posted by jack .
I Wanted To Find The Square Root Of (1-2i)
But I Didn't Know How To Solve It.
My Solution :
x+yi = sqr(1-2i)
x2-y2+2xyi = 1- 2i
x2-y2 = 1
2xy = -2
y = - 1/x
x2 - (-1/x) = 1 ] * x2
x4-x2-1 = 0
How To Evaluated (x4-x2-1) ?
I Tried To Sol
I would convert the complex number to polar.
1-2i= sqrt5 @arctan-2=sqrt 5 @-63.4349488 degrees or @(360-63.43)deg figure that angle out.
sqrt (1-2i)= sqrt(sqrt5)@1/2 (that angle)
now convert back to polar.
lets do it so it can be checked.
that angle= = 296.565051
1/2 that angle= = 148.282525
sqrt(1-2i)=1.4953cos 148.282525+i(1.4953sin 148.282525)
Now, just for fun sake,lets square that.
= 1.61-.617 -i 1.90=.99-1.9i
Ok, you can work it with more accuracy to get it.