Answer :
let z=x+iy
[tex] \frac{y}{x}=tan( \frac{ \pi }{6})= \frac{1}{ \sqrt{3} } [/tex]
[tex]x= y\sqrt{3} [/tex]
now as
[tex]|z-2 \sqrt{3}i |=p \\ x^2+(y-2 \sqrt{3} )^2=p^2=d[/tex]
sub. [tex]x= y\sqrt{3} [/tex]
[tex]d=3y^2+(y-2 \sqrt{3} )^2 \\ d'=6y+2(y-2 \sqrt{3} )[/tex]
put d'=0
[tex]y= \frac{ \sqrt{3} }{2} [/tex]
then by second derivative test
[tex]d"=8>0[/tex]
so if [tex]y= \frac{ \sqrt{3} }{2} [/tex]
[tex]d=3( \frac{3}{4} )+( \frac{ \sqrt{3} }{2}-2 \sqrt{3} )^2= \frac{9+27}{4}=9 [/tex]
so [tex]p= \sqrt{d}=3 [/tex]
[tex] \frac{y}{x}=tan( \frac{ \pi }{6})= \frac{1}{ \sqrt{3} } [/tex]
[tex]x= y\sqrt{3} [/tex]
now as
[tex]|z-2 \sqrt{3}i |=p \\ x^2+(y-2 \sqrt{3} )^2=p^2=d[/tex]
sub. [tex]x= y\sqrt{3} [/tex]
[tex]d=3y^2+(y-2 \sqrt{3} )^2 \\ d'=6y+2(y-2 \sqrt{3} )[/tex]
put d'=0
[tex]y= \frac{ \sqrt{3} }{2} [/tex]
then by second derivative test
[tex]d"=8>0[/tex]
so if [tex]y= \frac{ \sqrt{3} }{2} [/tex]
[tex]d=3( \frac{3}{4} )+( \frac{ \sqrt{3} }{2}-2 \sqrt{3} )^2= \frac{9+27}{4}=9 [/tex]
so [tex]p= \sqrt{d}=3 [/tex]