How would you find the argument of the following number. $$\cos{2}-i\sin{2}$$

I'm aware that complex numbers in the form $r(\cos{\theta}+i\sin{\theta})$ have an argument of $\theta$, but what do you do with the $-$ sign?


Regardless of method, you will demonstrate some knowledge of Euler's identity and/or trigonometric identities.

Method I: Arctangent of slope

The argument is the arctangent of the ratio of the imaginary component to the real component, accounting for quadrant. If you forget to account for quadrant, this gives arguments in the interval $(-\pi/2, \pi/2)$, "half" of which are wrong. (Recall that arctangent is not a single-valued inverse. This is obvious when one remembers that tangent is periodic.)

Since $\cos 2 < 0$ and $-\sin 2 < 0$, this is in quadrant III.
\begin{align*} \arctan \frac{-\sin 2}{\cos 2} &= \arctan(-\tan 2)) \\ &= - \arctan \tan 2 &&\text{$\arctan$ is an odd function}\\ &= - \arctan \tan (2 - \pi) &&\text{$\tan$ has period $\pi$}\\ &= -(2 - \pi +\pi k), k \in \mathbb{Z} &&-\pi/2 < 2 - \pi < \pi/2 \text{.} \end{align*} Choosing $k=1$, this lands in quadrant III, giving the argument $-2$. If we had just mechanically evaluated, say with a calculator, \begin{align*} \arctan \frac{-\sin 2}{\cos 2} &= \arctan(2.18504\dots) \\ &= 1.14159\dots{} + \pi k, k \in \mathbb{Z} \text{.} \end{align*} The subset of these in quadrant III have $k$ odd. Picking $k = -1$, we get $-2$.

Of all the $k$ that give a result in the correct quadrant, which do you pick? You pick the one that conforms to your conventions for the range of value of an argument, if you have such a convention. If you do not, pick the one that makes your subsequent steps easier (or don't pick and leave the result as an equivalence class $\mod 2 \pi$).

Method II: Conjugation

If your only problem is the wrong sign of imaginary component, use conjugation. \begin{align*} \arg(\cos 2 - \mathrm{i} \sin 2) &= \arg(\overline{\cos 2 + \mathrm{i} \sin 2}) \\ &= \arg(\overline{\mathrm{e}^{\mathrm{i} 2}}) \\ &= -\arg(\mathrm{e}^{\mathrm{i} 2}) \\ &= -2 \text{.} \end{align*}

Method III: Even-odd identities

Sine is odd. Cosine is even. This is expressed in the even-odd identities. So $\cos(-2) = \cos(2)$ and $\sin(-2) = -\sin(2)$. Consequently, \begin{align*} \cos 2 - \mathrm{i} \sin 2 &= \cos -2 + \mathrm{i} \sin -2 \\ &= \mathrm{e}^{\mathrm{i}(-2)} \text{,} \end{align*} having argument $-2$.

Brahadeesh's and MrYouMath's answers use this method, without identifying what was done. Michael Rozenberg's answer combines this with the periodicity (by $2\pi$) identity, again without identifying what was done.

  • $\begingroup$ Thank you very much. This has really cleared things up for me! $\endgroup$ – Ewan Miller Sep 9 '17 at 20:53

You can write this complex number as $\cos(-2) + i \sin(-2)$. Now you can see that the argument is $-2$.


$z=r(\sin\theta+i\sin\theta)$, where $r\geq0$ and $\theta=\arg{z}\in[0,2\pi)$.


Thus, $\arg(\cos2-i\sin2)=2\pi-2$.

  • $\begingroup$ Thank you. Am I right in thinking then that there is a general rule that r(cosA-isinA)=r(cos(2π-A)+isin(2π-A))? $\endgroup$ – Ewan Miller Sep 9 '17 at 16:05
  • $\begingroup$ @Ewan Miller It depends on definition. Sometimes $\arg{z}\in[0,2\pi)$ and sometimes $\arg{z}\in[-\pi,\pi)$. You need to ask your teacher. I like the first versa. $\endgroup$ – Michael Rozenberg Sep 9 '17 at 16:13
  • $\begingroup$ In the case where $\arg{z}\in[-\pi,\pi)$, would it be r(cosA-isinA)=r(cos(-A)+isini(-A))? $\endgroup$ – Ewan Miller Sep 9 '17 at 16:17
  • $\begingroup$ Also, I haven't got a teacher as I have just left school and am teaching this to myself in my gap year $\endgroup$ – Ewan Miller Sep 9 '17 at 16:19
  • $\begingroup$ @Ewan Miller It's not so easy. If $A\in(-\pi,\pi)$ then you are right:$r(\cos(-A)+i\sin(-A))$ , but for $A=-\pi$ we get $r(\cos{A}+i\sin{A})$. All this depend on the level of your teacher. Maybe he don't see here problems. $\endgroup$ – Michael Rozenberg Sep 9 '17 at 16:23

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