# Math Help - What is the magnitude & argument of...

1. ## What is the magnitude & argument of...

What is the magnitude & argument of (-1 + sqr(3i))^2?

2. Can you use DeMoivre? It's only a square, so it's not necessary, but it may be easier.

3. Originally Posted by katchat64
What is the magnitude & argument of (-1 + sqr(3i))^2?
$z=-1+\sqrt{3} i$

The magnitude is this complex number's length...

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

The argument is the angle made with the positive Real axis in the anticlockwise direction... Since z is in the second quadrant, it's evaluated by $\pi -\arctan (\frac{y}{x})$

$\arg{z}= \pi - \arctan (\frac{1}{\sqrt{3}}) = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$

So we can write z as

$z=2 cis (\frac{5\pi}{6})$.

Then by using De Moivre's Theorem...

$z^2 = 2^2 cis (2 \times \frac{5\pi}{6}) = 4 cis (\frac{5\pi}{3})$

So $|z^2|= 4$ and $\arg z = \frac{5\pi}{3}$

4. Close. $\pi\;-\;\arctan(\sqrt{3})\;=\;\frac{2\pi}{3}$

5. Originally Posted by TKHunny
Close. $\pi\;-\;\arctan(\sqrt{3})\;=\;\frac{2\pi}{3}$
Oops, typo, I did $\frac{x}{y}$ instead of $\frac{y}{x}$. Thanks.