# My Favorite One-Liners: Part 67

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Here are a couple of similar problems that arise in Precalculus:

1. Convert the point $(5,-5)$ from Cartesian coordinates into polar coordinates.
2. Convert the complex number $5 - 5i$ into trigonometric form.

For both problems, a point is identified that is 5 steps to the right of the origin and then 5 steps below the $x-$axis (or real axis). To make this more kinesthetic, I’ll actually walk 5 paces in front of the classroom, turn right face, and then walk 5 more paces to end up at the point.

I then ask my class, “Is there a faster way to get to this point?” Naturally, they answer: Just walk straight to the point. After some work with the trigonometry, we’ll establish that

1. $(5,-5)$ in Cartesian coordinates is equivalent to $(5\sqrt{2}, -\pi/4)$ in polar coordinates, or
2. $5-5i$ can be rewritten as $5\sqrt{2} [ \cos(-\pi/4) + i \sin (-\pi/4)]$ in trigonometric form.

Once this is obtained, I’ll walk it out: I’ll start at the origin, turn clockwise by 45 degrees, and then take $5\sqrt{2} \approx 7$ steps to end up at the same point as before.

Continuing the lesson, I’ll ask if the numbers $5\sqrt{2}$ and $-\pi/4$, or if some other angle and/or distance could have been chosen. Someone will usually suggest a different angle, like $7\pi/4$ or $15\pi/4$. I’ll demonstrate these by turning 315 degrees counterclockwise and walking 7 steps and then turning 675 degrees and walking 7 steps (getting myself somewhat dizzy in the process).

Finally, I’ll suggest turning only 135 degrees clockwise and then taking 7 steps backwards. Naturally, when I do this, I’ll do a poor man’s version of the moonwalk: