How do you convert $- 5 - 5 i$ $-5-5i$ to polar form?

Question

How do you convert $- 5 - 5 i$ $-5-5i$ to polar form?

1 Answer

Impact of this question

14038 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-19T16:10:17

$(5 \sqrt{2}, \frac{π}{4})$ $(5sqrt(2), pi/4)$

Explanation:

Polar form: $(r, θ)$ $(r, theta)$

$a + b i$ $a + bi$

$a = - 5$ $a = -5$
$b = - 5$ $b=-5$

$r = \sqrt{a^{2} + b^{2}} = \sqrt{{(- 5)}^{2} + {(- 5)}^{2}} = \sqrt{50} = 5 \sqrt{2}$ $r = sqrt(a^2 + b^2) = sqrt((-5)^2 + (-5)^2) = sqrt(50)= 5sqrt(2)$

$θ = tan (\frac{b}{a}) = tan (\frac{- 5}{- 5}) = tan (1) = \frac{π}{4}$ $theta = tan(b/a) = tan((-5)/-5) = tan(1) = pi/4$

$(5 \sqrt{2}, \frac{π}{4})$ $(5sqrt(2), pi/4)$

Science

Math

Humanities

... and beyond

How do you convert $- 5 - 5 i$ $-5-5i$ to polar form?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you convert −5−5i-5-5i to polar form?

1 Answer

Explanation:

Related questions

Impact of this question

How do you convert $- 5 - 5 i$ $-5-5i$ to polar form?