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

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- Oct 11th 2010, 10:43 PMprasumcomplex 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 12th 2010, 12:11 AMCaptainBlack
- Oct 12th 2010, 12:45 AMprasum
can yu please help me solve the problem

do we have to use concept of rotation in it - Oct 12th 2010, 03:40 AMCaptainBlack
The reason I have not attempted to solve this is because I do not know how you are expected to solve it, and suspect how I would approach this is not acceptable for you to submit.

You could start by writing it in cartesians so , and , then plug these into the proposed equation and show that it is a straight line which passes through and has gradient

CB - Oct 12th 2010, 07:13 AMprasum
can yu please elaborate yur solution