How do you convert #1 - 3i# to polar form?

Question

Zoe Austin · Accepted Answer

#color(purple)("Polar form "= (r, theta) = (3.1623, 288.43^@# #z = 1 - i 3# #r = sqrt (1^2 + (-3)^2) = sqrt 10 = 3.1623# #theta = tan ^(-1) (-3/1) = -71.57^@ = 288.43^@, " as point lies in IV quadrant"# #color(purple)("Polar form "= (r, theta) = (3.1623, 288.43^@#

Caroline Adams · Answer

undefined To convert the complex number $1 - 3i$ to polar form, we need to find its magnitude (absolute value) and argument (angle).

The magnitude $r$ of a complex number $a + bi$ is given by the formula:

$$ r = |a + bi| = \sqrt{a^2 + b^2} $$

In this case, $a = 1$ and $b = -3$. Substituting these values into the formula:

$$ r = \sqrt{1^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10} $$

So, the magnitude of $1 - 3i$ is $ \sqrt{10} $.

The argument $\theta$ of a complex number $a + bi$ can be found using the formula:

$$ \theta = \text{arctan}\left(\frac{b}{a}\right) $$

In this case, $a = 1$ and $b = -3$. Substituting these values into the formula:

$$ \theta = \text{arctan}\left(\frac{-3}{1}\right) = \text{arctan}(-3) $$

Using a calculator, we find that $ \text{arctan}(-3) \approx -1.249 $ radians.

Therefore, the polar form of $1 - 3i$ is $ \sqrt{10} \, \text{cis}(-1.249) $, where cis represents the complex exponential form.