modulus and argument

• Jun 14th 2009, 06:43 AM
z1llch
modulus and argument
find the modulus and arguments of the complex number $(1+\sqrt 3i)^3$.

my solution;
$(1+\sqrt 3i)(1+\sqrt 3i)(1+\sqrt 3i)$
$= 1+2\sqrt3i+3i^2(1+\sqrt 3i)$
$= -2+2\sqrt3i(1+\sqrt 3i)$
$= -2-2\sqrt3i+2\sqrt3i+2(3)i^2$
$= -2+6i^2$
$= -8$

is there anything wrong with my working above?
to find modulus and arguments of the complex number i need to express it in a+bi form first right? or can i assume b=0? being -8+0i
then |-8+0i| = 8
and arg |-8+0i| = 0
• Jun 14th 2009, 06:51 AM
yeongil
Quote:

Originally Posted by z1llch
find the modulus and arguments of the complex number $(1+\sqrt 3i)^3$.

my solution;
$(1+\sqrt 3i)(1+\sqrt 3i)(1+\sqrt 3i)$
$= {\color{red}(}1+2\sqrt3i+3i^2{\color{red})}(1+\sqr t 3i)$
$= {\color{red}(}-2+2\sqrt3i{\color{red})}(1+\sqrt 3i)$
$= -2-2\sqrt3i+2\sqrt3i+2(3)i^2$
$= -2+6i^2$
$= -8$

is there anything wrong with my working above?

Well technically, you need the parentheses I added above.

Quote:

to find modulus and arguments of the complex number i need to express it in a+bi form first right? or can i assume b=0? being -8+0i
then |-8+0i| = 8
and arg |-8+0i| = 0
If you look at the trig form of the complex number: $r(\cos \theta + i\sin \theta)$, r is the modulus and theta is the argument. Anyway, your modulus is right, but the argument is not. The argument should be pi.

01
• Jun 14th 2009, 06:53 AM
Quote:

Originally Posted by z1llch
find the modulus and arguments of the complex number $(1+\sqrt 3i)^3$.

my solution;
$(1+\sqrt 3i)(1+\sqrt 3i)(1+\sqrt 3i)$
$= 1+2\sqrt3i+3i^2(1+\sqrt 3i)$
$= -2+2\sqrt3i(1+\sqrt 3i)$
$= -2-2\sqrt3i+2\sqrt3i+2(3)i^2$
$= -2+6i^2$
$= -8$

is there anything wrong with my working above?
to find modulus and arguments of the complex number i need to express it in a+bi form first right? or can i assume b=0? being -8+0i
then |-8+0i| = 8
and arg |-8+0i| = 0

• Jun 14th 2009, 07:04 AM
z1llch
i do not understand the argument part.. can please explain on it a little bit?
• Jun 14th 2009, 07:22 AM
dhiab
Quote:

Originally Posted by z1llch
find the modulus and arguments of the complex number $(1+\sqrt 3i)^3$.

my solution;
$(1+\sqrt 3i)(1+\sqrt 3i)(1+\sqrt 3i)$
$= 1+2\sqrt3i+3i^2(1+\sqrt 3i)$
$= -2+2\sqrt3i(1+\sqrt 3i)$
$= -2-2\sqrt3i+2\sqrt3i+2(3)i^2$
$= -2+6i^2$
$= -8$

is there anything wrong with my working above?
to find modulus and arguments of the complex number i need to express it in a+bi form first right? or can i assume b=0? being -8+0i
then |-8+0i| = 8
and arg |-8+0i| = 0

ANATHOR RESOLUTION :
http://www.mathramz.com/xyz/latexren...426c615543.png
THANKS
• Jun 14th 2009, 07:57 AM
Plato
Here are some simple facts the would have made this an easy problem.
$\begin{gathered}
\left| {z^n } \right| = \left| z \right|^n \; \Rightarrow \;\left| {\left( {1 + \sqrt 3 i} \right)^3 } \right| = \left| {1 + \sqrt 3 i} \right|^3 \hfill \\
\arg \left( {z^n } \right) = n\arg (z) \hfill \\
\end{gathered}$
• Jun 14th 2009, 02:16 PM
yeongil
Quote:

Originally Posted by z1llch
i do not understand the argument part.. can please explain on it a little bit?

Have you learned about the trig form of the complex number? Plotting the number -8 + 0i on the complex plane is like plotting the point (-8, 0) on the Cartesian plane. You got an angle in standard position (where the initial side is the positive x-axis and the terminal side is a ray where the point is on), and this angle is pi radians. So the argument is pi. We've already figured that the modulus is 8, so
$-8 + 0i = 8(\cos \pi + i\sin \pi)$.

01