# complex problem

• October 11th 2010, 09:43 PM
prasum
complex problem
show that the equation of the tangent to the circle mod(z)=r where r>0 at point z0 is given by ((conjugate) (z0))*z+z0*((conjugate)(z)=2*r^2

not understanding how to do this as moving point on the circumference of circle is z but what is z0
• October 11th 2010, 11:11 PM
CaptainBlack
Quote:

Originally Posted by prasum
show that the equation of the tangent to the circle mod(z)=r where r>0 at point z0 is given by ((conjugate) (z0))*z+z0*((conjugate)(z)=2*r^2

not understanding how to do this as moving point on the circumference of circle is z but what is z0

$z_0$ is a given point on the circle and $z$ is an arbitrary point on the tangent.

CB
• October 11th 2010, 11:45 PM
prasum

do we have to use concept of rotation in it
• October 12th 2010, 02:40 AM
CaptainBlack
Quote:

Originally Posted by prasum
You could start by writing it in cartesians so $z=x+i y$, and $z_0=r (\cos(\theta)+i \sin(\theta))$, then plug these into the proposed equation and show that it is a straight line which passes through $z_0$ and has gradient $-1/\tan(\theta)$