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
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
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