# How do you find the length and direction of vector -4 - 3i?

Nov 21, 2015

Length $= 5$

Direction $= \left({\tan}^{- 1} \left(\frac{3}{4}\right) - \pi\right)$ rad counterclockwise from the Real axis.

#### Explanation:

Let $z = - 4 - 3 i$. $z$ represents a vector on an Argand diagram.

The magnitude of the vector is the modulus of $z$, which is found using the Pythagoras theorem.

$| z | = \sqrt{{\left(- 4\right)}^{2} + {\left(- 3\right)}^{2}} = 5$

The direction of the vector the principal argument of $z$, which is found using trigonometry.

The basic angle, $\alpha = {\tan}^{- 1} \left(\frac{3}{4}\right)$.

Since $\text{Re} \left(z\right) < 0$ and $\text{Im} \left(z\right) < 0$, the angle lies in the third quadrant.

$\text{arg} \left(z\right) = - \left(\pi - \alpha\right)$

$= {\tan}^{- 1} \left(\frac{3}{4}\right) - \pi$