Finding point of intersection of tangents through complex numbers 
If $P$ and $Q$ be two points on the circle $|w|=r$ represented by $w_1$ and $w_2$ respectively, then find the complex number representing the point of intersection of the tangents at $P$ and $Q$.

My attempt:
Well frankly i didnt even know how to start solving this problem. I tried converting the problem into vectors and I found the vector equations of the tangents and tried solving them but my answer was nowhere close to the one given in my book.
Could anyone please suggest a method on how to proceed with this question.
Thanks. 
 A: Let $p$ and $q$ be the complex numbers representing $P$ and $Q$. Since $P$ and $Q$ lie on the circle $\lvert w\rvert = r$, it holds $\lvert p\rvert = \lvert q\rvert = r$, which can be written as $p\bar{p} = q\bar{q} = r^2$ (since $z\bar{z} = \lvert z\rvert^2$ for every complex number $z$).
Now we want to find the equation of the tangent at $P$. Let $X$ be an arbitrary point in the plane and $x$ the complex number representing $X$. Also, let $O$ be the origin. $X$ belongs to the tangent iff $OP \perp PX$, that is to say, iff
$$\begin{align*}
& \Re\left(\frac{x - p}{p - 0}\right) = 0 \\
\iff & \frac{x - p}{p} = -\overline{\frac{x - p}{p}} = -\frac{\bar{x} - \bar{p}}{\bar{p}} = -\frac{\bar{x} - \frac{r^2}{p}}{\frac{r^2}{p}} = \frac{r^2 - p\bar{x}}{r^2} \\
\iff & r^2x + p^2\bar{x} - 2pr^2 = 0 \tag{1}
\end{align*}$$
So $(1)$ is the equation we were looking for.
In a similar way we get that the equation of the tangent at $Q$ is $$r^2x + q^2\bar{x} - 2qr^2 = 0 \tag{2}$$
Now, in order to determine the intersection of the two tangents, we just need to solve for $x$ (and $\bar{x}$) the system given by $(1)$ and $(2)$. Subtractiong $(2)$ from $(1)$ gives $(p^2 - q^2)\bar{x} - 2r^2(p - q) = 0$ which, cancelling a factor $p - q \ne 0$, becomes $(p + q)\bar{x} - 2r^2 = 0$, whence $\displaystyle \bar{x} = \frac{2r^2}{p + q}$. Subtitution in $(1)$ finally gives $$x = 2p - \frac{p^2\bar{x}}{r^2} = 2p - \frac{2p^2}{p + q} = \boxed{\frac{2pq}{p + q}}$$
Edit: Ok, I just read that actually $p$ and $q$ were already defined as $w_1$ and $w_2$...
A: Given points $A(z_1)$ and $B(z_2)$ in plane, all points $z$ on line $AB$ have this property that $\arg(z-z_1)=\arg(z_2-z_1)$ then the equation of the line $AB$ is
$$\arg\dfrac{z-z_1}{z_2-z_1}=0\tag{1}$$

Let $MP$ and $MQ$ are lines tangent to the circle $C$ with center at origin. From geometry we know $M(z)$ the point of intersection of the tangents at $P(w_1)$ and $Q(w_2)$ makes bisector $MO$ and perpendicular to $PQ$. 
With vectors $\vec{OP}+\vec{OQ}=\vec{OM}$ or $w_1+w_2=z$. 

Thus from (1), the locus of all $M$ (passing through origin $O(0)$) is
$$\arg\dfrac{z-0}{w_1+w_2-0}=0$$
or
$$\arg\dfrac{z}{w_1+w_2}=0$$
