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The Brainliest Answer!
2014-05-02T12:07:13+05:30
Let z=x+iy
 \frac{y}{x}=tan( \frac{ \pi }{6})= \frac{1}{ \sqrt{3} }

x= y\sqrt{3}
now as 
|z-2 \sqrt{3}i |=p \\ x^2+(y-2 \sqrt{3} )^2=p^2=d
sub. x= y\sqrt{3}
d=3y^2+(y-2 \sqrt{3} )^2 \\ d'=6y+2(y-2 \sqrt{3} )
put d'=0
y= \frac{ \sqrt{3} }{2}
then by second derivative test
d"=8>0
so if y= \frac{ \sqrt{3} }{2}
d=3( \frac{3}{4} )+( \frac{ \sqrt{3} }{2}-2 \sqrt{3}  )^2= \frac{9+27}{4}=9
so p= \sqrt{d}=3
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