What the Poly Q (1 Viewer)

[3.1415926535897]

It seems so easy.....

Solve for z
2z^2 = 1 + i

and the answers have...

+/-(x + iy) where x = 1/2.sqrt. [sqrt(2) + 1] , y = 1/2.sqrt.[sqrt(2) - 1]

hmmmm....

 

[martin]

trying to solve z^2 = 1/2 + 1/2 i

as is standard (I think) let z=x+iy then z^2 = x^2 - y^2 +2xyi

so x^2-y^2 = 1/2 and 2xy = 1/2. From second eqn x = 1/(4y) then in first eqn:

1/(16y^2) - y^2 = 1/2
so mult by 16y^2
1-16y^4 = 8y^2

so 16*y^4 + 8*y^2 - 1 =0. This is a quadratic in y^2 (you can let u=y^2 if you want)
so y^2 = (-8 ± sqrt(64+64))/(2*16)
= (-8 ± 8sqrt(2))/32
= (-1/4 ± 1/4*sqrt(2))
y^2 is positive so only take positive value so
y^2 = 1/4*(sqrt(2)-1)
y= ± 1/2 sqrt(sqrt(2)-1)

This is the y you want. Now x=1/(4y) and you'll have to rationalise the donominator (twice I think) to get to the x value given.

 

[3.1415926535897]

thanks matey