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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 $$z + a^2 \overline{z} = 2a$$

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(b) Use the result from Part (a) to show that the intersection of the tangents to the unit circle at $a$ and $b$ is $$\frac{2ab}{a + b}$$

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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!

Veliko
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1 Answers1

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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 $a\overline{a} = b\overline{b} = 1$.

Veliko
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