# Modulus and Argument

• Dec 15th 2008, 07:02 AM
r_maths
Modulus and Argument
Find modulus and argument of: $\displaystyle - 2 - \sqrt {12} i$

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

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

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

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

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

Thanks
• Dec 15th 2008, 07:34 AM
Moo
Hello,
Quote:

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

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

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

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

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

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

Thanks

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

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

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

You also could've written :
$\displaystyle z=-(2+\sqrt{12} i)=e^{i \pi} \left(4 \left(\frac 12+ \frac{\sqrt{12}}{2} i \right)\right)$
• Dec 15th 2008, 08:08 AM
Soroban
Hello, r_maths!

Quote:

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

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

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

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

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

Plot the complex number on the Argand diagram.
Code:

                |                 |             -2  |       - - + - - + - - - -           |    /|       __ |  / |     -√12 |  /  |           | /  |           |/    |           *    |                 |

The argument is an angle in Quadrant 3.