9

I see a lot of talk about complex numbers, and yet
for all they are worth, I have not once been
able to find an explaination on how to calculate
them by hand, namely the real component.
For example
(-5)**0.5
(1.3691967456605067e-16+2.23606797749979j)

2.23606 is obviously just the square root of 5, but where the hell did 1.369 come from?

Apparently no one fucking knows, and no site I've found gives a simple explanation for someone new to math in general.

"use a calculator", "hit a button",

How about no.

Comments
  • 7
    Great topic for Christmas morning! I couldn't help looking it up. One way is first converting the base to polar coordinates with a radius and a angle (you can do this by hand: radius is just 5, angle is 2*pi since it's pure imaginary. use arctangent for other cases which is also hand computable by Taylor expansion). Then apply power to radius which is now a real number. Then multiply the exponent to the angle which is now a real number. Then expand the trigonometry again by calculating the sin and cos which are also hand computable by Taylor expansion. You calculator probably encountered floating point rounding error while calculating cos(pi) which produced the tiny real part 1.3e-16.

    https://math.stackexchange.com/ques...
  • 2
    And I didn't even have to read much to tell you this is original Wisecrack content.
    I approve!
  • 2
    Just realized I gave wrong angles. Should have been half that. I kept thinking a pi is 90 degrees...
  • 9
    "but where the hell did 1.369 come from"

    From nowhere. That's a rounding error - take a look at the "E-16" at the end.
  • 0
    That's just sqrt(5) * sqrt(-1) = sqrt(5)*j
  • 1
    I am pretty sure that's wrong, it's just j√5
  • 0
    @Ranchonyx you just looked at the tag lol!
  • 1
    @bettercallshao thank you so much bettercallshao.

    teacher gave me an A for this answer.
  • 0
    For me, this task is also not easy, because I am not well versed in math. I can advise you to look for information from https://plainmath.net/secondary and maybe you will find an answer to your question. I always find useful information there and answers to questions.
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