Complex Numbers Geometry 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 $\frac{2ab}{a + b}.$

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