Complex number help! (1 Viewer)

Glyde · Oct 16, 2015

InteGrand said:
$You need to add on $\pi$ to the inverse tan answer because the complex number is in the second quadrant. The rule for finding the principal argument $\theta$ of a complex number $z=x+iy$ ($x,y$ real numbers) is \textbf{not} actually just $\theta = \tan ^{-1}\left(\frac{y}{x}\right)$ in general; this rule is correct if and only if $x$ is positive. If $x$ is negative, then instead, you need to compute $ \tan ^{-1}\left(\frac{y}{x}\right)$ and then add or subtract $\pi$, depending on the sign of $y$ (if $y$ is positive, add $\pi$, and if $y$ is negative, subtract $\pi$). If $x$ is 0, then the argument is $\frac{\pi}{2}$ if $y>0$ and $-\frac{\pi}{2}$ if $y<0$. If $x$ and $y$ are both 0, then $z=0$, and the argument is undefined. (All these rules are described by the $\mathrm{atan2}$ function; see the link below. Basically, that is the complete rule for finding the (principal) argument of a general complex number $x+iy$ )$

https://en.wikipedia.org/wiki/Atan2#Definition_and_computation

$It is always true that $\tan (\mathrm{arg}(z)) = \frac{y}{x}$ (provided $x\neq 0$), but due to the fact that inverse tan is between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, we need to make adjustments to get the argument (to get to the right quadrant) and can't just say $\theta = \tan ^{-1}\left(\frac{y}{x}\right)$ in general, so check the $\mathrm{atan2}$ page.$

Omg I had no idea about that !! Thanks !!

Glyde · Oct 16, 2015

So in visual terms we are adding or minusing pi to find the angle of the modulis that is closest to zero degrees ?

Glyde · Oct 16, 2015

InteGrand said:
$So in general, to find the (principal) argument $\theta$ of a complex number $z=x+iy$, where $x$ and $y$ are real numbers and $x\neq 0$, you will \textsl{always} need to calculate $\tan^{-1} \left(\frac{y}{x}\right)$; if $x$ is positive, you stop there, but if $x$ is \textsl{negative}, then you need to add or subtract $\pi$; add $\pi$ if $y$ is positive, and subtract $\pi$ if $y$ is negative. If $x=0$, it is easy to find the argument by inspection.$

Haha you have awesome teaching skills !! You could imagine how stressful these questions have been for me not knowing this rule

InteGrand · Oct 16, 2015

Glyde said:
So in visual terms we are adding or minusing pi to find the angle of the modulis that is closest to zero degrees ?

$No, not closest to zero degrees. Basically, we always have $\tan \theta = \frac{y}{x}$ (assuming $x\neq 0$), where $\theta$ is the principal argument of $z=x+iy$. So we are solving the equation $\tan \theta = \frac{y}{x}$ with $-\pi < \theta \leq \pi$. Solving this type of equation is something you've probably done a lot of before, with quadrants diagrams etc. Basically, $\tan^{-1}$ gets us an angle that's in quadrant 1 or 4, because the inverse tan function is between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$. So $\tan^{-1}\left(\frac{y}{x}\right)$ will give us the right angle (argument) provided that the complex number is in quadrant 1 or 4 (which happens precisely when $x>0$, as these quadrants are where $x>0$ on the $x$-axis).$

$If $x$ is negative instead, we know the angle should be in quadrants $2$ or $3$, since this is where $x<0$ on the $x$-axis. If $y$ is positive, we are in quadrant 2, and $\frac{y}{x}$ is negative, as $y$ and $x$ have opposite signs; this means $\tan^{-1}\left(\frac{y}{x}\right)$ gets us into the fourth quadrant (since inverse tan of a negative number is between 0 and $-\frac{\pi}{2}$), so we need to add $\pi$ to this to get to quadrant 2. A similar reasoning is used for the other case.$

Glyde · Oct 17, 2015

Could someone show me the most simplified answer for z1-z2??

InteGrand · Oct 17, 2015

Glyde said:
Could someone show me the most simplified answer for z1-z2??

$$z_1 = 1+\sqrt{3}i$, $z_2 = \sqrt{3}+i$. Hence $z_1 - z_2 = \left( 1-\sqrt{3}\right) + \left(\sqrt{3}-1 \right)i$ (Cartesian form).$

Complex number help! (1 Viewer)

Glyde

Member

Glyde

Member

Glyde

Member

InteGrand

Well-Known Member

Glyde

Member

InteGrand

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)