I'm not sure where to begin on this problem – do I plug in for a and solve for z?
I was also given a hint:
Let z be a point on the line we're trying to describe. We have good tools in complex numbers for collinearity and perpendicularity. Which would be useful here.
Here is the problem:
Let a and b be two complex numbers on the unit circle, i.e. $|a| = |b| = 1.$
(a) Show that the equation of the tangent to the unit circle at a is given by $ z + a^2 \overline{z} = 2a.$
(b) Show that the intersection of the tangents to the unit circle at a and b is $\frac{2ab}{a + b}.$
Best Answer
I don't know how to use the hints you've been given, but you could solve these problems as follows.
First, note that the equation in (a) is equivalent to $\bar{a}z+a\bar{z}=2$ and the one in (b) to $(a+b)\bar{z}=2$ since $\bar{a}=1/a$ for any $a$ on the unit circle (I assume $a+b\ne0$).
If $a=e^{i\phi}$, then the tangent line is given by $x\cos\phi+y\sin\phi=1$. You can use this to prove (a) by plugging $z=x+iy$ and $a=\cos\phi+i\sin\phi$ into my version of (a) given above and taking the products.
For (b), assume $\bar{a}z+a\bar{z}=2$ and $\bar{b}z+b\bar{z}=2$ and take the sum $$\bar{a}z+a\bar{z}+\bar{b}z+b\bar{z}=(a+b)\bar{z}+(\bar{a}+\bar{b})z=4.$$ Note that $(a+b)\bar{z}$ is a real number since $a+b$ and $z$, which is the intersection point, have the same argument and the argument of $\bar{z}$ is the negative of that. Therefore, $(a+b)\bar{z}=2$.