document.write( "Question 621949: If square root of -1 is j, what is square root of j? \n" ); document.write( "

Algebra.Com's Answer #391306 by jsmallt9(3758) $\"\"$ $\"About$

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.

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

\n" ); document.write( "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:
\n" ); document.write( " $\"z+=+r%2A%28cos%28theta%29+%2B+i%2Asin%28theta%29%29\"$
\n" ); document.write( "that
\n" ); document.write( " $\"z%5En+=+r%5En%2A%28cos%28n%2Atheta%29+%2B+i%2Asin%28n%2Atheta%29%29\"$

\n" ); document.write( "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:
\n" ); document.write( "i = 0 + 1*i
\n" ); document.write( "This makes a = 0 and b = 1. To write this in polar form we use:
\n" ); document.write( " $\"r+=+sqrt%28a%5E2%2Bb%5E2%29\"$ and $\"theta+=+tan%5E%28-1%29%28b%2Fa%29\"$
\n" ); document.write( "Inserting our a and b into these we get:
\n" ); document.write( " $\"r+=+sqrt%280%5E2+%2B+1%5E2%29+=+sqrt%280%2B1%29+=+sqrt%281%29+=+1\"$ and
\n" ); document.write( " $\"theta+=+tan%5E%28-1%29%281%2F0%29+=+90\"$
\n" ); document.write( "So \"i\" in polar form:
\n" ); document.write( " $\"i+=+1%28cos%2890%29+%2B+i%2Asin%2890%29%29\"$

\n" ); document.write( "Now we can use the theorem:
\n" ); document.write( "

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

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

\n" ); document.write( "Simplifying:
\n" ); document.write( " $\"2%2F4+%2B+2%2A%282%2F4%29%2Ai+%2B+%282%2F4%29%2Ai%5E2%29\"$
\n" ); document.write( "Since $\"i%5E2+=+-1\"$ :
\n" ); document.write( " $\"2%2F4+%2B+2%2A%282%2F4%29%2Ai+%2B+%282%2F4%29%2A%28-1%29\"$
\n" ); document.write( " $\"2%2F4+%2B+2%2A%282%2F4%29%2Ai+%2B+%28-2%2F4%29\"$
\n" ); document.write( " $\"0+%2B+1%2Ai\"$
\n" ); document.write( "i
\n" ); document.write( "Check! \n" ); document.write( "

\n" );