# complex problem

• Oct 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
• Oct 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

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

CB
• Oct 11th 2010, 11:45 PM
prasum

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

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