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Mathematics 9 Online
OpenStudy (anonymous):

Convert the complex number -5+5√3 i into polar form. Can someone explain to me the process of doing this? Thanks.

hartnn (hartnn):

if you have a complex number as \(\Large x+iy \) then to convert it into polar form \(\Large r\angle \theta \) we use \(\Large r = \sqrt{x^2+y^2}\) and \(\Large \theta = tan^{-1}(\dfrac{y}{x})\)

hartnn (hartnn):

here, \(\Large x+iy = -5+5\sqrt 3 i\ \\ \Large so, x = -5 , y = 5\sqrt 3 \) just plug in!

OpenStudy (loser66):

@hartn I don't get it \[r=\sqrt{x^2+y^2}=\sqrt{(-5)^2+(5\sqrt{3})^2}=10\] and \(\large tan^- ({\dfrac{y}{x}})= tan^-(\dfrac{5\sqrt{3}}{-5})=tan^-(-\sqrt{3}) \rightarrow \theta =-60 ^0\) So, I have the polar form is ( \(10, -60^0\)) However, when I test it, x = r cos \(\theta\) = 10 cos (-60) = 5 not -5 . And \( y = r sin\theta = 10*sin (-60) = -5\sqrt{3}\) which are not the given information. What's wrong, Please explain me.

OpenStudy (loser66):

@hartnn

hartnn (hartnn):

tan inverse (-sqrt 3) will have 2 angles as answers which one would u pick ?

hartnn (hartnn):

one in 2nd quadrant, one in 4th quadrant now see the co-ordinates in the question, to which quadrant does they belong ?

hartnn (hartnn):

r is most definitely 10

OpenStudy (loser66):

Ok, got you!! I forgot the signs of x, y and I used calculator to find tan inverse, so that it gave me -60 degree Thanks for make it clear.

hartnn (hartnn):

welcome :)

OpenStudy (anonymous):

how does tangent inverse 3 have two angles as answers? All i get is -60 degrees

hartnn (hartnn):

tan is periodic with 180 degrees. so each value of tan repeats after 180 degrees so, in 0 to 360, there will always be 2 angles for any value of tan

hartnn (hartnn):

if one of them is theta other will be theta + 180 or theta - 180

hartnn (hartnn):

now -60 is in 4th quadrant add 180 to it, and you will get the angle in 2nd quadrant

hartnn (hartnn):

we need angle in 2nd quadrant because the the co-ordinates in the question are in 2nd quadrant

OpenStudy (anonymous):

ohh alright thanks

hartnn (hartnn):

welcome ^_^

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