Confusion in least argument of z

Question

$a$ and $b$ are ordered pairs of a point on the Argand plane. For tan $\theta = b/a$. It’s $z = a+ib$

My sir taught me that unique value of theta = $\tan ^{-1}|b/a|$ where theta belongs to $0,\pi/2$. Since then only it is present in the first quadrant. And tan inverse function has ranged from $-\pi/2$ to $\pi/2$.

Then , for the least + argument of $z$:

unique value of theta = $\tan ^{-1}|b/a|$ such that $0<\theta<2\pi$.

Why is the theta different for least + argument of z and unique value of theta?

Q Q Find least + argument for z = -1-i$\sqrt{3}$ .

Here , mod of z = 2. So ,angle theta = 60 degree

But least + argument is in -pi/3 which we write as -pi+3 + 2pi = 5pi/3.

Here theta is less than 60 but isn’t it a coincidence.

What are $a$, $b$ and $\theta$ here? How do they relate to $z$? Please provide the necessary context. — Christoph, Jan 24 '21 at 10:51
@Christoph it is just that when you solve a question , theta has only one value . That’s why unique value — Srijan, Jan 24 '21 at 10:57
The two statements are actually the same, because the smallest $\theta$ in $0 <\theta<2\pi$ is in the first quadrant. However they don't match tan$\theta=b/a$ unless $b/a$ is positive. — Peter, Jan 24 '21 at 11:04
My comment was not quite right, but I ran out of time to fix it. The question hasalso been edited. — Peter, Jan 24 '21 at 11:20

score 0 · Accepted Answer · answered Jan 24 '21 at 12:03

On the Argand plane a complex number is given by $z=a+ib$, and also by $z=r\ cis\theta$. The two expressions are connected by $a=r\ cos\theta$ and $b=r\ sin\theta$. We also have $r=\sqrt{a^2+b^2}$ and $tan\theta=b/a$, but this last expression does not give us enough information to find $\theta$.

One way to find $\theta$ is to find $\alpha=tan^{-1}|b/a|$ and then use the quadrant to decide which of the possible values for $\theta$ is most appropriate. The value for $\theta$ in $0\le\theta\lt2\pi $ (the least positive value) will be one of $\alpha,\ $ $\pi-\alpha,\ $ $\pi+\alpha,\ $ $2\pi-\alpha.\ $ For your example we have $a=-1$ and $b=-\sqrt{3}$, so we find $\alpha=tan^{-1}\sqrt{3}=\pi/3$ ($60$ degrees). The point $z$ is in the 3rd quadrant, so $\theta=\pi+\alpha=\pi+\sqrt{3}=4\pi/3$.

You have found $\theta=5\pi/3$. This is a 4th quadrant angle, and would be correct for $z=1-i\sqrt{3}$.

@user102532, cis theta is correct. It is short for cos + i sin theta. Note $cis\theta=e^{i\theta}$ — Peter, Jan 24 '21 at 12:52

Confusion in least argument of z

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