How do you convert #(1, sqrt3)# into polar form?

2 Answers
F. Javier B.
Mar 12, 2018

#2_(pi/3)#

See below

Explanation:

The pole #rho# is the module of complex number #z=1+sqrt3i#. So

#rho=sqrt(1+3)=2#

#theta=arctan(sqrt3/1)=60º=pi/3#.

Jim G.
Mar 12, 2018

#(2,pi/3)#

Explanation:

#"to convert from "color(blue)"cartesian to polar form"#

#"that is " (x,y)to(r,theta)" using"#

#•color(white)(x)r=sqrt(x^2+y^2)#

#•color(white)(x)theta=tan^-1(y/x)#

#"here "x=1" and "y=sqrt3#

#rArrr=sqrt(1^2+(sqrt3)^2)=2#

#rArrtheta=tan^-1(sqrt3)=pi/3#

#rArr(1,sqrt3)to(2,pi/3)#

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