What is wrong with the statement #1 = sqrt(1) = sqrt((-1)*(-1)) = sqrt(-1) * sqrt(-1) = -1# ?

1 Answer
Apr 9, 2016

Answer:

#sqrt((-1)*(-1)) != sqrt(-1)*sqrt(-1)#

Explanation:

The rule #sqrt(ab) = sqrt(a)sqrt(b)# only holds if at least one of #a, b >= 0#.

Why is this so?

Every number (Real or Complex) apart from #0# has two square roots.

In order to tell them apart, we call one of them the principal square root and denote it by #sqrt(x)#. The other, non-principal square root is then #-sqrt(x)#.

Which one is principal?

If #x >= 0# then #sqrt(x)# denotes the non-negative square root and lies on the non-negative part of the Real line.

If #x < 0# then #sqrt(x)# denotes the square root that lies on the "positive" part of the imaginary axis. That is, #sqrt(x) = sqrt(-x)*i#.

If that sounds a little arbitrary, it is.

When you think about the square root of Complex numbers you start to see the problem better:

Imagine a point moving slowly anticlockwise around the unit circle in the Complex plane, starting from the point #1# on the Real axis.

Its principal square root also starts to move around the unit circle anticlockwise, but at half the speed.

When the original point has completed one full revolution, where is the point tracking the principal square root?

If we had just let it proceed smoothly on its journey then it would have only completed half a revolution, so would be at #-1#. Here's our problem.

In order that the principal square root be well defined, we need to choose somewhere to put a discontinuity - called a branch cut.

Most people put it just below the negative part of the Real axis. Some people prefer to put it just below the positive part of the Real axis.

Once we do this, the point representing the principal square root abruptly jumps from one side of the unit circle to the opposite side as the point representing the original number crosses the cut.