How do you convert #1+jsqrt3# to polar form?

Question

3519 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-23T06:40:23

#1+jsqrt3=2(cos(pi/3)+jsin(pi/3))#

To write the complex number, say #a+jb#, where #j^2=-1# in polar coordinates, we write them as

#z=r(costheta+jsintheta)#.

Now as #rcostheta=a# and #rsintheta=b#.

#r=sqrt(a^2+b^2)#, #theta=tan^(-1)(b/a)#

Hence here for complex number #1+jsqrt3#

#r=sqrt(1^2+(sqrt5)^2)=sqrt4=2# and #theta=tan^(-1)(sqrt3/1)=tan^(-1)sqrt3=pi/3#

Hence #1+jsqrt3=2(cos(pi/3)+jsin(pi/3))#