math
posted by jack .
I Wanted To Find The Square Root Of (12i)
But I Didn't Know How To Solve It.
My Solution :
x+yi = sqr(12i)
x2y2+2xyi = 1 2i
x2y2 = 1
2xy = 2
y =  1/x
x2  (1/x) = 1 ] * x2
x4x21 = 0
How To Evaluated (x4x21) ?
I Tried To Sol

I would convert the complex number to polar.
12i= sqrt5 @arctan2=sqrt 5 @63.4349488 degrees or @(36063.43)deg figure that angle out.
sqrt (12i)= sqrt(sqrt5)@1/2 (that angle)
now convert back to polar.
lets do it so it can be checked.
(5^.25)=1.49534878
that angle= = 296.565051
1/2 that angle= = 148.282525
sqrt(12i)=1.4953cos 148.282525+i(1.4953sin 148.282525)
= 1.27i*0.786
Now, just for fun sake,lets square that.
(1.27^2)(..786^2)2i(1.21*.786)
= 1.61.617 i 1.90=.991.9i
Ok, you can work it with more accuracy to get it.
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