# What is the correct solution of $\arg{z}\leq\frac{\pi}{4}$?

Find the solution of the following inequality. $$\arg{z}\leq\frac{\pi}{4}$$

My first answer: Any number $$z=x+iy$$ in $$y\leq x$$, $$y\geq 0$$ and $$x\not=0$$ satisfies $$\arg{z}\leq\frac{\pi}{4}$$.

I am confused in determining whether or not we are allowed to subtract an integer multiple of $$2\pi$$ from an angle when comparing two angles. If it is allowed then for example, $$z=i$$ can also be the solution because $$\frac{\pi}{2} - 2\pi =-\frac{3\pi}{2}<\frac{\pi}{4}$$. Thus the solution is all complex number excluding $$z=0$$.

### Question

What is the correct solution of $$\arg{z}\leq\frac{\pi}{4}$$?

The "correct" answer to this question depends on the definition of $$\arg$$. If the convention is that it is single valued with an answer in $$[0,2\pi)$$ then your first answer is correct. If it's multivalued as you describe it, then the second answer is correct.
• Another convention is $(-\pi,\pi]$. Feb 27 at 15:39