SOLUTION: If square root of -1 is j, what is square root of j?

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Question 621949: If square root of -1 is j, what is square root of j?
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Pretty much everyone uses "i" as a name for sqrt%28-1%29. So I'm a little confused by the problem's use of "j" instead. I'm going to use "i". If you're really supposed to be using "j", then just replace my i's with j's.

The "trick" to this problem is to realize that square roots can be expressed with an exponent of 1/2. IOW:
sqrt%28i%29+=+i%5E%281%2F2%29

To find a power of a complex number, like i, we use what I believe is called DeMoivre's Theorem. This theorem states that for a complex number written in polar form:
z+=+r%2A%28cos%28theta%29+%2B+i%2Asin%28theta%29%29
that
z%5En+=+r%5En%2A%28cos%28n%2Atheta%29+%2B+i%2Asin%28n%2Atheta%29%29

So we need to write the complex number, i, in polar form. And, unless you are clever enough to figure this out on your own, we first need to write it in standard a + b*i form:
i = 0 + 1*i
This makes a = 0 and b = 1. To write this in polar form we use:
r+=+sqrt%28a%5E2%2Bb%5E2%29 and theta+=+tan%5E%28-1%29%28b%2Fa%29
Inserting our a and b into these we get:
r+=+sqrt%280%5E2+%2B+1%5E2%29+=+sqrt%280%2B1%29+=+sqrt%281%29+=+1 and
theta+=+tan%5E%28-1%29%281%2F0%29+=+90
So "i" in polar form:
i+=+1%28cos%2890%29+%2B+i%2Asin%2890%29%29

Now we can use the theorem:

Simplifying:
i%5E%281%2F2%29+=+1%28cos%2845%29+%2B+i%2Asin%2845%29%29
i%5E%281%2F2%29+=+cos%2845%29+%2B+i%2Asin%2845%29
i%5E%281%2F2%29+=+sqrt%282%29%2F2+%2B+i%2A%28sqrt%282%29%2F2%29
or, in a + bi form:
i%5E%281%2F2%29+=+sqrt%282%29%2F2+%2B+%28sqrt%282%29%2F2%29%2Ai
Note: This is the positive square root. The negative square root out be:
-i%5E%281%2F2%29+=+-sqrt%282%29%2F2+%2B+%28-sqrt%282%29%2F2%29%2Ai


To check, we can square our answer and see if we get i:
%28sqrt%282%29%2F2+%2B+%28sqrt%282%29%2F2%29%2Ai%29%5E2
To square this we can use FOIL or the %28a%2Bb%29%5E2+=+a%5E2%2B2ab%2Bb%5E2 pattern. I like using the pattern:

Simplifying:
2%2F4+%2B+2%2A%282%2F4%29%2Ai+%2B+%282%2F4%29%2Ai%5E2%29
Since i%5E2+=+-1:
2%2F4+%2B+2%2A%282%2F4%29%2Ai+%2B+%282%2F4%29%2A%28-1%29
2%2F4+%2B+2%2A%282%2F4%29%2Ai+%2B+%28-2%2F4%29
0+%2B+1%2Ai
i
Check!