Question

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

Answer 1

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.
(5^.25)=1.49534878

that angle= = 296.565051
1/2 that angle= = 148.282525

sqrt(1-2i)=1.4953cos 148.282525+i(1.4953sin 148.282525)
= -1.27-i*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=.99-1.9i

Ok, you can work it with more accuracy to get it.