How do you convert # sqrt(3)+i# to polar form?

2 Answers
Jun 23, 2016

I found: #2/_30^@#

Explanation:

Have a look:

enter image source here

Jun 23, 2016

#(2,pi/6)#

Explanation:

To convert cartesian coordinates (x ,y) to Polar coordinates #(r,theta)#
Use the formulae which link them.

#•r=sqrt(x^2+y^2)#

and since #sqrt3+i=(sqrt3,1)# is in the 1st quadrant , then.

#•theta=tan^-1(y/x)#

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

#rArrr=sqrt((sqrt3)^2+1^2)=sqrt4=2#

and #theta=tan^-1(1/sqrt3)=pi/6" or "30^@#

#rArr(sqrt3,1)to(2,pi/6)#