Find the set of Arg(z)

Question

Find the set Arg(z) of z such that $z=x+yi$ is on the curve $xy=1$.

So i know $-π<Arg(z)\leq π$ and $z=\frac{1}{y}+yi$ but dont know how to continue.

Answer 1

Hint:$Arg(z) = \arctan(y/x) = \arctan(y^2)$ where $y$ is an arbitrary real number.

Answer 2

Think of the function $f(x) = \frac1x $. You know how to plot it, getting one hyperbole in the first quadrant and one in the third quadrant. Any point on the graph of $f $ can be thought of as a complex number of real part $x $ and imaginary part $\frac1x $. Thinking this way, the complex numbers on the first hyperbole have argument between $0$ and $\frac\pi2$ while the complex numbers on the second hyperbole (the one in the 3rd quadrant) have argument between $-\pi $ and $-\frac\pi2$.

Answer 3

The simple way to do it is to write $z$ in polar form, $z=re^{i\phi}$, where $r=|z|$, and $\phi=Arg(z)$. You can now write $x=r\cos\phi$ and $y=r\sin\phi$. You know $xy=1$ so $r^2\sin\phi\cos\phi=1$. Depending on the value of $\phi$, the equation $$r^2=\frac{1}{\sin\phi\cos\phi}$$ sometimes has a real solution for $r$. It should be easy for you to find when $\sin\phi\cos\phi>0$