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Given $|x|= 1$, show that the equation of a tangent to the unit circle at point $x$ is given by $y + x^2 \overline{y} = 2x$

Not sure how to start. I know that $y + \overline{y} = 2x$ must be true, but I am not sure what to do after that.

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    "*I know that $y + \overline{y} = 2x$ must be true*" Why? You may be thinking at the textbook $z=x+iy$ with $x,y \in \mathbb{R}$ so that $z+\bar z = 2x\,$, but here both $x$ and $y$ look to be assumed to be complex numbers. – 2017-01-27
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    Great, now I am back to square one. I have no idea where to start @dxiv – 2017-01-27
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    For a starting point, $|x|^2=x \bar x=1$ so dividing by $x \ne 0$ gives $x \bar y + \bar x y = 2\,$. – 2017-01-27
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    @dxiv I tried messing with that, but am unable to solve it. Should I be using the general formula for finding tangents? – 2017-01-27

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Prerequisite: We know that the equation of tangent through a point $c$ lying on a circle $|z|=r$ is given by $$z\bar c +\bar z c =2r^2\tag {1}$$

From the given equation, dividing by $x $ gives us (as $x \neq 0$) gives us: $$\frac {1}{x}y + x\bar y =2$$ $$\Rightarrow \frac {\bar x}{x\bar x}y + x\bar y =2$$ $$\Rightarrow \bar x y + x \bar y =2\tag {2}$$

Can you now relate $(1)$ and $(2)$? Hope it helps.

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    I am actually a little lost, how does $\overline{x}y + x \overline{y} = 2 = z\overline{c} +\overline{z}c =2r^2$? If I am correct $z$ is a circle, however $x$ is a point on a circle. – 2017-01-30
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Lemma. Given $4$ points $a \ne b, c\ne d$ in the complex plane, segments $ab$ and $cd$ are perpendicular iff $\operatorname{Re}\left((a-b)(\bar c- \bar d)\right) = 0\,$.

Let $z = (a-b)(\bar c- \bar d)$ then $\arg(z) = arg(a-b) + \arg(\bar c - \bar d) = \arg(a-b) - \arg(c-d)\,$. It follows that $\operatorname{Re}(z) = 0 \iff \arg(a-b) - \arg(c-d) = \pm \pi/2\;$ i.e. $\;ab \perp cd\,$.

Proof. Given that $|x|^2=x \bar x =1$ it follows that $\bar x = 1/x$. Then dividing the given relation by $x \ne 0$ results in $x \bar y + \bar x y = 2$.

Now, the segments between $(0,x)$ and $(y,y-x)$ are perpendicular by the previous Lemma since:

$$ \require{cancel} 2 \operatorname{Re}((x-0)(\bar y-\bar x))= x(\bar y-\bar x)+\bar x(y - x) = x \bar y - |x|^2+\bar x y -|x|^2 = 2 - 1 - 1 =0 $$

Since $(0,x)$ is a radius of the unit circle, it follows that $y$ lies on the tangent to the circle at $x$.