How does $\arctan(0)$ equal $\pi?$
I was recently doing a problem, and apparently got it wrong.
The question asked: what is the polar form of the sum: $(5+3i)$ + $(-10-3i)$.
From my understanding, the sum would equal: $-5+0i$
Now,
the magnitude would equal: $\sqrt{-5^2 + 0^2} = 5$
And,
the angle would equal: $\tan^{-1}(\frac{0}{-5}) = 0$
Instead, the answer is actually $5 (\pi)$, instead of $5 (0)$.
I check on this website: 2pif.com (under the "Complex Numbers" tool), and it also says that 5 (0) is the answer.
So I'm wondering, how in the world does $\tan^{-1}(\frac{0}{-5}) = \pi$
$\endgroup$ 46 Answers
$\begingroup$This is a common error among new people into the complex numbers. The argument of a complex number $z = a + ib$ is not $\arctan (b/a).$ It is given by $$\arg z = \begin{cases} \arctan (b/a) &\mbox{if } a>0 &\\ \arctan (b/a)+\pi &\mbox{if } a<0,b\geq 0 \\ \arctan (b/a)-\pi &\mbox{if } a<0,b< 0 \\ \pi/2 &\mbox{if } a=0,b> 0 \\ -\pi/2 &\mbox{if } a=0,b< 0 \\ \mathrm{undefined} &\mbox{if } a=0,b= 0 \\ \end{cases} $$
$\endgroup$ 3 $\begingroup$It is an oversimplification to say the argument of $x+iy$ is given by $\theta =\arctan \dfrac yx$. Actually you have to solve the system \begin{cases}\cos\theta=\dfrac x{\sqrt{x^2+y^2}}\\ \sin\theta=\dfrac y{\sqrt{x^2+y^2}}\end{cases} which implies $\tan\theta=\dfrac yx$, but is not equivalent to $\theta=\arctan\dfrac yx$.
The general solution is $\theta\equiv\arctan\dfrac yx\mod \pi$, and you have to choose the value of $\theta$ (mod $2\pi$) in function of the signs of $\cos\theta$ and $\sin\theta$, i.e. of $x$ and $y$.
$\endgroup$ $\begingroup$To add to the other answers here, the argument is the angle the corresponding vector on the Complex Plane would make with that of $1+0i$
In this particular case, the angle is $180^{\circ}$ or $\pi$ radians as in the picture.
$\endgroup$ 1 $\begingroup$$\tan(x)=0$ has the solutions $x=n\pi$, so $\pi$ is also a solution. Which solution you choose, depends on the signs of $\sin(x)$ and $\cos(x)$. This is why computers have a special function atan2, which calculates the solutions of $\tan(x)=\frac{u}{v}$ and returns the value in the proper quadrant, by accounting for the signs of $u$ and $v$ (as well as for the values where $\tan(x)=\frac{u}{v}$ might be ambiguous/undefined).
$\endgroup$ $\begingroup$because the angle of a complex number is not given by the ordinary artan, is necessary make an adjust of Pi if the real part is negative, add pi and ist done
$\endgroup$ $\begingroup$$\arctan(0)=0,$ not $\pi$, but $\tan(\pi)=\tan(0)=0$ As $x$ ranges from $-\frac \pi 2$ to $\frac \pi 2$, $\tan x$ ranges from $-\infty$ to $+\infty$. As we want $\arctan$ to be a function, we have to choose a range of output values. When you use the $\arctan$ to get the polar angle, there is an uncertainty of $\pi$ because $\tan (x+\pi)=\tan x$. You have to use the real and imaginary parts to decide the multiple of $\pi$ to add to the $\arctan$ output to get the proper polar angle you want. You might even need to add $2\pi$. For example, if you want the polar form of $1-i$, the $\arctan$ will give you $-\frac \pi 4$ but you probably want to report the polar angle as $\frac {7 \pi}4$
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