complex problem

**prasum** · October 11th 2010, 09:43 PM

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

**CaptainBlack** · October 11th 2010, 11:11 PM

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

**prasum** · October 11th 2010, 11:45 PM

can yu please help me solve the problem

do we have to use concept of rotation in it

**CaptainBlack** · October 12th 2010, 02:40 AM

Originally Posted by prasum

can yu please help me solve the problem

do we have to use concept of rotation in it

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 $z=x+i y$ , 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 $-1/\tan(\theta)$

CB

**prasum** · October 12th 2010, 06:13 AM

can yu please elaborate yur solution

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