Complex Numbers Geometry

Question

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 $\frac{2ab}{a + b}.$

Answer 1

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, $(a+b)\bar{z}=2$.

Answer 2

Note that if $z$ is a point of the tangent line then: $$z=\lambda a i +a \quad (1)$$ and $$\bar{z}=-\lambda \bar{a} i + \bar{a} \quad (2)$$ where $\lambda$ is a real number.

Recall that if $a$ is in unit circle then: $$a \bar{a}=1$$ Let's calculate $z+ \bar{z} a^2$: $$z+ \bar{z} a^2=(\lambda a i +a)+(-\lambda \bar{a} i + \bar{a})a^2=2a$$ and we conclude that the statement (a) is true.

For statement (b) use twice statement (a) and solve the system, you will find that (b) is also true.

Answer 3

(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*}

(b) Let's set $w$ as the intersecting point. Using out solution from part a, we can say: $$w+a^2\overline{w}-2a=0=w+b^2\overline{w}-2b$$ Let's manipulate this equation: \begin{align*} w+a^2\overline{w}-2a &= w+b^2\overline{w}-2b \\ a^2\overline{w}-b^2\overline{w} &= 2a-2b \\ \overline{w}(a^2-b^2) & = 2(a-b) \\ \overline{w}(a+b) &= 2 \\ \overline{w} &= \frac{2}{a+b} \\ w &= \frac{2}{\overline{a}+\overline{b}} \end{align*} Let's take a step back and figure $\overline{a}$ and $\overline{b}$ \begin{align*} a\overline{a} = |a|^2 \\ a\overline{a} = 1 \\ \overline{a} = \frac{1}{a} \end{align*} Using same logic, we can also derive $\overline{b} = \frac{1}{b}$ Now let's substitute these values in: \begin{align*} w &= \frac{2}{\overline{a}+\overline{b}} \\ &= \frac{2}{\frac{1}{a}+\frac{1}{b}} \\ &= \frac{2}{\frac{a+b}{ab}} \\ &= \frac{2ab}{a+b} \end{align*}