Square of "i"
I read one of the answers to the question about i, but I
am not completely sure if I understand. What is i^2?! Using two
different methods, both mathematically correct, render you with two
different answers, 1 and -1.
You can follow the rules where i^2 can be broken down into
(-1^(1/2))^2. And the exponents multiply to give you -1^1 which is -1.
But if you break it down into two separate radicals, you have
(sqrt(-1))*(sqrt(-1)), which can then be simplified by putting the product
under the same radical, meaning sqrt(-1*-1). Thus rendering you with the
sqrt(1) which is just 1. And compared with the first method, you then
have that -1=1. How is this so?
You have to be careful about which rules you apply.
There are no real numbers which, when you square them, result in a negative
But you can leave the "realm" of real numbers and enter the world of complex
numbers which have a real part and a complex part (in the form a + bi).
This is a different mathematical model which is useful for many "real world"
applications. Electrical engineers use complex numbers as a way of
describing vectors of electrical power (where the length of the vector
represents the magnitude of a voltage and the angle represents its phase,
for example). In this realm of complex numbers, "i" represents some
quantity that, when squared, gives you negative 1. You can represent any
negative number using this defined quantity (a quantity squared that gives
you -2 is 2i, a quantity squared that gives you -pi is (pi)i, etc).
So, by definition, i^2 is -1 in the realm of complex numbers. If you
rewrite i^2 as sqrt(-1)*sqrt(-1), then you may NOT apply the rule (which
works only in the realm of real numbers) that this is equal to
sqrt((-1)*(-1)) -- because in the realm of real numbers, there is NO number
that you can square to give you a negative number (so sqrt (-1) is
This can be confusing to students because they want to apply rules that work
in the realm of real numbers to complex numbers. But different rules apply.
How do you multiply (a + bi) * (c + di)? How do you add (a +bi) + (c + di)?
The rules for manipulating complex numbers are different than the rules for
Todd Clark, Office of Science
US Department of Energy
You have to be careful when you "square" a number, because there are
two solutions (+N) and (-N) that both have the same value of (N)^2 --
(+N)*(+N) and (-N)*(-N). So in the case of "i": (+ i)*(+ i) = (i)^2 = -1
and (-i)*(-i) = (-1* i)* (-1*i) = [(-1)*(-1)]*[i * i] = [+1]*[i*i] = (i)^2
= -1. The confusion arises (and you are by no means alone) because using
the notation for complex numbers (a + i*b) obscures the fundamental
definition of multiplication of complex numbers, which is: Complex numbers
are ordered pairs (a,b) of real numbers "a" and "b", that is (a,b) is not
equal to (b,a). The multiplication of two complex numbers (a,b) and (c,d)
is defined as an ordered pair of real numbers: (a,b)*(c,d) = (ac-bd,
ad+bc). That is how mathematicians define complex numbers. The rest of
us use the shorthand: (a+ib)*(c+id) = (a*c-(i)^2b*d) + i*(b*c+ a*d) and
multiply using binomial multiplication, as we would do for real numbers.
Where you have to be careful is when "squaring" a complex number. Usually,
when one speaks about "squaring" a complex number, what is actually meant
is multiply the complex number and its complex conjugate. The complex
conjugate of (a+ ib) is defined as (a - ib), that is, the "i"'s are turned
into (-"i"'s). So (a+ ib)*(a - ib) = a^2 + b^2 which is a real number.
This is not the same quantity as (a+ ib)*(a+ ib) = (a^2- b^2, 2*a*b) or in
the "i" notation: (a^2 - b^2) + i(2*a*b).
When dealing with numbers that are not real and positive, the relation
sqrt(a)*sqrt(b)=sqrt(ab) is not necessarily true. When you go beyond
positive numbers, "root" functions become more involved. When you have
complex numbers, you cannot assume the answer will be positive. The square
root of 1 can be either +1 or -1. The fourth root of 1 can be +1, +i, -1,
or -i. All are fourth roots of 1 when working with complex numbers.
When working in only real numbers, sqrt(-1)*sqrt(-1) doesn't have a value.
The value of i-squared is -1. This quantity allows us to work with square
roots of negative numbers. It also provides a mathematical system that can
provide some useful shortcuts when dealing with vectors.
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Update: June 2012