# How to correctly get phase of a complex number

If $$z = -99.75 + 3i$$ is a complex number.

How can I calculate the Argument of $$z$$?

I tried $$\arctan\left(\frac{3}{99.75}\right) = 1.72°$$ but if I try to compute it on wolfram the result is 178.712° (that should be correct) how do I get here?

• The argument of a complex number, $\theta$, satisfies the two relations$$|z|\cos{(\theta)}=\Re{(z)}\quad|z|\sin{(\theta)}=\Im{(z)}$$which implies that $\tan{(\theta)}=\Im{(z)}/\Re{(z)}$ but isn't equivalent to this statement. Sep 1, 2020 at 18:52
• See this Wikipedia entry: atan2 Sep 1, 2020 at 19:07
• @stevengregory but wikipedai says to do arctan(y/x) + pi = 1.72+180 = 181.72 that is not the correct result Sep 1, 2020 at 19:09
• Not quite: the arctangent is negative.
– J.G.
Sep 1, 2020 at 19:10
• Of course, the complex number $-2+3\iota$ corresponds to the point $(-2,3)$ in the $x-y$ plane, but $2+3\iota$ corresponds to a different point $(2,3)$. Sep 1, 2020 at 19:16

$$z=-99.75+3\iota$$ is the point $$P(-99.75,3)$$ in the $$x-y$$ plane, which is in the $$2^{nd}$$ quadrant, so the angle $$AP$$ ($$A$$ is the origin) is making with the positive $$x-$$axis (which is defined as $$\arg(z)$$), is $$90^\circ+\angle CAP=90^\circ+\arctan\left(\dfrac{CP}{CA}\right)=90^\circ+\arctan\left(\dfrac{99.75}3\right)\approx 178.28^\circ$$ $$\arg(z)$$ is usually defined modulo $$180^\circ$$, i.e. depending on the location of $$z$$, the angle the vector $$\vec{Az}$$ (formed by joining the origin to the point $$z$$) makes with the positive $$x-$$axis, measured in the counter-clockwise direction (positive angle) or clockwise direction (negative angle), whichever gives the least absolute value.
• $\arg(z)\in(-\pi,\pi]$ is defined as the angle made by the vector $\vec{Oz}$, where $O$ is the origin, with the positive $x-$axis. In this case, $\arg(z)$ is a positive angle (measured counterclockwise) that is $<180^\circ$. We want to measure that angle. After that it's trigonometry in the right-angled $\triangle CAP$. This is a safe process to arrive at the right answer because the range of $\arctan(x)$ is $\left[-\frac{\pi}2,\frac{\pi}2\right]$, but the range of $\arg{z},z\in\Bbb{C}$ is $(-\pi,\pi]$. The link in the comment by @steven gregory up in your post will help you too. Sep 1, 2020 at 19:09
• $$\arg{z}\in(-\pi,0)\implies z\in\text{ lower quadrants} \\ \arg{z}\in(0,\pi)\implies z\in\text{ upper quadrants} \\ \arg{z}=0\implies z\in\text{ positive x-axis}\\ \arg{z}=\pi \implies z\in\text{ negative x-axis}$$ so all possible positions of $z$ are considered. Sep 1, 2020 at 19:14