Yesterday, i encountered a question:
$z=a+bi$
$Arg(z-\overline z + 4) = {4\pi \over 3}$
$b=?$
I solved the question using basic method: $$\overline z = a-bi$$ $$ w = z - \overline z + 4$$ $$ w = a+bi-(a-bi)+4 $$ $$ w = 4 + 2bi $$ $$ Arg(w) = {4\pi \over 3}$$ $$ \frac{Im(w)}{Re(w)} = \tan {4\pi \over 3} $$ $$ {2bi \over 4} = \sqrt 3 $$ $$ b = 2 \sqrt 3 $$ But I think there is a problem here. $4\pi\over 3$ is on the third region, so a complex number in 3rd region have both Real an Imaginary parts negative. So I'm thinking $Re(w), Im(w)<0$ The teacher said $Arg(z)$ and $Arg(z-z_0)$ are different things, but I really can't figure out that.
Am I right or the question is true?