Shortest distance formula in complex numbers

Question

Let $L$ be a line in $\mathbb{C}$ that makes an angle $\alpha$ with the real axis. Let $z=x+iy$ be any point on the line $L$. Let $d$ be the shortest distance from $L$ to origin. Prove geometrically that $d=|\text{Im}(ze^{-i\alpha})|$.

My solution uses both geometry and trigonometry but I can not come with a purely geometrical solution. Is using both geometry and trigonometry in my solution allowed or I only have to use geometry? Nonetheless I would be interested to see a pure geometrical solution.

Answer 1

What do you mean by "pure" geometry? For one thing trigonometry ultimately has its root in geometry (ratio of parts of a triangle whose sides are radii of a circle).

That said, to solve this problem you just need a geometric understanding of what the map $f_{\theta}(z) = e^{i\theta}z$ does to points on the complex plane. Maybe you need Euler's formula and some trigonometry to gain that understanding -- there are perhaps other ways of going about it (i.e. remarking that $f_{\theta}$ is the exponential map on the Lie algebra of 2D rotations) but I don't see anything "impure" about the obvious approach.

Answer 2

It would be difficult to have a "purely" geometrical solution, since the desired expression for the minimum distance $ \ d \ $ involves a complex exponential function, so trigonometric functions are present implicitly. Nonetheless, there are ways to hold off invoking trigonometry until one is near the end of the work.

Here is a diagram of the geometry on the complex plane described in the problem statement. While the figure abounds in similar right triangles, after appropriate marking, I didn't succeed in extracting from the resulting proportionality equations any simple, clean way of developing the minimal distance equation sought. So I will describe a solution that forestalls the use of trigonometry until that becomes unavoidable.

Since Student doesn't indicate what their solution was, and user7530 doesn't say what the "obvious approach" is, I will give a solution which uses some of the features I want to apply in the "nearly-geometrical" solution (and is perhaps the one being alluded to). Here, we will label the point $ \ z \ = \ x_0 \ + \ iy_0 \ $ as $ \ P \ $ and mark the point on the line $ \ L \ $ that is closest to the origin as $ \ Q \ $ . We will also just use the familiar fact that the line segment $ \ \overline{OQ} \ $ is perpendicular to $ \ L \ $ (we will prove this in the next solution). It only requires a simple argument through the use of similar triangles in the first diagram above that the angle between the segment $ \ \overline{OQ} \ $ and the $ \ y-$ axis is $ \ \alpha \ $ . Thus, we have $ \ m( \angle QOP) \ = \ \alpha \ + \ (\frac{\pi}{2} \ - \ \theta) \ $ . Now we bring in trigonometry for the right triangle $ \ \Delta QOP \ $ , which tells us that $ \ d \ = \ \rho \ \cos(\alpha \ + \ \frac{\pi}{2} \ - \ \theta) \ = \ \rho \ \cos( \ \frac{\pi}{2} \ - \ [ \ \theta \ - \ \alpha \ ] \ ) \ $ .

We now use polar form to write $ \ z \ = \ \rho \ cis \ \theta \ $ ; $ \ e^{-i \alpha} \ $ has unit modulus and so may be written as $ \ 1 \ \cdot \ cis \ (- \alpha) \ $ . This gives us the product $ \ ze^{-i \alpha} \ = \ \rho \ \cdot \ 1 \ \cdot \ cis( \ \theta \ + \ [- \alpha] \ ) \ = \ \rho \ cis( \ \theta \ - \alpha \ ) \ $ , of which the imaginary part is $ \ \mathfrak{Im} (ze^{-i \alpha}) \ = \ \rho \ \sin( \ \theta \ - \alpha \ ) \ $ . But since $ \ \ \cos( \ \frac{\pi}{2} \ - \ [ \ \theta \ - \ \alpha \ ] \ ) \ = \ \sin( \ \theta \ - \alpha \ ) \ $ , this is the expression we found above for $ \ d \ $ .

We used a line with positive slope and placed the point $ \ z \ $ in the first quadrant in this argument. If we generalize this result to cover other arrangements of point and line, we must add absolute value brackets, giving us $ \ d \ = \ | \ \mathfrak{Im} (ze^{-i \alpha}) \ | \ $ .

$ \ \ $

As to the "more geometrical" solution, we will write the slope of the line $ \ L \ $ as $ \ m \ $ through most of the discussion (knowing that $ \ m \ = \ \tan \alpha \ $ ) . This places the point we call $ \ Q \ $ on the line

$$ \ y \ - \ y_0 \ = \ m \ ( x \ - \ x_0 ) \ \ \ \text{or} \ \ \ y \ - \ mx \ + \ mx_0 \ - \ y_0 \ = \ 0 \ \ . $$

We wish to minimize the distance between $ \ Q \ (x, \ y) \ $ and the origin; since this distance is non-negative, we can minimize the function $ \ x^2 \ + \ y^2 \ $ instead. We have a choice of techniques: we will use Lagrange-multipliers here to save some writing. So we wish to minimize $ \ f(x, \ y) \ = \ x^2 \ + \ y^2 \ $ , subject to the constraint function $ \ g(x, \ y) \ = \ y \ - \ mx \ + \ mx_0 \ - \ y_0 \ $ . We obtain the "Lagrange equations"

$$ 2 \ x \ = \ \lambda \ (-m) \ \ \ , \ \ \ 2 \ y \ = \ \lambda \ \cdot \ 1 \ \ \Rightarrow \ \ y \ = \ - \frac{x}{m} \ \ . $$

Inserting this into the equation for the line $ \ L \ $ yields

$$ \left( \frac{1}{m} \ + \ m \right) x \ = \ m \ x_0 \ - \ y_0 \ $$ $$ \Rightarrow \ \ x \ = \ \frac{m}{m^2 + 1} \ (m \ x_0 \ - \ y_0) \ \ , \ \ y \ = \ -\frac{1}{m^2 + 1} \ (m \ x_0 \ - \ y_0) \ \ . $$

The distance of this point from the origin is $ \ d \ = \ \frac{1}{\sqrt{m^2 + 1}} \ (m \ x_0 \ - \ y_0) \ $ .

We've held off using trigonometry as long as could be managed. We now insert the value for the slope $ \ m \ = \ \tan \alpha \ $ to produce

$$ d \ = \ \frac{1}{\sqrt{(\tan^2 \alpha) + 1}} \ (\ [ \tan \alpha ] \ x_0 \ - \ y_0) \ = \ \cos \alpha \ (\ [ \tan \alpha ] \ x_0 \ - \ y_0) $$ $$ = \ (\sin \alpha ) \ x_0 \ - \ (\cos \alpha ) \ y_0 \ \ . $$

We will now interpret this expression in the following manner. We may define a unit vector $ \ \hat{u} \ $ which points in the direction $ \ \vec{OQ} \ $ . This vector makes an angle $ \ \alpha \ $ to the positive $ \ y-$ axis in the second quadrant, so it points in the direction $ \ \alpha \ + \ \frac{\pi}{2} \ $ . Thus,

$$ \ \hat{u} \ = \ \langle \ \cos(\alpha \ + \ \frac{\pi}{2}) \ , \ \sin(\alpha \ + \ \frac{\pi}{2}) \ \rangle \ = \ \langle -\sin \alpha \ , \ \cos \alpha \rangle \ \ . $$

Our result above for $ \ d \ $ can thus be understood as the scalar projection of the vector $ \ \vec{OP} \ $ , extending from the origin to the point for the complex number $ \ z \ $ , onto $ \ \hat{u} \ $ . This is $ \ \hat{u} \ \centerdot \ \vec{OP} \ = \ \langle -\sin \alpha \ , \ \cos \alpha \rangle \ \centerdot \ \langle \ x_0 \ , \ y_0 \ \rangle \ $ . The "included angle" between the vectors is $ \ \frac{\pi}{2} \ + \ \alpha \ - \ \theta \ $ , so the value of this dot product is given by

$$ \ 1 \ \cdot \ \rho \ \cdot \ \cos(\frac{\pi}{2} \ + \ \alpha \ - \ \theta ) \ = \ \rho \ \cos( \ \frac{\pi}{2} \ - \ [ \ \theta \ - \ \alpha \ ] \ ) \ = \ \rho \ \sin( \ \theta \ - \alpha \ ) \ \ . $$

We will want the absolute value of this scalar projection, $ \ | \ \rho \ \sin( \ \theta \ - \alpha \ ) \ | \ $ , which then conforms with the expression we found in the previous solution for $ \ d \ $ .

We could also have handled the problem of locating the point $ \ Q \ $ by parametrizing the line $ \ L \ $ as $ \ x \ = \ x_0 \ - \ t \ \ , \ y \ = \ \ y_0 \ - \ mt \ $ . The distance from the origin is minimized for the parameter value $ \ t \ = \ \frac{x_0 + my_0}{m^2 + 1} \ $ and the rest of the argument proceeds as described.