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) Use the result from Part (a) to show that the intersection of the tangents to the unit circle at $a$ and $b$ is
$$\frac{2ab}{a + b}$$
I got part $a$ like this:
(a) Since we know the tangent line and the radius from the origin to $a$ are perpendicular, we can say $$z-a = a(e^{\pi i/2}k) = aki$$ where $k$ is some integer.
Manipulating the equation we get:
\begin{align*}
z-a &= aki \\
z &= a(1+ki) \\
\overline{z} &= \overline{a}(1-ki) \\
a^2\overline{z} &= a^2\overline{a}(1-ki)
\end{align*}
Let's simplify the right side: $$a^2\overline{a}(1-ki) = a\cdot a\overline{a}(1-ki) = a\cdot |a|^2(1-ki) = a(1-ki)$$
Now we can continue:
\begin{align*}
a^2\overline{z} &= a^2\overline{a}(1-ki) \\
a^2\overline{z} &= a(1-ki) \\
a^2\overline{z} &= a-aki \\
a^2\overline{z} &= 2a – a – aki \\
a^2\overline{z} &= 2a – z \\
z+a^2\overline{z} &= 2a
\end{align*}
However, I need help on part $b$. I assume that we would need to use the answer from part $a$ for 2 lines and set them equal, but I'm not sure. Could someone help?
Thank you!
Best Answer
Suppose $z_0$ is the intersection. Then it satisfies both equations and so $$ z_0 +a^2 \overline{z}_0 -2a = z_0 +b^2 \overline{z}_0 -2b $$ which leads to $(a^2-b^2)\overline{z}_0 = 2(a-b)$. Now because obviously $a \neq b$ (in order for the problem to be consistent) we can cancel $a-b$ and get $$ \overline{z_0} = \dfrac{2}{a+b} \Rightarrow z_0 = \dfrac{2}{\overline{a}+\overline{b}} = \dfrac{2}{\dfrac{1}{a} + \dfrac{1}{b}} = \dfrac{2ab}{a+b}\cdot $$ Note: since $|a|=|b| = 1$ we have $a\overline{a} = b\overline{b} = 1$.