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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?

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  • $\begingroup$ 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. $\endgroup$ Commented Sep 1, 2020 at 18:52
  • $\begingroup$ See this Wikipedia entry: atan2 $\endgroup$ Commented Sep 1, 2020 at 19:07
  • $\begingroup$ @stevengregory but wikipedai says to do arctan(y/x) + pi = 1.72+180 = 181.72 that is not the correct result $\endgroup$
    – Marià
    Commented Sep 1, 2020 at 19:09
  • $\begingroup$ Not quite: the arctangent is negative. $\endgroup$
    – J.G.
    Commented Sep 1, 2020 at 19:10
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    $\begingroup$ 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 $(2,3)$. $\endgroup$ Commented Sep 1, 2020 at 19:16

1 Answer 1

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$z=-99.75+3\iota$ 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 $$90^\circ+\angle CAP=90^\circ+\arctan\left(\dfrac{CP}{CA}\right)=90^\circ+\arctan\left(\dfrac{99.75}3\right)\approx 178.28^\circ$$

enter image description here

$\arg(z)$ 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.

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  • $\begingroup$ Why are we doin 90+arctan ? $\endgroup$
    – Marià
    Commented Sep 1, 2020 at 19:01
  • $\begingroup$ $\arg(z)\in(-\pi,\pi]$ 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. $\endgroup$ Commented Sep 1, 2020 at 19:09
  • $\begingroup$ But when we said is ∈(−π,π) it means that is only defined in the lower quadrants ? $\endgroup$
    – Marià
    Commented Sep 1, 2020 at 19:11
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    $\begingroup$ $$\arg{z}\in(-\pi,0)\implies z\in\text{ lower quadrants} \\ \arg{z}\in(0,\pi)\implies z\in\text{ upper quadrants} \\ \arg{z}=0\implies z\in\text{ positive x-axis}\\ \arg{z}=\pi \implies z\in\text{ negative x-axis} $$ so all possible positions of $z$ are considered. $\endgroup$ Commented Sep 1, 2020 at 19:14

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