What is wrong in Method $3$? 
Let $z,w$ be complex numbers such that $\bar z+i\bar w=0$ and $\arg(zw)=\pi$, find $\arg(z)$

Method $1:$
$\bar z=-i\bar w\implies z=iw$
Also, $\arg z+\arg w=\pi\implies\arg z+\arg\frac zi=\pi$
Therefore, $2\arg z-\frac\pi2=\pi\implies\arg z=\frac{3\pi}4 $
Method $2:$
$zw=-k, k\gt0$
Also, $w=\frac zi$
Therefore, $z^2=-ik$
It means, $\arg z^2=\frac{3\pi}2$
Therefore, $\arg z=\frac{3\pi}4$
Method $3:$
Let $z=x+iy, w=a+ib$
Given, $\bar z+i\bar w=0\implies x-iy+i(a-ib)=0$
Therefore, $x=-b, y=a$
Also, $\arg(zw)=\pi\implies\arg(x+iy)(a+ib)=\pi$
It means, $\arg (ax-by+i(bx+ay))=\pi$
Therefore, $\tan\pi=\frac{bx+ay}{ax-by}\implies ay=-bx$
Putting $b=-x, a=y,$ I get, $y^2=x^2$
Therefore, $y=x$ or $y=-x$
$y=x$ gives $\frac{\pi}4$. Is this incorrect? If yes, how to eliminate this?

(After reading the comments)
I got $\arg (ax-by+i(bx+ay))=\pi$
Putting $b=-x, a=y,$ I get, $$\arg(2xy+i(y^2-x^2))=\pi$$
Making $x^2=y^2$ could mean the argument is zero or $\pi$.
For it to be necessarily $\pi$, $2xy$ should be negative i.e. $x,y$ are of opposite sign.
Therefore, the answer is $\frac{3\pi}4$ and not $\frac\pi4$
