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

1 Answer
Shwetank Mauria
May 4, 2016

#sqrt3+i# in polar form is written as #2cos(pi/6)+2isin(pi/6)# or #2e^(i(pi/6)#

Explanation:

Complex number #a+ib# is written as #rcostheta+irsintheta# or #re^(itheta)# in polar form.

where, #r=sqrt(a^2+b^2)3 and #tantheta=b/a# or #theta=arctan(b/a)#

Hence for #sqrt3+i#

#r=sqrt((sqrt3)^2+1^2)=sqrt4=2# and #theta=arctan(1/sqrt3)=pi/6#

Hence #sqrt3+i# in polar form is written as #2cos(pi/6)+2isin(pi/6)#

or #2e^(i(pi/6)#

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