How do you write the trigonometric form of #sqrt3+i#?

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Answer 1 · 2016-11-19T16:58:55

#sqrt3+i=2cos(pi/6)+2isin(pi/6)#

A number #a+ib# can be written in trigonometric form as

#rcostheta+irsintheta#.

As #rcostheta=a# and #rsintheta=b#, squaring and adding them we get #r^2=a^2+b^2#.

As such for #sqrt3+i#, #r=sqrt((sqrt3)^2+1^2)=sqrt(3+1)=sqrt4=2#

and #costheta=sqrt3/2# and #sintheta=1/2#

i.e. #theta=pi/6#

Hence in trigonometric form

#sqrt3+i=2cos(pi/6)+2isin(pi/6)#