Question

What is the least value of p for which two curves
arg(z)=π/6  and  |z-2√3i|=p intersect ?

Answer 1

let z=x+iy
[tex] \frac{y}{x}=tan( \frac{ \pi }{6})= \frac{1}{ \sqrt{3} } [/tex]
$x= y\sqrt{3}$
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 $p= \sqrt{d}=3$