What's the ROI? If 15% of wild caught fish are used to support fish farms that produce twice as much, it's not as obviously a bad thing. There'd need to be another food source though.

We're back to "crud" and "shucks" now boomer

If you have a lot of semantic breakpoints (like the end of a concept) that don't line up with syntactic breakpoints (like the end of a method or expression body) your code probably needs to be refactored. If you don't, then automatic code formatting is probably all you need.

Why would you need to set up a postgres db...? Unions are a fundamental set theoretic operation that are applicable to all set-like collections. You may as well say "an in memory hash map / list is absurdly inefficient compared to a relational db." Is it efficient, to you, to spin up a postgres instance to hold a dozen key value pairs?

I really don't think it's that bad. The only weird thing is .NET Core becoming just .NET in version 5.

Counterpoint: "NieR: Automata"

Only if it's snowing a lot. Otherwise you just bundle up and try not to go outside for more than a couple minutes.

"master" in version control has no corresponding "slave," but nevertheless the "master/slave" terminology is the reason why GitHub switched to "main" and everyone else followed suit

"Please sir, step down while there's still a chance to replace you with someone reasonable"?

Step 1, to play the heffer you will need to adjust your bones with a large mallet

Midwest US it's more like $10 for a single crappy fake crab roll, $20 for a good roll. But hey, we have, uh, corn.

If you think of complex numbers in their polar form, everything is much simpler. If you know basic calculus, it can be intuitive.

Instead of z = + iy, write z = (r, t) where r is the distance from the origin and t is the angle from the positive x-axis. Now addition is trickier to write, but multiplication is simple: (a,b) * (c,d) = (ab, b + d). That is, the lengths multiply and the angles add. Multiplication by a number (1, t) simply adds t to the angle. That is, multiplying a point by (1, t) is the same as rotating it counterclockwise about the origin by an angle t.

The function f(t) = (1, t) then parameterizes a circular motion with a constant radial velocity t. The tangential velocity of a circular motion is perpendicular to the current position, and so the derivative of our function is a constant 90 degree multiple of itself. In radians, that means f'(t) = (1, pi/2)f(t). And now we have one of the simplest differential equations whose solution can only be f(t) = k * e^(t* (1, pi/2)) = ke^(it) for some k. Given f(0) = 1, we have k = 1.

All that said, we now know that f(t) = e^(it) is a circular motion passing through f(0) = 1 with a rate of 1 radian per unit time, and e^(i pi) is halfway through a full rotation, which is -1.

If you don't know calculus, then consider the relationship between exponentiation and multiplication. We learn that when you take an interest rate of a fixed annual percent r and compound it n times a year, as you compound more and more frequently (i.e. as n gets larger and larger), the formula turns from multiplication (P(1+r/n)^(nt)) to exponentiation (Pe^(rt)). Thus, exponentiation is like a continuous series of tiny multiplications. Since, geometrically speaking, multiplying by a complex number (z, z^(2), z^(3), ...) causes us to rotate by a fixed amount each time, then complex exponentiation by a continuous real variable (z^t for t in [0,1]) causes us to rotate continuously over time. Now the precise nature of the numbers e and pi here might not be apparent, but that is the intuition behind why I say e^(it) draws a circular motion, and hopefully it's believable that e^(i pi) = -1.

All explanations will tend to have an algebraic component (the exponential and the number e arise from an algebraic relationship in a fundamental geometric equation) and a geometric component (the number pi and its relationship to circles). The previous explanations are somewhat more geometric in nature. Here is a more algebraic one.

The real-valued function e^(x) arises naturally in many contexts. It's natural to wonder if it can be extended to the complex plane, and how. To tackle this, we can fall back on a tool we often use to calculate values of smooth functions, which is the Taylor series. Knowing that the derivative of e^(x) is itself immediately tells us that e^(x) = 1 + x + x^(2)/2! + x^(3)/3! + ..., and now can simply plug in a complex value for x and see what happens (although we don't yet know if the result is even well-defined.)

Let x = iy be a purely imaginary number, where y is a real number. Then substitution gives e^x = e^(iy) = 1 + iy + i^(2)y^(2)/2! + i^(3)y^(3)/3! + ..., and of course since i^(2) = -1, this can be simplified:

e^(iy) = 1 + iy - y^(2)/2! - iy^(3)/3! + y^(4)/4! + iy^(5)/5! - y^(6)/6! + ...

So we're alternating between real/imaginary and positive/negative. Let's factor it into a real and imaginary component: e^(iy) = a + bi, where

a = 1 - y^(2)/2! + y^(4)/4! - y^(6)/6! + ...

b = y - y^(3)/3! + y^(5)/5! - y^(7)/7! + ...

And here's the kicker: from our prolific experience with calculus of the real numbers, we instantly recognize these as the Taylor series a = cos(y) and b = sin(y), and thus conclude that if anything, e^(iy) = a + bi = cos(y) + i sin(y). Finally, we have e^(i pi) = cos(pi) + i sin(pi) = -1.