What is the square root of -49?

2 Answers
Jan 1, 2017

#sqrt(-49) = 7i#

Explanation:

A square root of a number #n# is a number #x# such that #x^2 = n#

Note that if #x# is a Real number then #x^2 >= 0#.

So any square root of #-49# is not a Real number.

In order to be able to take square roots of negative numbers, we need Complex numbers.

That's where the mysterious number #i# comes into play. This is called the imaginary unit and has the property:

#i^2 = -1#

So #i# is a square root of #-1#. Note that #-i# is also a square root of #-1#, since:

#(-i)^2 = (-1*i)^2 = (-1)^2*i^2 = 1*(-1) = -1#

Then we find:

#(7i)^2 = 7^2*i^2 = 49*(-1) = -49#

So #7i# is a square root of #-49#. Note that #-7i# is also a square root of #-49#.

What do we mean by the square root of #-49#

For positive values of #n#, the square root is usually taken to mean the principal square root #sqrt(n)#, which is the positive one.

For negative values of #n#, the square roots are both multiples of #i#, so neither positive nor negative, but we can define:

#sqrt(n) = i sqrt(-n)#

With this definition, the principal square root of #-49# is:

#sqrt(-49) = i sqrt(49) = 7i#

#color(white)()#
Footnote

The question remains: Where does #i# come from?

It is possible to define Complex numbers formally, as pairs of Real numbers with rules for arithmetic like this:

#(a, b) + (c, d) = (a+c, b+d)#

#(a, b) * (c, d) = (ac-bd, ab+cd)#

These rules for addition and multiplication work as expected with commutativity, distributivity, etc.

Then Real numbers are just Complex numbers of the form #(a, 0)# and we find:

#(0, 1)*(0, 1) = (-1, 0)#

That is #(0, 1)# is a square root of #(-1, 0)#

Then we can define #i = (0, 1)# and:

#(a, b) = a(1, 0) + b(0, 1) = a+bi#

Jan 1, 2017

I'm not altogether happy with the #7i# being described as the square root of #-49# or even the principal value of the multi-valued function, though it seems to be common practice.

Explanation:

The notation doesn't really extend to #n#th roots, nor roots of complex numbers. What is the fourth root of #1#? (#1#, #i#, #-1#, #-i#). In the special case of square roots of negative numbers it seems to be common practice to take the root to be the complex number with a positive imaginary part: here, #[0,7]# (or #7i# in the more common notation. You also get problems with sudden jumps in the #n# root of #z# if #z# moves, say, along the unit circle, if you use the concept of principal root.

Decades ago at school we learnt a lot about the #n#th complex roots of unity (and hence of any complex number including the pure real negative ones), but as far as I recall we never used the notation #sqrt(x)# (or other roots) if #x# might be negative or complex.

However, we did use the notion of primitive roots of a complex number #z#, from which all others can be derived by raising the primitive root to integral powers. So #i# is a primitive fourth root of #1# but #-1# is not. I am sure we did not use the concept of principal root, analogous to principal values of inverse trig functions.

I prefer the notion that the #n#th root is a function which maps a complex number on to a set of #n# complex numbers, where each such number raised to the #n#th power is the original number. I don't consider "multi-valued function" to be an oxymoron!