How do you express the complex number in standard form: #sqrt3(cos π/6 + i sin π/6)#?

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Answer 1 · 2017-08-04T19:15:13

The standard form is #=3/2+sqrt3/2i#

We need

#cos(pi/6)=sqrt3/2#

#sin(pi/6)=1/2#

The complex number is

#z=sqrt3(cos(pi/6)+isin(pi/6))#

Therefore,

#z=sqrt3(sqrt3/2+1/2i)#

#=3/2+sqrt3/2i#