Let $a,b$ be fixed complex numbers and let $L$ be the line
$$L=\{a+bt:t\in\Bbb R\}.$$
Let $w\in\Bbb C\setminus L$. Let's calculate $$d(w,L)=\inf\{|w-z|:z\in L\}=\inf_{t\in\Bbb R}|w-(a+bt)|.$$
The first observation is
$$d(w,L)=\inf_{t\in\Bbb R}|w-(a+bt)|=\inf_{t\in\Bbb R}|(w-a)-bt|=d(w-a,L-a).$$
So without loss of generality we can assume that $a=0$, so
$$L=b\Bbb R.$$
And then, let's assume that $|b|=1$.
Assume
$$b=x+iy,\qquad x,y\in\Bbb R,$$
and let
$$b'=y-ix.$$
so that $b'\bar{b}$ is purely imaginary and $|b'|=|b|=1$.
Consider the line
$$L'=w+b'\Bbb R.$$
Then $L$ and $L'$ are orthogonal and since $w=w+b'\cdot 0$ we get $w\in L'$. Let $w'$ the point in which these lines meet. Let's calculate it
$$\begin{align*}
w+b's &= bs\\
s &= \frac{w}{b-b'}.
\end{align*}$$
so $w'=w+b's$. The distance $d(w,L)$ must be achieved at $w'$ (because $L'$ is the perpendicular line to $L$ through $w$), so
$$d(w,L)=|w-w'|=|b's|=1|s|=\left|\frac{w}{b-b'}\right|=\frac{|w|}{|b-b'|}.$$
Let's compute $|b-b'|$
$$|b-b'|^2=|x+iy-(y-ix)|^2=(x-y)^2+(x+y)^2=2|b|^2=2$$
therefore
$$d(w,L)=\frac{|w|}{\sqrt{2}}.$$
What I have just proved is equivalent to say that the triangle with vertex $0$, $w$, $w'$ is isosceles rectangle, regardless of the line through the origin and the $w$. Of course this is absurd but,
What is wrong with the above?
Best Answer
When calculating where the points meet, you set $w+b's=bs$. This a mistake, you should set that $w+b's=bt$ where $s$ does not necessarily equal $t$, but both are real. The point you calculate is a point equidistant from $0$ and $w$ of distance $|s|$ . Because, for example, $$d(w,w+b's) = |b's|=|s|=|bs|=d(0,bs).$$
Note that doing it your original way, $s$ is not necessarily real, so the point may not even lie on either of the lines! (And probably won't!)
Note: To solve $w+b's=bt$, equate real and imaginary parts to get a pair of simultaneous equations to solve.
In addition, doing the problem this way seems a little overcomplicated. Further, you assert that the distance is minimised by connecting $w$ to $L$ perpendicularly, which should be rigorously proven. A much quicker way is to simply calculate $$\inf_{t\in\mathbb{R}}\{|w-bt|\}=\inf_{t\in\mathbb{R}}\{((w_1-xt)^2+(w_2-yt)^2)^\frac12\}$$ which is achieved by finding the value of $t$ minimising the function $$f(t) = (w_1-xt)^2+(w_2-yt)^2,$$ which is a quadratic.