ---------------------
2a

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

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

but

(-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
plane.

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!  http://mathforum.org/dr.math/
```
Associated Topics:
College Imaginary/Complex Numbers
High School Imaginary/Complex Numbers
High School Square & Cube Roots
Middle School Square Roots

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