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, 09:43 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 11th 2010, 11:11 PMCaptainBlack
- Oct 11th 2010, 11:45 PMprasum
can yu please help me solve the problem

do we have to use concept of rotation in it - Oct 12th 2010, 02:40 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 $\displaystyle z=x+i y$, 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 $\displaystyle -1/\tan(\theta)$

CB - Oct 12th 2010, 06:13 AMprasum
can yu please elaborate yur solution