Question

The square root of $1-i$ is $\sqrt{\frac{\sqrt{2}+1}{2}}-i\sqrt{\frac{\sqrt{2}-1}{2}}$. If this is true enter 1, else enter 0.

Answer 1

$\sqrt{(1-i)}=\pm (x-iy)$ say. .....(1)

Squaring, we get
$1-i=x^2-y^2-i2xy$
$\therefore x^2-y^2=1,2xy=1$ ...(2)
$\therefore {(x^2+y^2)}^2={(x^2-y^2)}^2+4x^2{y}^2=1+1=2$
$\therefore x^2+y^2=\sqrt{2}$ and $x^2-y^2=1$ by (2)

On adding and subtracting the above equations , we get
$\therefore x^2=\frac{\sqrt{2}+1}{2}$ and $y^2=\frac{\sqrt{2}-1}{2}.$

$\therefore x=\pm \sqrt{\frac{\sqrt{2}+1}{2}}$and$y=\pm \sqrt{\frac{\sqrt{2}-1}{2}}$
So, by equation (1),
$\sqrt{1-i}=\sqrt{\frac{\sqrt{2}+1}{2}}-i\sqrt{\frac{\sqrt{2}-1}{2}}$
Ans: 1