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

Question

5189 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-21T03:31:56

Length $= 5$

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

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

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

$|z|=sqrt((-4)^2+(-3)^2)=5$

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

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

Since $"Re"(z)<0$ and $"Im"(z)<0$ , the angle lies in the third quadrant.

$"arg"(z)=-(pi-alpha)$

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

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