1. ## Modulus and Argument

Find modulus and argument of: $- 2 - \sqrt {12} i$

$\left| z \right| = \sqrt {( - 2)^2 + ( - \sqrt {12} )^2 } = 4$

$
\tan (Argz) = \frac{y}
{x}
$

$Argz = \tan ^{ - 1} \frac{y}
{x} = \tan ^{ - 1} \frac{{ - \sqrt {12} }}
{{ - 2}} = \tan ^{ - 1} \sqrt 3 = \frac{\pi }
{6}
$

The correct answer is: $- \frac{{2\pi }}
{3}$

Could someone explain to me why I have to subtract $\pi$ from my final answer.

Thanks

2. Hello,
Originally Posted by r_maths
Find modulus and argument of: $- 2 - \sqrt {12} i$

$\left| z \right| = \sqrt {( - 2)^2 + ( - \sqrt {12} )^2 } = 4$

$
\tan (Argz) = \frac{y}
{x}
$

$Argz = \tan ^{ - 1} \frac{y}
{x} = \tan ^{ - 1} \frac{{ - \sqrt {12} }}
{{ - 2}} = \tan ^{ - 1} \sqrt 3 = \frac{\pi }
{6}
$

The correct answer is: $- \frac{{2\pi }}
{3}$

Could someone explain to me why I have to subtract $\pi$ from my final answer.

Thanks
I don't like using the arctan function. It makes problems because of the restrictions of the domain ><

$z=4 \left(-\frac 12- i \frac{\sqrt{3}}{2}\right)=4(\cos \theta+i \sin \theta)$
Note that the cosine and the sine are both negative. So $\theta$ will be in the third quadrant. That is $\theta$ lies between $[3 \pi/2;2 \pi] +2k \pi$

With the specific trig values you should know, $\theta=-\frac{2 \pi}{3} +2k \pi$

You also could've written :
$z=-(2+\sqrt{12} i)=e^{i \pi} \left(4 \left(\frac 12+ \frac{\sqrt{12}}{2} i \right)\right)$

3. Hello, r_maths!

Find modulus and argument of: $-2 -\sqrt{12}i$

$\tan (\text{Arg}z) \:= \:\frac{y}{x}$

$\text{Arg}z \:= \:\tan^{-1}\left(\frac{-\sqrt{12}}{-2}\right) \:=\: \tan^{-1}\sqrt{3} \:=\: \frac{\pi }{6}$ . . . . no, you mean: ${\color{blue}\frac{\pi}{3}}$

The correct answer is: . $-\frac{2\pi}{3}$ . . . . or ${\color{blue}\frac{4\pi}{3}}$

Could someone explain to me why I have to subtract $\pi$ from my final answer.

Plot the complex number on the Argand diagram.
Code:
                |
|
-2  |
- - + - - + - - - -
|    /|
__ |   / |
-√12 |  /  |
| /   |
|/    |
*     |
|

The argument is an angle in Quadrant 3.