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

**Viral** · October 11th 2009, 04:15 PM

What is the principal argument of the complex number $-1+i\sqrt{3}$ ?

My attempt:

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

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

**TheEmptySet** · October 11th 2009, 04:46 PM

Originally Posted by Viral

What is the principal argument of the complex number $-1+i\sqrt{3}$ ?

My attempt:

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

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

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

$\frac{4}{3}\pi-2\pi=-\frac{2}{3}\pi$

**Viral** · October 11th 2009, 04:51 PM

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

**TheEmptySet** · October 11th 2009, 05:04 PM

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

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

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

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

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

**Viral** · October 11th 2009, 05:07 PM

Just to make sure, $2\pi$ is one whole circle?

**TheEmptySet** · October 11th 2009, 05:11 PM

Originally Posted by Viral

Just to make sure, $2\pi$ is one whole circle?

yes that is correct.

**Viral** · October 11th 2009, 05:31 PM

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

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

**mr fantastic** · October 11th 2009, 05:35 PM

Originally Posted by Viral

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

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

$r(\cos (-\theta) + i \sin (-\theta)) = r(\cos\theta - \sin\theta)$ . The convention is to use $r(\cos\theta+i \sin\theta)$ .

**Viral** · October 11th 2009, 05:37 PM

Hmm, so the one I stated first is the only one I can use. This webpage has used both of my examples >< .

**Plato** · October 12th 2009, 03:04 AM

If $z = a + bi,\,a \ne 0$
$\arg (z) = \left\{ {\begin{array}{rl} {\arctan \left( {\frac{b} {a}} \right),} & {a > 0} \\ {\arctan \left( {\frac{b} {a}} \right) + \pi ,} & {a < 0\;\& \,b > 0} \\ {\arctan \left( {\frac{b} {a}} \right) - \pi ,} & {a < 0\;\& \,b < 0} \\ \end{array} } \right.$

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