# Super Code from Digital Knife Monkey Productions

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#### Equation of the Week #5: Squareroots of complex numbers Posted by DarthWho on May 8, 2011 at 12:00 AM

Any one who is sufficiently well versed in number theory understands that the squareroot of a negative number produces what is known as an imaginary number:designated by the symbol i but then what is the squareroot of i or for that matter any of the general group of the complex numbers (a+b*i)?

√(A+B*i)=√((A + √(A2 + B2))/2) + SGN(B) * √((-A + √(A2 + B2))/2) * i

which for i produces about 0.707106781 + 0.707106781 * i

Next Week: Basic Complex Number Rules (addition subtration multiplication and division)

Categories: Equation of the Week

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#### 1 Comment

DarthWho
11:56 AM on July 21, 2011
PhilOfPerth says...
I'm not really into heavy math, but numbers still interest me. For instance, the number you came up with in your discussion on the square root of negative numbers (i), is used in electrical theory in relation to the peak voltage of a sine wave. Peak to peak is 1.414 (twice your number) of the RMS value.

that may very well be the case however (I am admittedly presuming you have an understanding of the imaginary numbers. I must point out that 1,-1, i and -i are all at an absolute distance from the origin of 1 the square root function brings values toward that ring as well as toward 1 so finding the square-root of i requires that it must be the same distance from the origin that 1 is. if You apply the pythagorean theorem to the two values the resulting distance is 1

SQRT(0.707106781^2 + 0.707106781^2)=1

if you do not believe the results output by the formula above then you are welcome to type the following into Google: sqrt(i) and test using another source.