# magnitude and argument

• Jul 15th 2008, 06:53 AM
cruxkitty
magnitude and argument
I dont understand what this problem wants so can someone please help explain it... sqrt means square root of ...
what is the magnitude and argument of (-1+sqrt3i)^2
• Jul 15th 2008, 07:38 AM
earboth
Quote:

Originally Posted by cruxkitty
I dont understand what this problem wants so can someone please help explain it... sqrt means square root of ...
what is the magnitude and argument of (-1+sqrt3i)^2

As far as I understand the problem you are asked to simplify the term. Doing this you are supposed to know that

$\sqrt{i} = \frac12 \cdot \sqrt{2} + \frac12 \cdot \sqrt{2} \cdot i$ and

$\sqrt{3i} = \sqrt{3} \cdot \sqrt{i}$

Plug in these terms and calculate the square. Collect like terms.
• Jul 15th 2008, 08:26 AM
Isomorphism
Quote:

Originally Posted by cruxkitty
I dont understand what this problem wants so can someone please help explain it... sqrt means square root of ...
what is the magnitude and argument of (-1+sqrt3i)^2

I assume you mean $(-1+i\sqrt{3})^2 =1 + (\sqrt{3}i)^2 - 2i\sqrt{3} = -2 - 2i\sqrt{3}$

MAGNITUDE:

Magnitude of a complex number $x+iy$ is denoted by $|x+iy|$ and is defined as $\sqrt{x^2 + y^2}$

Geometrically, that is, on the Argand plane, this denotes the distance of the complex number from the origin.

$|(-1+i\sqrt{3})^2| = |-2 - 2i\sqrt{3}| = \sqrt{(-2)^2 + (2\sqrt{3})^2} = \sqrt{4+12} = \sqrt{16} = 4$

ARGUMENT:

Argument of a complex number $x+iy$ is denoted by $\text{arg}(x+iy)$ and is defined as

$\arg(x+iy)= \begin{matrix} \phi\cdot \text{sgn}(y) & x > 0 \\
\frac{\pi}{2}\cdot \text{sgn}(y) & x = 0 \\
(\pi - \phi)\cdot \text{sgn}(y) & x < 0 \\
\end{matrix}$

Where $\phi = \tan^{-1} \left|\frac{y}{x}\right|$ and sgn is the signum function.

Geometrically, that is, on the Argand plane, this denotes the angle made by the line joining the origin and the complex number with the positive direction of the x axis.

$\text{arg}(-2 - 2i\sqrt{3}) = \tan^{-1}\left(\frac{2\sqrt{3}}{2}\right) - \pi =-\frac{2\pi}3$

Aliter:

If you know cube roots of unity and their positions in th argand plane.

Then observe that

$\omega = -\frac12 + \frac{i\sqrt{3}}2$

$2\omega = -1 + i\sqrt{3}$

$(2\omega)^2 = (-1 + i\sqrt{3})^2$

Thus the complex number is $4\omega^2$.

But we know that the cube roots of unity are on the unit circle and that they are distributed over the unit circle at $120^{\circ}$ to each other. 1 is on the real line. $\omega$ is at an angle $120^{\circ}$ to the x-axis. $\omega^2$ is at an angle $-120^{\circ}$ to the x-axis.

Mathematically they mean $|\omega^2| = 1, \arg{\omega^2} = -\frac{2\pi}{3}$

Lets use these facts:

$(-1 + i\sqrt{3})^2 = 4\omega^2$

$|4\omega^2| = |4||\omega^2| = 4$

$\arg(4\omega^2) = \arg(4) + \arg(\omega^2) = 0 -\frac{2\pi}{3} = -\frac{2\pi}{3}$
• Jul 15th 2008, 08:40 AM
Plato
Quote:

Originally Posted by Isomorphism
$(-1+i\sqrt{3})^2 =\color{red}-2 - 2i\sqrt{3}$

Note the correction above.

I am sure that the question deals with this number: $z = - 1 + \sqrt 3 i$.
“Multitude” must mean absolute value (distance from origin): $\left| {a + bi} \right| = \sqrt {a^2 + b^2 }$.
The argument is the angle the ray from the origin through the number makes with the positive real axis.
In doing this problem here are two useful facts: $\left| {z^2 } \right| = \left| z \right|^2 \,\& \,\arg (z^2 ) = 2\arg (z)$.