Proving the tangent of the unit circle is given by the equation $z+a^2\overline{z}= 2a$ Let $a$ and $b$ be 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 of part (a) to show that the intersection of the tangents to the unit circle at $a$ and $b$ is:
$\frac{2ab}{a+b}$.
My thoughts: Given that the point $a$ is on the unit circle we know that $|a|=1$. In addition, we know that from the center of the circle to $a$ will be perpendicular with the tangent. The general formula for proving that two lines are perpendicular is: $(z_3-z_1)=e^{i\theta}(z_2-z_1)$ for some complex numbers $z_1$, $z_2$, and $z_3$. 
We could use this equation to find a point on the line that is tangent to the circle and then find the slope of the equation. But, I don't quite understand how this would take the form of $z+a^2\overline{z}= 2a$? Thus, I cannot continue on (a) or start on (b). Any hints/help would be much appreciated! 
 A: With $z = x + i y$ and $a = u + i v$ we would require
$$
(z-a) \bot a \iff \\
0 
= (x-u) u + (y-v) v 
= \mbox{Re}(a (\overline{{z-a}})) = 
\frac{a(\overline{z-a}) + \bar{a} (z-a)}{2} \iff \\
0 
= a^2 (\bar{z}-\bar{a}) + a \bar{a} (z - a) 
= a^2\bar{z}- a(a\bar{a})+ z - a \iff \\
a^2\bar{z} + z = 2a
$$
because $a\bar{a} = |a|^2 = 1$.
A: now as the line is tangent , the line joining $a$ to center is perpendicular to tangent ,therefore $$\frac{z-a}{a}=ki$$ 
$$z=a(1+ki)$$
$$\overline{z}=\overline{a}(1-ki)$$
$$a^2\overline{z}=a^2*\overline{a}(1-ki)=a*1*(1-ki)$$
thus $$z+a^2\overline{z}=2a$$
A: The (A) part is done in another answer. 
(B)$$z_1=a(1+it_1)$$$$z_2=b(1+it_2)$$at $P$, they meet. Then 
$$t_1=-t_2$$ from here you can get the answer. 
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*}
