# Principal Argument of the Complex Number -1 + i sqrt(3)

• Oct 11th 2009, 04:15 PM
Viral
Principal Argument of the Complex Number -1 + i sqrt(3)
What is the principal argument of the complex number $\displaystyle -1+i\sqrt{3}$?

My attempt:

$\displaystyle \begin{array}{rcrcrc} tan\alpha=\frac{\sqrt{3}}{-1}\\ \alpha=\tan^{-1}(\frac{\sqrt{3}}{-1})\\ =\frac{-1}{3}\pi\\ arg(z)=\pi-\alpha\\ =\frac{4}{3}\pi \end{array}$

However, I know this is wrong because it does not fit the inequality $\displaystyle -\pi<\theta\leq\pi$.
• Oct 11th 2009, 04:46 PM
TheEmptySet
Quote:

Originally Posted by Viral
What is the principal argument of the complex number $\displaystyle -1+i\sqrt{3}$?

My attempt:

$\displaystyle \begin{array}{rcrcrc} tan\alpha=\frac{\sqrt{3}}{-1}\\ \alpha=\tan^{-1}(\frac{\sqrt{3}}{-1})\\ =\frac{-1}{3}\pi\\ arg(z)=\pi-\alpha\\ =\frac{4}{3}\pi \end{array}$

However, I know this is wrong because it does not fit the inequality $\displaystyle -\pi<\theta\leq\pi$.

remember you can add(or subtact) multiples of $\displaystyle 2\pi$ to find a co-terminal angle that falls where you want it

$\displaystyle \frac{4}{3}\pi-2\pi=-\frac{2}{3}\pi$
• Oct 11th 2009, 04:51 PM
Viral
Hmm, I'm not quite sure what you mean. Can you explain it in more details and cite proof?
• Oct 11th 2009, 05:04 PM
TheEmptySet
Quote:

Originally Posted by Viral
Hmm, I'm not quite sure what you mean. Can you explain it in more details and cite proof?

ummm...

Well lets take an example.

Say you have the angle

$\displaystyle \frac{\pi}{4}$ and lets compare that with the angle.

$\displaystyle \frac{9\pi}{4}$.

$\displaystyle \frac{9\pi}{4}=2\pi+\frac{\pi}{4}$.

So to get to the angle $\displaystyle \frac{9\pi}{4}$ I would rotate all the way through $\displaystyle 2\pi$ radians (one around the circle so I am back where I started) then go $\displaystyle \frac{\pi}{4}$ radians.

Moral of the story.

I can go around any multiple of 2\pi radians as many times as I want and still end up in the same place.
• Oct 11th 2009, 05:07 PM
Viral
Just to make sure, $\displaystyle 2\pi$ is one whole circle?
• Oct 11th 2009, 05:11 PM
TheEmptySet
Quote:

Originally Posted by Viral
Just to make sure, $\displaystyle 2\pi$ is one whole circle?

yes that is correct.
• Oct 11th 2009, 05:31 PM
Viral
I don't think this warrants another topic as it's a simple question.

The modulus argument form is $\displaystyle r(cos\theta+isin\theta)$. However, it could also be $\displaystyle r(cos\theta-isin\theta)$. How do I know which to use?
• Oct 11th 2009, 05:35 PM
mr fantastic
Quote:

Originally Posted by Viral
I don't think this warrants another topic as it's a simple question.

The modulus argument form is $\displaystyle r(cos\theta+isin\theta)$. However, it could also be $\displaystyle r(cos\theta-isin\theta)$. How do I know which to use?

$\displaystyle r(\cos (-\theta) + i \sin (-\theta)) = r(\cos\theta - \sin\theta)$. The convention is to use $\displaystyle r(\cos\theta+i \sin\theta)$.
• Oct 11th 2009, 05:37 PM
Viral
Hmm, so the one I stated first is the only one I can use. This webpage has used both of my examples >< .
• Oct 12th 2009, 03:04 AM
Plato
If $\displaystyle z = a + bi,\,a \ne 0$
$\displaystyle \arg (z) = \left\{ {\begin{array}{rl} {\arctan \left( {\frac{b} {a}} \right),} & {a > 0} \\ {\arctan \left( {\frac{b} {a}} \right) + \pi ,} & {a < 0\;\& \,b > 0} \\ {\arctan \left( {\frac{b} {a}} \right) - \pi ,} & {a < 0\;\& \,b < 0} \\ \end{array} } \right.$