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

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Answer 1 · 2015-11-21T03:31:56

Length $= 5$ $= 5$

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

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

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

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

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

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

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

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

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

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